#set document(title: "4.6 Solve Systems of Equations Using Determinants", author: "OpenStax / XYZ Homework")
#set page(width: 8.5in, height: auto, margin: 1in)
#import "@preview/cetz:0.5.2"
#set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en")
#show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math"))
#set par(justify: true, leading: 0.62em, spacing: 0.9em)
#set enum(spacing: 1.1em)   // room between list items so tall inline fractions don't collide
#set list(spacing: 1.1em)
#set table(stroke: 0.5pt + rgb("#c7ccd3"))

#let BLUE = rgb("#183B6F")     // brand navy — section bars + example/solution labels (white on navy 11.09:1)
#let ORANGE = rgb("#A94509")   // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text)
#let RED = rgb("#DC2626")      // brand error-600
#let GREEN = rgb("#059669")    // brand success-600 (decoration only; small green text uses green-text #007942)

#show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt,
  fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt,
  text(fill: white, weight: "bold", size: 19pt, it.body))
#show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt,
  fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt,
  text(fill: white, weight: "bold", size: 12pt, it.body))
#show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body)
#show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body)

#let examplebox(label, title, body) = block(width: 100%, breakable: true,
  fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt,
  above: 12pt, below: 12pt)[
  #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt,
    text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]]
  #body]

// rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade
#let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint,
    stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt),
    above: 11pt, below: 11pt)[
    #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)]
    #linebreak() #body]

#let solutionbox(body) = block(above: 4pt, below: 8pt)[
  #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body]

#let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"),
  stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[
  #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))]

// Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a
// consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the
// print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a
// degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero.
#let _STD_W = 3.5
#let _WIDE_W = 5.6
#let _MAX_H = 3.4
#let _ASPECT_WIDE = 2.2
#let _UPSCALE_MAX = 1.15
#let stdfig(body) = context {
  let m = measure(body)
  let w = calc.max(m.width / 1in, 0.01)
  let h = calc.max(m.height / 1in, 0.01)
  let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W }
  let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX)
  align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body)))
}

#show figure: set block(breakable: false)
#set figure(gap: 8pt)
#show figure.caption: set text(size: 8.5pt, fill: rgb("#555"))


== 4.6#h(0.6em)Solve Systems of Equations Using Determinants

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Before you get started, take this readiness quiz.

Simplify: #math.equation(block: false, alt: "5 open parenthesis −2 close parenthesis minus open parenthesis −4 close parenthesis open parenthesis 1 close parenthesis .")[$5 ( −2 ) − ( −4 ) ( 1 ) .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "minus 6")[$− 6$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Simplify: #math.equation(block: false, alt: "−3 open parenthesis 8 minus 10 close parenthesis plus open parenthesis −2 close parenthesis open parenthesis 6 minus 3 close parenthesis minus 4 open parenthesis −3 minus open parenthesis −4 close parenthesis close parenthesis .")[$−3 ( 8 − 10 ) + ( −2 ) ( 6 − 3 ) − 4 ( −3 − ( −4 ) ) .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "minus 4")[$− 4$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Simplify: #math.equation(block: false, alt: "the fraction −12 over −8 .")[$frac(−12, −8) .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "the fraction 3 over 2")[$frac(3, 2)$]
]
]

In this section we will learn of another method to solve systems of linear equations called Cramer’s rule. Before we can begin to use the rule, we need to learn some new definitions and notation.


=== Evaluate the Determinant of a #math.equation(block: false, alt: "2 times 2")[$2 #h(0.2em) × #h(0.2em) 2$] Matrix

If a matrix has the same number of rows and columns, we call it a #strong[square matrix]. Each square matrix has a real number associated with it called its #strong[determinant]. To find the determinant of the square matrix #math.equation(block: false, alt: "[ a b; c d ] ,")[$[ a #h(1.5em) b \ c #h(1.5em) d ] ,$] we first write it as #math.equation(block: false, alt: "| a b; c d | .")[$| a #h(1.5em) b \ c #h(1.5em) d | .$] To get the real number value of the determinant we subtract the products of the diagonals, as shown.

#figure(figph[A 2 by 2 determinant is show, with its first row being a, b and second one being c, d. These values are written between two vertical lines instead of brackets as in the case of matrices. Two arrows are shown, one from a to d, the other from c to b. This determinant is equal to ad minus bc.], alt: "A 2 by 2 determinant is show, with its first row being a, b and second one being c, d. These values are written between two vertical lines instead of brackets as in the case of matrices. Two arrows are shown, one from a to d, the other from c to b. This determinant is equal to ad minus bc.", caption: none)

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Determinant]

The determinant of any square matrix #math.equation(block: false, alt: "[ a b; c d ] ,")[$[ a #h(1.5em) b \ c #h(1.5em) d ] ,$] where #emph[a, b, c,] and #emph[d] are real numbers, is

#math.equation(block: true, alt: "| a b; c d | equals a d minus b c")[$| a #h(1.5em) b \ c #h(1.5em) d | = a d − b c$]
]

#examplebox("Example 1")[][
Evaluate the determinant of ⓐ #math.equation(block: false, alt: "[ 4 −2; 3 −1 ]")[$[ 4 #h(1.5em) −2 \ 3 #h(1.5em) −1 ]$] ⓑ #math.equation(block: false, alt: "[ −3, −4; −2, 0 ] .")[$[ −3 &  &  & −4 \ −2 &  &  & 0 ] .$]

#solutionbox[
ⓐ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A 2x2 matrix is shown, containing the numbers 4, -2, 3, and -1. The first row elements are 4 and -2, while the second row elements are 3 and -1.], alt: "A 2x2 matrix is shown, containing the numbers 4, -2, 3, and -1. The first row elements are 4 and -2, while the second row elements are 3 and -1.", caption: none)]),
  [Write the determinant.], [#figure(figph[A 2x2 matrix with values 4, -2, 3, -1, illustrating diagonal multiplication with blue arrows, often used for determinants.], alt: "A 2x2 matrix with values 4, -2, 3, -1, illustrating diagonal multiplication with blue arrows, often used for determinants.", caption: none)],
  [Subtract the products of the diagonals.], [#figure(figph[A mathematical expression showing 4 multiplied by -1, minus 3 multiplied by -2. The expression is 4(-1) - 3(-2).], alt: "A mathematical expression showing 4 multiplied by -1, minus 3 multiplied by -2. The expression is 4(-1) - 3(-2).", caption: none)],
  [Simplify.], [#figure(figph[The mathematical expression -4 + 6 is displayed in a dark gray font against a white background, showing the addition of a negative number and a positive number.], alt: "The mathematical expression -4 + 6 is displayed in a dark gray font against a white background, showing the addition of a negative number and a positive number.", caption: none)],
  [Simplify.], [#figure(figph[The numeral 2 is prominently displayed in the center of a plain white background, rendered in a dark gray hue with a slight gradient effect.], alt: "The numeral 2 is prominently displayed in the center of a plain white background, rendered in a dark gray hue with a slight gradient effect.", caption: none)],
))

#linebreak() 
ⓑ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A 2x2 matrix is shown, containing the elements -3, -4 in the first row and -2, 0 in the second row, enclosed within square brackets.], alt: "A 2x2 matrix is shown, containing the elements -3, -4 in the first row and -2, 0 in the second row, enclosed within square brackets.", caption: none)]),
  [Write the determinant.], [#figure(figph[A mathematical diagram showing cross-multiplication with numbers -3, -4, -2, and 0 arranged in a grid, featuring diagonal blue arrows indicating pairings.], alt: "A mathematical diagram showing cross-multiplication with numbers -3, -4, -2, and 0 arranged in a grid, featuring diagonal blue arrows indicating pairings.", caption: none)],
  [Subtract the products of the diagonals.], [#figure(figph[An algebraic expression featuring multiplication and subtraction of integers: -3(0) - (-2)(-4).], alt: "An algebraic expression featuring multiplication and subtraction of integers: -3(0) - (-2)(-4).", caption: none)],
  [Simplify.], [#figure(figph[The image displays the numbers '0-8' in a simple, grayscale font on a plain white background, likely representing a range or identifier.], alt: "The image displays the numbers '0-8' in a simple, grayscale font on a plain white background, likely representing a range or identifier.", caption: none)],
  [Simplify.], [#figure(figph[A vertical representation of the number -8, with the minus sign positioned to the left of the numeral eight, against a plain white background.], alt: "A vertical representation of the number -8, with the minus sign positioned to the left of the numeral eight, against a plain white background.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Evaluate the determinant of ⓐ #math.equation(block: false, alt: "[ 5 −3; 2 −4 ]")[$[ 5 #h(1.5em) −3 \ 2 #h(1.5em) −4 ]$] ⓑ #math.equation(block: false, alt: "[ −4, −6; 0, 7 ] .")[$[ −4 &  &  & −6 \ 0 &  &  & 7 ] .$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "−14 ;")[$−14 ;$] ⓑ #math.equation(block: false, alt: "−28")[$−28$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Evaluate the determinant of ⓐ #math.equation(block: false, alt: "[ −1 3; −2 4 ]")[$[ −1 #h(1.5em) 3 \ −2 #h(1.5em) 4 ]$] ⓑ #math.equation(block: false, alt: "[ −7, −3; −5, 0 ] .")[$[ −7 &  &  & −3 \ −5 &  &  & 0 ] .$]

#solutionbox[
ⓐ 2 ⓑ #math.equation(block: false, alt: "−15")[$−15$]
]
]


=== Evaluate the Determinant of a #math.equation(block: false, alt: "3 times 3")[$3 #h(0.2em) × #h(0.2em) 3$] Matrix

To evaluate the determinant of a #math.equation(block: false, alt: "3 times 3")[$3 #h(0.2em) × #h(0.2em) 3$] matrix, we have to be able to evaluate the #strong[minor of an entry] in the determinant. The minor of an entry is the #math.equation(block: false, alt: "2 times 2")[$2 #h(0.2em) × #h(0.2em) 2$] determinant found by eliminating the row and column in the #math.equation(block: false, alt: "3 times 3")[$3 #h(0.2em) × #h(0.2em) 3$] determinant that contains the entry.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Minor of an entry in #math.equation(block: false, alt: "3 times 3")[$3 #h(0.2em) × #h(0.2em) 3$] a Determinant]

The #strong[minor of an entry] in a #math.equation(block: false, alt: "3 times 3")[$3 #h(0.2em) × #h(0.2em) 3$] determinant is the #math.equation(block: false, alt: "2 times 2")[$2 #h(0.2em) × #h(0.2em) 2$] determinant found by eliminating the row and column in the #math.equation(block: false, alt: "3 times 3")[$3 #h(0.2em) × #h(0.2em) 3$] determinant that contains the entry.
]

To find the minor of entry #math.equation(block: false, alt: "a sub 1 ,")[$a_(1) ,$] we eliminate the row and column which contain it. So we eliminate the first row and first column. Then we write the #math.equation(block: false, alt: "2 times 2")[$2 #h(0.2em) × #h(0.2em) 2$] determinant that remains.

#figure(figph[The first row of the 3 by 3 determinant is a1, b1, c1. Row 2 is a2, b2, c2. Row 3 is a3, b3, c3. a1 is highlighted. Lines strike out the first row and the first column. What remains is called minor of a1. It is shown as a separate determinant whose first row is b2, c2 and second row is b3, c3.], alt: "The first row of the 3 by 3 determinant is a1, b1, c1. Row 2 is a2, b2, c2. Row 3 is a3, b3, c3. a1 is highlighted. Lines strike out the first row and the first column. What remains is called minor of a1. It is shown as a separate determinant whose first row is b2, c2 and second row is b3, c3.", caption: none)

To find the minor of entry #math.equation(block: false, alt: "b sub 2 ,")[$b_(2) ,$] we eliminate the row and column that contain it. So we eliminate the 2#super[nd] row and 2#super[nd] column. Then we write the #math.equation(block: false, alt: "2 times 2")[$2 #h(0.2em) × #h(0.2em) 2$] determinant that remains.

#figure(figph[The first row of the 3 by 3 determinant is a1, b1, c1. Row 2 is a2, b2, c2. Row 3 is a3, b3, c3. b2 is highlighted. Lines strike out the second row and second column. What remains is minor of b2. It is written as a separate determinant whose first row is a1, c1 and second row is a3, c3.], alt: "The first row of the 3 by 3 determinant is a1, b1, c1. Row 2 is a2, b2, c2. Row 3 is a3, b3, c3. b2 is highlighted. Lines strike out the second row and second column. What remains is minor of b2. It is written as a separate determinant whose first row is a1, c1 and second row is a3, c3.", caption: none)

#examplebox("Example 2")[][
For the determinant #math.equation(block: false, alt: "| 4, −2, 3; 1, 0, −3; −2, −4, 2 | ,")[$| 4 &  &  & −2 &  &  & 3 \ 1 &  &  & 0 &  &  & −3 \ −2 &  &  & −4 &  &  & 2 | ,$] find and then evaluate the minor of ⓐ #math.equation(block: false, alt: "a sub 1")[$a_(1)$] ⓑ #math.equation(block: false, alt: "b sub 3")[$b_(3)$] ⓒ #math.equation(block: false, alt: "c sub 2 .")[$c_(2) .$]

#solutionbox[
ⓐ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A 3x3 matrix with elements \[4, -2, 3; 1, 0, -3; -2, -4, 2\] enclosed by vertical bars, typically denoting a determinant or the absolute value of the matrix.], alt: "A 3x3 matrix with elements [4, -2, 3; 1, 0, -3; -2, -4, 2] enclosed by vertical bars, typically denoting a determinant or the absolute value of the matrix.", caption: none)]),
  [Eliminate the row and column that contains #math.equation(block: false, alt: "a sub 1 .")[$a_(1) .$]], [#figure(figph[A 3x3 matrix showing numerical values: top row is 4, -2, 3; middle row is 1, 0, -3; and bottom row is -2, -4, 2. Vertical and horizontal lines are drawn over some of the numbers.], alt: "A 3x3 matrix showing numerical values: top row is 4, -2, 3; middle row is 1, 0, -3; and bottom row is -2, -4, 2. Vertical and horizontal lines are drawn over some of the numbers.", caption: none)],
  [Write the #math.equation(block: false, alt: "2 times 2")[$2 #h(0.2em) × #h(0.2em) 2$] determinant that remains.], [#figure(figph[A mathematical expression showing the minor of a1 as a 2x2 matrix with elements 0, -3 in the first row and -4, 2 in the second row, enclosed by vertical lines indicating a determinant.], alt: "A mathematical expression showing the minor of a1 as a 2x2 matrix with elements 0, -3 in the first row and -4, 2 in the second row, enclosed by vertical lines indicating a determinant.", caption: none)],
  [Evaluate.], [#figure(figph[A mathematical expression 0(2) - (-3)(-4) is displayed, requiring calculation of a product of zero and two, minus the product of negative three and negative four.], alt: "A mathematical expression 0(2) - (-3)(-4) is displayed, requiring calculation of a product of zero and two, minus the product of negative three and negative four.", caption: none)],
  [Simplify.], [#figure(figph[The image displays the number -12.], alt: "The image displays the number -12.", caption: none)],
))

#linebreak() 
ⓑ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Eliminate the row and column that contains #math.equation(block: false, alt: "b sub 3 .")[$b_(3) .$]], [#figure(figph[A 3x3 matrix with values: 4, -2, 3 in the first row; 1, 0, -3 in the second; and 2, 4, 2 in the third. Light brown grid lines intersect between the numbers.], alt: "A 3x3 matrix with values: 4, -2, 3 in the first row; 1, 0, -3 in the second; and 2, 4, 2 in the third. Light brown grid lines intersect between the numbers.", caption: none)]),
  [Write the #math.equation(block: false, alt: "2 times 2")[$2 #h(0.2em) × #h(0.2em) 2$] determinant that remains.], [#figure(figph[The image shows the mathematical notation for the minor of b\_3, which is represented by a 2x2 determinant with the first row containing 4 and 3, and the second row containing 1 and -3.], alt: "The image shows the mathematical notation for the minor of b_3, which is represented by a 2x2 determinant with the first row containing 4 and 3, and the second row containing 1 and -3.", caption: none)],
  [Evaluate.], [#figure(figph[A mathematical expression displaying 4 multiplied by -3, followed by a subtraction sign, and then the product of 1 and 3.], alt: "A mathematical expression displaying 4 multiplied by -3, followed by a subtraction sign, and then the product of 1 and 3.", caption: none)],
  [Simplify.], [#figure(figph[The number -15 is centrally displayed in a bold, sans-serif font against a plain white background, appearing as simple numerical text.], alt: "The number -15 is centrally displayed in a bold, sans-serif font against a plain white background, appearing as simple numerical text.", caption: none)],
))

#linebreak() 
ⓒ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A 3x3 matrix with integer entries. The first row is 4, -2, 3. The second row is 1, 0, -3. The third row is -2, -4, 2.], alt: "A 3x3 matrix with integer entries. The first row is 4, -2, 3. The second row is 1, 0, -3. The third row is -2, -4, 2.", caption: none)]),
  [Eliminate the row and column that contains #math.equation(block: false, alt: "c sub 2 .")[$c_(2) .$]], [#figure(figph[A 3x3 mathematical matrix is shown with elements 4, -2, 3; 1, 0, -3; -2, -4, 2. Red horizontal and vertical lines highlight specific rows and columns, indicating a calculation step.], alt: "A 3x3 mathematical matrix is shown with elements 4, -2, 3; 1, 0, -3; -2, -4, 2. Red horizontal and vertical lines highlight specific rows and columns, indicating a calculation step.", caption: none)],
  [Write the #math.equation(block: false, alt: "2 times 2")[$2 #h(0.2em) × #h(0.2em) 2$] determinant that remains.], [#figure(figph[The image displays the mathematical expression 'minor of c₂' next to a 2x2 matrix. The matrix contains the elements 4, -2 in the first row and -2, -4 in the second row, enclosed by vertical bars.], alt: "The image displays the mathematical expression 'minor of c₂' next to a 2x2 matrix. The matrix contains the elements 4, -2 in the first row and -2, -4 in the second row, enclosed by vertical bars.", caption: none)],
  [Evaluate.], [#figure(figph[A mathematical expression showing the calculation: 4 multiplied by -4, minus the product of -2 and -2.], alt: "A mathematical expression showing the calculation: 4 multiplied by -4, minus the product of -2 and -2.", caption: none)],
  [Simplify.], [#figure(figph[The number negative twenty (-20) isolated on a white background.], alt: "The number negative twenty (-20) isolated on a white background.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
For the determinant #math.equation(block: false, alt: "| 1, −1, 4; 0, 2, −1; −2, −3, 3 | ,")[$| 1 &  &  & −1 &  &  & 4 \ 0 &  &  & 2 &  &  & −1 \ −2 &  &  & −3 &  &  & 3 | ,$] find and then evaluate the minor of ⓐ #math.equation(block: false, alt: "a sub 1")[$a_(1)$] ⓑ #math.equation(block: false, alt: "b sub 2")[$b_(2)$] ⓒ #math.equation(block: false, alt: "c sub 3 .")[$c_(3) .$]

#solutionbox[
ⓐ 3 ⓑ 11 ⓒ 2
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
For the determinant #math.equation(block: false, alt: "| −2, −1, 0; 3, 0, −1; −1, −2, 3 | ,")[$| −2 &  &  & −1 &  &  & 0 \ 3 &  &  & 0 &  &  & −1 \ −1 &  &  & −2 &  &  & 3 | ,$] find and then evaluate the minor of ⓐ #math.equation(block: false, alt: "a sub 2")[$a_(2)$] ⓑ #math.equation(block: false, alt: "b sub 3")[$b_(3)$] ⓒ #math.equation(block: false, alt: "c sub 2 .")[$c_(2) .$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "−3")[$−3$] ⓑ 2 ⓒ 3
]
]

We are now ready to evaluate a #math.equation(block: false, alt: "3 times 3")[$3 #h(0.2em) × #h(0.2em) 3$] determinant. To do this we expand by minors, which allows us to evaluate the #math.equation(block: false, alt: "3 times 3")[$3 #h(0.2em) × #h(0.2em) 3$] determinant using #math.equation(block: false, alt: "2 times 2")[$2 #h(0.2em) × #h(0.2em) 2$] determinants—which we already know how to evaluate!

To evaluate a #math.equation(block: false, alt: "3 times 3")[$3 #h(0.2em) × #h(0.2em) 3$] determinant by expanding by minors along the first row, we use the following pattern:

#figure(figph[A 3 by 3 determinant is equal to a1 times minor of a1 minus b1 times minor of b1 plus c1 times minor of c1.], alt: "A 3 by 3 determinant is equal to a1 times minor of a1 minus b1 times minor of b1 plus c1 times minor of c1.", caption: none)

Remember, to find the minor of an entry we eliminate the row and column that contains the entry.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Expanding by Minors along the First Row to Evaluate a #math.equation(block: false, alt: "3 times 3")[$3 #h(0.2em) × #h(0.2em) 3$] Determinant]

To evaluate a #math.equation(block: false, alt: "3 times 3")[$3 #h(0.2em) × #h(0.2em) 3$] determinant by #strong[expanding by minors along the first row], the following pattern:

#figure(figph[A 3 by 3 determinant is equal to a1 times minor of a1 minus b1 times minor of b1 plus c1 times minor of c1.], alt: "A 3 by 3 determinant is equal to a1 times minor of a1 minus b1 times minor of b1 plus c1 times minor of c1.", caption: none)
]

#examplebox("Example 3")[][
Evaluate the determinant #math.equation(block: false, alt: "| 2, −3, −1; 3, 2, 0; −1, −1, −2 |")[$| 2 &  &  & −3 &  &  & −1 \ 3 &  &  & 2 &  &  & 0 \ −1 &  &  & −1 &  &  & −2 |$] by expanding by minors along the first row.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A 3x3 matrix with numerical entries. The first row, in red, is 2, -3, -1. The second row is 3, 2, 0. The third row is -1, -1, -2.], alt: "A 3x3 matrix with numerical entries. The first row, in red, is 2, -3, -1. The second row is 3, 2, 0. The third row is -1, -1, -2.", caption: none)]),
  [Expand by minors along the first row], [#figure(figph[Calculation of a 3x3 matrix determinant using cofactor expansion, with minors of 2, -3, and -1 explicitly shown.], alt: "Calculation of a 3x3 matrix determinant using cofactor expansion, with minors of 2, -3, and -1 explicitly shown.", caption: none)],
  [Evaluate each determinant.], [#figure(figph[A mathematical expression reads 2(4-0)+3(6-0)-1(-3-(-2)), presented in a horizontal line with red-orange coloring on the leading coefficients and plus/minus signs.], alt: "A mathematical expression reads 2(4-0)+3(6-0)-1(-3-(-2)), presented in a horizontal line with red-orange coloring on the leading coefficients and plus/minus signs.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical expression displays 2 multiplied by -4, added to 3 multiplied by -6, and then subtracting 1 multiplied by -1. The calculation is 2(-4) + 3(-6) - 1(-1).], alt: "A mathematical expression displays 2 multiplied by -4, added to 3 multiplied by -6, and then subtracting 1 multiplied by -1. The calculation is 2(-4) + 3(-6) - 1(-1).", caption: none)],
  [Simplify.], [#figure(figph[A mathematical expression displays the numbers -8, -18, and +1, representing an arithmetic problem involving addition and subtraction of integers.], alt: "A mathematical expression displays the numbers -8, -18, and +1, representing an arithmetic problem involving addition and subtraction of integers.", caption: none)],
  [Simplify.], [#figure(figph[The number -25 is displayed in a dark gray font on a plain white background.], alt: "The number -25 is displayed in a dark gray font on a plain white background.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Evaluate the determinant #math.equation(block: false, alt: "| 3, −2, 4; 0, −1, −2; 2, 3, −1 | ,")[$| 3 &  &  & −2 &  &  & 4 \ 0 &  &  & −1 &  &  & −2 \ 2 &  &  & 3 &  &  & −1 | ,$] by expanding by minors along the first row.

#solutionbox[
37
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Evaluate the determinant #math.equation(block: false, alt: "| 3, −2, −2; 2, −1, 4; −1, 0, −3 | ,")[$| 3 &  &  & −2 &  &  & −2 \ 2 &  &  & −1 &  &  & 4 \ −1 &  &  & 0 &  &  & −3 | ,$] by expanding by minors along the first row.

#solutionbox[
7
]
]

To evaluate a #math.equation(block: false, alt: "3 times 3")[$3 #h(0.2em) × #h(0.2em) 3$] determinant we can expand by minors using any row or column. Choosing a row or column other than the first row sometimes makes the work easier.

When we expand by any row or column, we must be careful about the sign of the terms in the expansion. To determine the sign of the terms, we use the following sign pattern chart.

#math.equation(block: true, alt: "| plus minus plus; minus plus minus; plus minus plus |")[$| + #h(1.5em) − #h(1.5em) + \ − #h(1.5em) + #h(1.5em) − \ + #h(1.5em) − #h(1.5em) + |$]#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Sign Pattern]

When expanding by minors using a row or column, the sign of the terms in the expansion follow the following pattern.

#math.equation(block: true, alt: "| plus minus plus; minus plus minus; plus minus plus |")[$| + #h(1.5em) − #h(1.5em) + \ − #h(1.5em) + #h(1.5em) − \ + #h(1.5em) − #h(1.5em) + |$]
]

Notice that the sign pattern in the first row matches the signs between the terms in the expansion by the first row.

#figure(figph[A 3 by 3 determinant has row 1: plus, minus, plus, row 2: minus, plus, minus and row 3: plus, minus, plus. The three signs in the first row each point to a minor determinant in the expansion of a 3 by 3 determinant. Plus points to minor of a1, minus to the minor of b1 and plus to the minor of c1.], alt: "A 3 by 3 determinant has row 1: plus, minus, plus, row 2: minus, plus, minus and row 3: plus, minus, plus. The three signs in the first row each point to a minor determinant in the expansion of a 3 by 3 determinant. Plus points to minor of a1, minus to the minor of b1 and plus to the minor of c1.", caption: none)

Since we can expand by any row or column, how do we decide which row or column to use? Usually we try to pick a row or column that will make our calculation easier. If the determinant contains a 0, using the row or column that contains the 0 will make the calculations easier.

#examplebox("Example 4")[][
Evaluate the determinant #math.equation(block: false, alt: "| 4, −1, −3; 3, 0, 2; 5, −4, −3 |")[$| 4 &  &  & −1 &  &  & −3 \ 3 &  &  & 0 &  &  & 2 \ 5 &  &  & −4 &  &  & −3 |$] by expanding by minors.

#solutionbox[
To expand by minors, we look for a row or column that will make our calculations easier. Since 0 is in the second row and second column, expanding by either of those is a good choice. Since the second row has fewer negatives than the second column, we will expand by the second row.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A 3x3 determinant with elements: first row (4, -1, -3), second row (3, 0, 2) in red, and third row (5, -4, -3).], alt: "A 3x3 determinant with elements: first row (4, -1, -3), second row (3, 0, 2) in red, and third row (5, -4, -3).", caption: none)]),
  [Expand using the second row.], [],
  [Be careful of the signs.], [#figure(figph[This image illustrates the process of calculating a 3x3 determinant using cofactor expansion. It shows the alternating sign pattern for cofactors and the terms for the expansion, where each term is a coefficient multiplied by its 2x2 minor.], alt: "This image illustrates the process of calculating a 3x3 determinant using cofactor expansion. It shows the alternating sign pattern for cofactors and the terms for the expansion, where each term is a coefficient multiplied by its 2x2 minor.", caption: none)],
  [Evaluate each determinant.], [#figure(figph[A mathematical expression featuring negative numbers, parentheses, and arithmetic operations: -3(3-12)+0(-12-(-15))-2(-16-(-5)).], alt: "A mathematical expression featuring negative numbers, parentheses, and arithmetic operations: -3(3-12)+0(-12-(-15))-2(-16-(-5)).", caption: none)],
  [Simplify.], [#figure(figph[A mathematical expression featuring negative numbers, multiplication, and addition/subtraction: -3(-9) + 0 - 2(-11).], alt: "A mathematical expression featuring negative numbers, multiplication, and addition/subtraction: -3(-9) + 0 - 2(-11).", caption: none)],
  [Simplify.], [#figure(figph[The image displays the mathematical expression '27 + 0 + 22' in bold, dark gray text against a plain white background.], alt: "The image displays the mathematical expression '27 + 0 + 22' in bold, dark gray text against a plain white background.", caption: none)],
  [Add.], [#figure(figph[The number 49 is displayed.], alt: "The number 49 is displayed.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Evaluate the determinant #math.equation(block: false, alt: "| 2, −1, −3; 0, 3, −4; 3, −4, −3 |")[$| 2 &  &  & −1 &  &  & −3 \ 0 &  &  & 3 &  &  & −4 \ 3 &  &  & −4 &  &  & −3 |$] by expanding by minors.

#solutionbox[
#math.equation(block: true, alt: "−11")[$−11$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Evaluate the determinant #math.equation(block: false, alt: "| −2, −1, −3; −1, 2, 2; 4, −4, 0 |")[$| −2 &  &  & −1 &  &  & −3 \ −1 &  &  & 2 &  &  & 2 \ 4 &  &  & −4 &  &  & 0 |$] by expanding by minors.

#solutionbox[
#math.equation(block: true, alt: "−12")[$−12$]
]
]


=== Use Cramer’s Rule to Solve Systems of Equations

Cramer’s Rule is a method of solving systems of equations using determinants. It can be derived by solving the general form of the systems of equations by elimination. Here we will demonstrate the rule for both systems of two equations with two variables and for systems of three equations with three variables.

Let’s start with the systems of two equations with two variables.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Cramer’s Rule for Solving a System of Two Equations]

For the system of equations #math.equation(block: false, alt: "{ a sub 1 x plus b sub 1 y equals k sub 1; a sub 2 x plus b sub 2 y equals k sub 2 ,")[$\{ a_(1) x + b_(1) y = k_(1) \ a_(2) x + b_(2) y = k_(2) ,$] the solution #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] can be determined by

#figure(figph[x is Dx upon D and y is Dy upon D where D is determinant with row 1: a1, b1 and row 2 a2, b2, use coefficients of the variables; Dx is determinant with row 1: k1, b1 and row 2: k2, b2, replace the x coefficients with the consonants; Dy is determinant with row 1: a1, k1 and row 2: a2, k2, replace the y coefficients with constants], alt: "x is Dx upon D and y is Dy upon D where D is determinant with row 1: a1, b1 and row 2 a2, b2, use coefficients of the variables; Dx is determinant with row 1: k1, b1 and row 2: k2, b2, replace the x coefficients with the consonants; Dy is determinant with row 1: a1, k1 and row 2: a2, k2, replace the y coefficients with constants", caption: none)
]

Notice that to form the determinant #emph[D], we use take the coefficients of the variables.

#figure(figph[The equations are a1x plus b1y equals k1 and a2x plus b2y equals k2. Here, a1, a2, b1, b2 are coefficients. The determinant is D with row 1: a1, b1 and row 2: a2, b2. Column 1 has coefficients of x and column 2 has coefficients of], alt: "The equations are a1x plus b1y equals k1 and a2x plus b2y equals k2. Here, a1, a2, b1, b2 are coefficients. The determinant is D with row 1: a1, b1 and row 2: a2, b2. Column 1 has coefficients of x and column 2 has coefficients of", caption: none)

Notice that to form the determinant #math.equation(block: false, alt: "D sub x")[$D_(x)$] and #math.equation(block: false, alt: "D sub y ,")[$D_(y) ,$] we substitute the constants for the coefficients of the variable we are finding.

#figure(figph[The equations are a1x plus b1y equals k1 and a2x plus b2y equals k2. Here, a1, a2, b1, b2 are coefficients. The determinant is Dx has row 1: k1, b1 and row 2: k2, b2. Here columns 1 and 2 re constants and coefficients of y respectively. Determinant Dy has row 1: a1, k1 and row 2: a2, k2. Here, columns 1 and 2 are coefficients of x and constants respectively.], alt: "The equations are a1x plus b1y equals k1 and a2x plus b2y equals k2. Here, a1, a2, b1, b2 are coefficients. The determinant is Dx has row 1: k1, b1 and row 2: k2, b2. Here columns 1 and 2 re constants and coefficients of y respectively. Determinant Dy has row 1: a1, k1 and row 2: a2, k2. Here, columns 1 and 2 are coefficients of x and constants respectively.", caption: none)

#examplebox("Example 5")[How to Solve a System of Equations Using Cramer’s Rule][
Solve using Cramer’s Rule: #math.equation(block: false, alt: "{ 2 x plus y equals −4; 3 x minus 2 y equals −6 .")[$\{ 2 x + y = −4 \ 3 x − 2 y = −6 .$]

#solutionbox[
#figure(figph[The equations are 2x plus y equals minus 4 and 3x minus 2y equals minus 6. Step 1. Evaluate the determinant D, using the coefficients of the variables. Determinant D has row 1: 2, 1 and row 2: 3, minus 2. So, D is minus 7.], alt: "The equations are 2x plus y equals minus 4 and 3x minus 2y equals minus 6. Step 1. Evaluate the determinant D, using the coefficients of the variables. Determinant D has row 1: 2, 1 and row 2: 3, minus 2. So, D is minus 7.", caption: none)

#figure(figph[Step 2. Evaluate the determinant Dx. Use the constants in place of the x coefficients. We replace the coefficients of x, 2 and 3, with the constants, negative 4 and negative 6. We get Dx equal to 14.], alt: "Step 2. Evaluate the determinant Dx. Use the constants in place of the x coefficients. We replace the coefficients of x, 2 and 3, with the constants, negative 4 and negative 6. We get Dx equal to 14.", caption: none)

#figure(figph[Step 3. Evaluate the determinant Dy. Use the constants in place of the y coefficients. We replace the coefficients of y, 1 and 2, with the constants, negative 4 and negative 6. We get Dy equal to 0.], alt: "Step 3. Evaluate the determinant Dy. Use the constants in place of the y coefficients. We replace the coefficients of y, 1 and 2, with the constants, negative 4 and negative 6. We get Dy equal to 0.", caption: none)

#figure(figph[Step 4. Find x and y. Substituting values of D, Dx and Dy in the equations x equal to Dx upon D and y equal to Dy upon D, we get x equal to minus 2 and y equal to 0.], alt: "Step 4. Find x and y. Substituting values of D, Dx and Dy in the equations x equal to Dx upon D and y equal to Dy upon D, we get x equal to minus 2 and y equal to 0.", caption: none)

#figure(figph[Step 5. Write the solution as an ordered pair minus 2, 0.], alt: "Step 5. Write the solution as an ordered pair minus 2, 0.", caption: none)

#figure(figph[Step 6. Check that the ordered pair is a solution to both original equations.], alt: "Step 6. Check that the ordered pair is a solution to both original equations.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve using Cramer’s rule: #math.equation(block: false, alt: "{ 3 x plus y equals −3; 2 x plus 3 y equals 6 .")[$\{ 3 x + y = −3 \ 2 x + 3 y = 6 .$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis minus the fraction 15 over 7 , the fraction 24 over 7 close parenthesis")[$( − frac(15, 7) , frac(24, 7) )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve using Cramer’s rule: #math.equation(block: false, alt: "{ − x plus y equals 2; 2 x plus y equals −4 .")[$\{ "−" x + y = 2 \ 2 x + y = −4 .$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis −2 , 0 close parenthesis")[$( −2 , 0 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Solve a system of two equations using Cramer’s rule.]

+ Evaluate the determinant #emph[D], using the coefficients of the variables.
+ Evaluate the determinant #math.equation(block: false, alt: "D sub x .")[$D_(x) .$] Use the constants in place of the #emph[x] coefficients.
+ Evaluate the determinant #math.equation(block: false, alt: "D sub y .")[$D_(y) .$] Use the constants in place of the #emph[y] coefficients.
+ Find #emph[x] and #emph[y]. #math.equation(block: false, alt: "x equals the fraction D sub x over D ,")[$x = frac(D_(x), D) ,$] #math.equation(block: false, alt: "y equals the fraction D sub y over D")[$y = frac(D_(y), D)$]
+ Write the solution as an ordered pair.
+ Check that the ordered pair is a solution to both original equations.
]

To solve a system of three equations with three variables with Cramer’s Rule, we basically do what we did for a system of two equations. However, we now have to solve for three variables to get the solution. The determinants are also going to be #math.equation(block: false, alt: "3 times 3")[$3 #h(0.2em) × #h(0.2em) 3$] which will make our work more interesting!

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Cramer’s Rule for Solving a System of Three Equations]

For the system of equations #math.equation(block: false, alt: "{ a sub 1 x plus b sub 1 y plus c sub 1 z equals k sub 1; a sub 2 x plus b sub 2 y plus c sub 2 z equals k sub 2; a sub 3 x plus b sub 3 y plus c sub 3 z equals k sub 3 ,")[$\{ a_(1) x + b_(1) y + c_(1) z = k_(1) \ a_(2) x + b_(2) y + c_(2) z = k_(2) \ a_(3) x + b_(3) y + c_(3) z = k_(3) ,$] the solution #math.equation(block: false, alt: "open parenthesis x , y , z close parenthesis")[$( x , y , z )$] can be determined by

#figure(figph[The image shows equations for x, y, and z expressed as determinants. Each determinant uses variables a through c and k through d. The first determinant's first column contains the variable x coefficients. The second determinant’s first column contains the variable y coefficients. The third determinant’s first column contains the variable z coefficients.], alt: "The image shows equations for x, y, and z expressed as determinants. Each determinant uses variables a through c and k through d. The first determinant's first column contains the variable x coefficients. The second determinant’s first column contains the variable y coefficients. The third determinant’s first column contains the variable z coefficients.", caption: none)
]

#examplebox("Example 6")[][
Solve the system of equations using Cramer’s Rule: #math.equation(block: false, alt: "{ 3 x minus 5 y plus 4 z equals 5; 5 x plus 2 y plus z equals 0; 2 x plus 3 y minus 2 z equals 3 .")[$\{ 3 x − 5 y + 4 z = 5 \ 5 x + 2 y + z = 0 \ 2 x + 3 y − 2 z = 3 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Evaluate the determinant #emph[D].], [#figure(figph[Matrix D with highlighted elements in the first column, showing their rows: (3, -5, 4), (5, 2, 1) and (2, 3, -2).], alt: "Matrix D with highlighted elements in the first column, showing their rows: (3, -5, 4), (5, 2, 1) and (2, 3, -2).", caption: none)]),
  [Expand by minors using column 1.], [],
  [#figure(figph[Be careful with the signs.], alt: "Be careful with the signs.", caption: none)], [#figure(figph[Expansion of a 3x3 determinant by cofactors to calculate D.], alt: "Expansion of a 3x3 determinant by cofactors to calculate D.", caption: none)],
  [Evaluate the determinants.], [#figure(figph[A mathematical equation involving operations with whole numbers, including multiplication and subtraction within parentheses to be solved.], alt: "A mathematical equation involving operations with whole numbers, including multiplication and subtraction within parentheses to be solved.", caption: none)],
  [Simplify.], [#figure(figph[The image shows the equation D = 3(-7) - 5(-2) + 2(-13), a mathematical operation with integers.], alt: "The image shows the equation D = 3(-7) - 5(-2) + 2(-13), a mathematical operation with integers.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation showing D is equal to negative 21 plus 10 minus 26.], alt: "A mathematical equation showing D is equal to negative 21 plus 10 minus 26.", caption: none)],
  [Simplify.], [#figure(figph[The image shows the equation D = -37.], alt: "The image shows the equation D = -37.", caption: none)],
  [Evaluate the determinant #math.equation(block: false, alt: "D sub x .")[$D_(x) .$] Use the #linebreak() constants to replace the coefficients of #emph[x].], [#figure(figph[A 3x3 Dx matrix with elements 5, -5, 4 in the first row; 0, 2, 1 in the second; and 3, 3, -2 in the third. The first column is highlighted in red.], alt: "A 3x3 Dx matrix with elements 5, -5, 4 in the first row; 0, 2, 1 in the second; and 3, 3, -2 in the third. The first column is highlighted in red.", caption: none)],
  [Expand by minors using column 1.], [#figure(figph[Calculation of the determinant D\_x using cofactor expansion, showing the algebraic sum of products of elements and subdeterminants.], alt: "Calculation of the determinant D_x using cofactor expansion, showing the algebraic sum of products of elements and subdeterminants.", caption: none)],
  [Evaluate the determinants.], [#figure(figph[A mathematical equation for Dx is shown on a white background: Dx = 5(-4-3) - 0(10-12) + 3(-5-8). The numbers 5, 0, and 3 are highlighted in red.], alt: "A mathematical equation for Dx is shown on a white background: Dx = 5(-4-3) - 0(10-12) + 3(-5-8). The numbers 5, 0, and 3 are highlighted in red.", caption: none)],
  [Simplify.], [#figure(figph[Mathematical expression: Dx is equal to 5 multiplied by -7, minus 0, plus 3 multiplied by -13. Numbers in red highlight specific terms of the operation.], alt: "Mathematical expression: Dx is equal to 5 multiplied by -7, minus 0, plus 3 multiplied by -13. Numbers in red highlight specific terms of the operation.", caption: none)],
  [Simplify.], [#figure(figph[The equation D subscript x equals -74 is displayed on a white background.], alt: "The equation D subscript x equals -74 is displayed on a white background.", caption: none)],
  [Evaluate the determinant #math.equation(block: false, alt: "D sub y .")[$D_(y) .$] Use the #linebreak() constants to replace the coefficients of #emph[y].], [#figure(figph[The 3x3 matrix D\_y with numeric elements, where the second column (5, 0, 3) is highlighted in red.], alt: "The 3x3 matrix D_y with numeric elements, where the second column (5, 0, 3) is highlighted in red.", caption: none)],
  [#figure(figph[Math instructions: 'Expand by minors using column 2. Be careful with the signs.' A 3x3 matrix is ​​shown with the pattern of alternating signs (+ - + / - + - / + - +) to remember.], alt: "Math instructions: 'Expand by minors using column 2. Be careful with the signs.' A 3x3 matrix is ​​shown with the pattern of alternating signs (+ - + / - + - / + - +) to remember.", caption: none)], [#figure(figph[Calculation of Dy through the expansion of a 3x3 determinant. The numbers in red indicate the elements used in the cofactor expansion for the first row.], alt: "Calculation of Dy through the expansion of a 3x3 determinant. The numbers in red indicate the elements used in the cofactor expansion for the first row.", caption: none)],
  [Evaluate the determinants.], [#figure(figph[A mathematical equation that calculates Dy: Dy = -5(-10-2) + 0(-10-12) - 3(3-20). Some numbers appear in red, possibly indicating key steps or coefficients.], alt: "A mathematical equation that calculates Dy: Dy = -5(-10-2) + 0(-10-12) - 3(3-20). Some numbers appear in red, possibly indicating key steps or coefficients.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation, Dy = -5(-12) + 0 - 3(-17), is shown in black text on a white background, detailing a calculation with positive and negative integers.], alt: "A mathematical equation, Dy = -5(-12) + 0 - 3(-17), is shown in black text on a white background, detailing a calculation with positive and negative integers.", caption: none)],
  [Simplify.], [#figure(figph[Equation showing Dy equal to the sum of 60, 0, and 51.], alt: "Equation showing Dy equal to the sum of 60, 0, and 51.", caption: none)],
  [Simplify.], [#figure(figph[D sub y is equal to 111.], alt: "D sub y is equal to 111.", caption: none)],
  [Evaluate the determinant #math.equation(block: false, alt: "D sub z .")[$D_(z) .$] Use the #linebreak() constants to replace the coefficients of #emph[z].], [#figure(figph[D\_z matrix with the third column highlighted in red, showing the values ​​5, 0 and 3.], alt: "D_z matrix with the third column highlighted in red, showing the values ​​5, 0 and 3.", caption: none)],
  [#figure(figph[Expand by minors using column 3. Be careful with signs. (A sign matrix for cofactor expansion is shown)], alt: "Expand by minors using column 3. Be careful with signs. (A sign matrix for cofactor expansion is shown)", caption: none)], [#figure(figph[Calculation of the determinant D\_x through expansion by cofactors.], alt: "Calculation of the determinant D_x through expansion by cofactors.", caption: none)],
  [Evaluate the determinants.], [#figure(figph[A mathematical equation is displayed: Dx = 5(15-4) - 0(9-(-10)) + 3(6-(-25)).], alt: "A mathematical equation is displayed: Dx = 5(15-4) - 0(9-(-10)) + 3(6-(-25)).", caption: none)],
  [Simplify.], [#figure(figph[A mathematical expression for D subscript x showing the calculation as 5 multiplied by 11, minus 0, plus 3 multiplied by 31.], alt: "A mathematical expression for D subscript x showing the calculation as 5 multiplied by 11, minus 0, plus 3 multiplied by 31.", caption: none)],
  [Simplify.], [#figure(figph[Mathematical equation with D sub x equal to fifty-five minus zero plus ninety-three, showing a simple arithmetic operation.], alt: "Mathematical equation with D sub x equal to fifty-five minus zero plus ninety-three, showing a simple arithmetic operation.", caption: none)],
  [Simplify.], [#figure(figph[A close-up view of the equation Dx = 148 against a white background.], alt: "A close-up view of the equation Dx = 148 against a white background.", caption: none)],
  [Find #emph[x], #emph[y], and #emph[z].], [#figure(figph[Formulas to calculate x, y, and z using the determinants Dx, Dy, Dz, and the principal determinant D, as in Cramer's Rule.], alt: "Formulas to calculate x, y, and z using the determinants Dx, Dy, Dz, and the principal determinant D, as in Cramer's Rule.", caption: none)],
  [Substitute in the values.], [#figure(figph[The equations show values ​​for x, y, and z as fractions: x = -74/-37, y = 111/-37, and z = 148/-37.], alt: "The equations show values ​​for x, y, and z as fractions: x = -74/-37, y = 111/-37, and z = 148/-37.", caption: none)],
  [Simplify.], [#figure(figph[The values ​​of the variables are shown: x is equal to 2, y is equal to -3, and z is equal to -4.], alt: "The values ​​of the variables are shown: x is equal to 2, y is equal to -3, and z is equal to -4.", caption: none)],
  [Write the solution as an ordered triple.], [#figure(figph[A set of three-dimensional coordinates, (2, -3, -4), is displayed in dark gray text against a plain white background.], alt: "A set of three-dimensional coordinates, (2, -3, -4), is displayed in dark gray text against a plain white background.", caption: none)],
  [Check that the ordered triple is a solution #linebreak() to #strong[all three] original equations.], [We leave the check to you.],
  [], [The solution is #math.equation(block: false, alt: "open parenthesis 2 , −3 , −4 close parenthesis .")[$( 2 , −3 , −4 ) .$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve the system of equations using Cramer’s Rule: #math.equation(block: false, alt: "{ 3 x plus 8 y plus 2 z equals −5; 2 x plus 5 y minus 3 z equals 0; x plus 2 y minus 2 z equals −1 .")[$\{ 3 x + 8 y + 2 z = −5 \ 2 x + 5 y − 3 z = 0 \ x + 2 y − 2 z = −1 .$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis −9 , 3 , −1 close parenthesis")[$( −9 , 3 , −1 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve the system of equations using Cramer’s Rule: #math.equation(block: false, alt: "{ 3 x plus y minus 6 z equals −3; 2 x plus 6 y plus 3 z equals 0; 3 x plus 2 y minus 3 z equals −6 .")[$\{ 3 x + y − 6 z = −3 \ 2 x + 6 y + 3 z = 0 \ 3 x + 2 y − 3 z = −6 .$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis −6 , 3 , −2 close parenthesis")[$( −6 , 3 , −2 )$]
]
]

Cramer’s rule does not work when the value of the #emph[D] determinant is 0, as this would mean we would be dividing by 0. But when #math.equation(block: false, alt: "D equals 0 ,")[$D = 0 ,$] the system is either inconsistent or dependent.

When the value of #math.equation(block: false, alt: "D equals 0")[$D = 0$] and #math.equation(block: false, alt: "D sub x , D sub y")[$D_(x) , D_(y)$] and #math.equation(block: false, alt: "D sub z")[$D_(z)$] are all zero, the system is consistent and dependent and there are infinitely many solutions.

When the value of #math.equation(block: false, alt: "D equals 0")[$D = 0$] and #math.equation(block: false, alt: "D sub x , D sub y")[$D_(x) , D_(y)$] and #math.equation(block: false, alt: "D sub z")[$D_(z)$] are not all zero, the system is inconsistent and there is no solution.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Dependent and Inconsistent Systems of Equations]

For any system of equations, where the #strong[value of the determinant] #math.equation(block: false, alt: "D equals 0 ,")[$D = 0 ,$] #linebreak() 
#math.equation(block: false, alt: "Value of determinants, Type of system, Solution; D equals 0 and D sub x , D sub y and D sub z are all zero, consistent and dependent, infinitely many solutions; D equals 0 and D sub x , D sub y and D sub z are not all zero, inconsistent, no solution")[$"Value of determinants" &  &  & "Type of system" &  &  & "Solution" \ D = 0 #h(0.2em) "and" #h(0.2em) D_(x) , D_(y) #h(0.2em) "and" #h(0.2em) D_(z) #h(0.2em) "are all zero" &  &  & "consistent and dependent" &  &  & "infinitely many solutions" \ D = 0 #h(0.2em) "and" #h(0.2em) D_(x) , D_(y) #h(0.2em) "and" #h(0.2em) D_(z) #h(0.2em) "are not all zero" &  &  & "inconsistent" &  &  & "no solution"$]
]

In the next example, we will use the values of the determinants to find the solution of the system.

#examplebox("Example 7")[][
Solve the system of equations using Cramer’s rule : #math.equation(block: false, alt: "{ x plus 3 y equals 4; −2 x minus 6 y equals 3 .")[$\{ x + 3 y = 4 \ −2 x − 6 y = 3 .$]

#solutionbox[
#math.equation(block: true, alt: "{ x plus 3 y equals 4; − 2 x minus 6 y equals 3; Evaluate the determinant D , using the; coefficients of the variables., D equals | 1, 3; −2, −6 |; D equals −6 minus open parenthesis minus 6 close parenthesis; D equals 0")[$&  &  &  &  & \{ x + 3 y = 4 \ "−" 2 x − 6 y = 3 \ "Evaluate the determinant" #h(0.2em) D , #h(0.2em) "using the" \ "coefficients of the variables." &  &  &  &  & D = | 1 &  &  & 3 \ −2 &  &  & −6 | \  &  &  &  &  & D = −6 − ( − 6 ) \  &  &  &  &  & D = 0$]

We cannot use Cramer’s Rule to solve this system. But by looking at the value of the determinants #math.equation(block: false, alt: "D sub x")[$D_(x)$] and #math.equation(block: false, alt: "D sub y ,")[$D_(y) ,$] we can determine whether the system is dependent or inconsistent.

#math.equation(block: true, alt: "Evaluate the determinant D sub x ., D sub x equals | 4, 3; 3, −6 |; D sub x equals −24 minus 9; D sub x equals -33")[$"Evaluate the determinant" #h(0.2em) D_(x) . &  &  &  &  & #h(3.5em) D_(x) = | 4 &  &  & 3 \ 3 &  &  & −6 | \  &  &  &  &  & #h(3.5em) D_(x) = −24 − 9 \  &  &  &  &  & #h(3.5em) D_(x) = -33$]

Since all the determinants are not zero, the system is inconsistent. There is no solution.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve the system of equations using Cramer’s rule: #math.equation(block: false, alt: "{ 4 x minus 3 y equals 8; 8 x minus 6 y equals 14 .")[$\{ 4 x − 3 y = 8 \ 8 x − 6 y = 14 .$]

#solutionbox[
no solution
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve the system of equations using Cramer’s rule: #math.equation(block: false, alt: "{ x equals −3 y plus 4; 2 x plus 6 y equals 8 .")[$\{ x = −3 y + 4 \ 2 x + 6 y = 8 .$]

#solutionbox[
infinite solutions
]
]


=== Solve Applications using Determinants

An interesting application of determinants allows us to test if points are collinear. Three points #math.equation(block: false, alt: "open parenthesis x sub 1 , y sub 1 close parenthesis ,")[$( x_(1) , y_(1) ) ,$] #math.equation(block: false, alt: "open parenthesis x sub 2 , y sub 2 close parenthesis")[$( x_(2) , y_(2) )$] and #math.equation(block: false, alt: "open parenthesis x sub 3 , y sub 3 close parenthesis")[$( x_(3) , y_(3) )$] are collinear if and only if the determinant below is zero.

#math.equation(block: true, alt: "| x sub 1 y sub 1 1; x sub 2 y sub 2 1; x sub 3 y sub 3 1 | equals 0")[$| x_(1) #h(1.5em) y_(1) #h(1.5em) 1 \ x_(2) #h(1.5em) y_(2) #h(1.5em) 1 \ x_(3) #h(1.5em) y_(3) #h(1.5em) 1 | = 0$]#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Test for Collinear Points]

Three points #math.equation(block: false, alt: "open parenthesis x sub 1 , y sub 1 close parenthesis ,")[$( x_(1) , y_(1) ) ,$] #math.equation(block: false, alt: "open parenthesis x sub 2 , y sub 2 close parenthesis")[$( x_(2) , y_(2) )$] and #math.equation(block: false, alt: "open parenthesis x sub 3 , y sub 3 close parenthesis")[$( x_(3) , y_(3) )$] are collinear if and only if

#math.equation(block: true, alt: "| x sub 1 y sub 1 1; x sub 2 y sub 2 1; x sub 3 y sub 3 1 | equals 0")[$| x_(1) #h(1.5em) y_(1) #h(1.5em) 1 \ x_(2) #h(1.5em) y_(2) #h(1.5em) 1 \ x_(3) #h(1.5em) y_(3) #h(1.5em) 1 | = 0$]
]

We will use this property in the next example.

#examplebox("Example 8")[][
Determine whether the points #math.equation(block: false, alt: "open parenthesis 5 , −5 close parenthesis ,")[$( 5 , −5 ) ,$] #math.equation(block: false, alt: "open parenthesis 4 , −3 close parenthesis ,")[$( 4 , −3 ) ,$] and #math.equation(block: false, alt: "open parenthesis 3 , −1 close parenthesis")[$( 3 , −1 )$] are collinear.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[Mathematical matrix or determinant showing three points (x1, y1), (x2, y2), (x3, y3) and a column of ones, commonly used in geometry for area calculations or collinearity tests.], alt: "Mathematical matrix or determinant showing three points (x1, y1), (x2, y2), (x3, y3) and a column of ones, commonly used in geometry for area calculations or collinearity tests.", caption: none)]),
  [Substitute the values into the determinant. #linebreak() #math.equation(block: false, alt: "open parenthesis 5 , −5 close parenthesis ,")[$( 5 , −5 ) ,$] #math.equation(block: false, alt: "open parenthesis 4 , −3 close parenthesis ,")[$( 4 , −3 ) ,$] and #math.equation(block: false, alt: "open parenthesis 3 , −1 close parenthesis")[$( 3 , −1 )$]], [#figure(figph[A 3x3 determinant is shown with values 5, -5, 1 in the first row; 4, -3, 1 in the second; and 3, -1, 1 in the third row.], alt: "A 3x3 determinant is shown with values 5, -5, 1 in the first row; 4, -3, 1 in the second; and 3, -1, 1 in the third row.", caption: none)],
  [Evaluate the determinant by expanding #linebreak() by minors using column 3.], [#figure(figph[A mathematical expression showing the cofactor expansion of a 3x3 determinant, represented as a sum and difference of three 2x2 determinants with coefficients 1, -1, and 1.], alt: "A mathematical expression showing the cofactor expansion of a 3x3 determinant, represented as a sum and difference of three 2x2 determinants with coefficients 1, -1, and 1.", caption: none)],
  [Evaluate the determinants.], [#figure(figph[A mathematical expression involving integers and operations is shown, featuring the sum and differences of parenthesized terms: 1(-4-(-9)) - 1(-5-(-15)) + 1(-15-(-20)).], alt: "A mathematical expression involving integers and operations is shown, featuring the sum and differences of parenthesized terms: 1(-4-(-9)) - 1(-5-(-15)) + 1(-15-(-20)).", caption: none)],
  [Simplify.], [#figure(figph[A mathematical expression 1(5) - 1(10) + 1(5) is shown on a white background, which simplifies to 0.], alt: "A mathematical expression 1(5) - 1(10) + 1(5) is shown on a white background, which simplifies to 0.", caption: none)],
  [Simplify.], [#figure(figph[A minimal black vertical line with a slight oval curvature is centered on a plain white background, creating a stark and simple abstract design.], alt: "A minimal black vertical line with a slight oval curvature is centered on a plain white background, creating a stark and simple abstract design.", caption: none)],
  [], [The value of the determinant is 0, so the #linebreak() points are collinear.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Determine whether the points #math.equation(block: false, alt: "open parenthesis 3 , −2 close parenthesis ,")[$( 3 , −2 ) ,$] #math.equation(block: false, alt: "open parenthesis 5 , −3 close parenthesis ,")[$( 5 , −3 ) ,$] and #math.equation(block: false, alt: "open parenthesis 1 , −1 close parenthesis")[$( 1 , −1 )$] are collinear.

#solutionbox[
yes
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Determine whether the points #math.equation(block: false, alt: "open parenthesis −4 , −1 close parenthesis ,")[$( −4 , −1 ) ,$] #math.equation(block: false, alt: "open parenthesis −6 , 2 close parenthesis ,")[$( −6 , 2 ) ,$] and #math.equation(block: false, alt: "open parenthesis −2 , −4 close parenthesis")[$( −2 , −4 )$] are collinear.

#solutionbox[
yes
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Access these online resources for additional instruction and practice with solving systems of linear equations using determinants or Cramer's Rule.

- #link("https://www.openstax.org/l/37sysdeterm")[Solving Systems of Linear Equations Using Determinants]
- #link("https://www.openstax.org/l/37syscramer")[Solving Systems of Linear Equations Using Cramer's Rule]
]


=== Key Concepts

- #strong[Determinant:] The determinant of any square matrix #math.equation(block: false, alt: "[ a b; c d ] ,")[$[ a #h(1.5em) b \ c #h(1.5em) d ] ,$] where #emph[a, b, c,] and #emph[d] are real numbers, is #linebreak() 
#math.equation(block: true, alt: "| a b; c d | equals a d minus b c")[$| a #h(1.5em) b \ c #h(1.5em) d | = a d − b c$]
- #strong[Expanding by Minors along the First Row to Evaluate a 3 × 3 Determinant:] To evaluate a #math.equation(block: false, alt: "3 times 3")[$3 #h(0.2em) × #h(0.2em) 3$] determinant by expanding by minors along the first row, the following pattern: #linebreak() 
#figure(figph[A 3 by 3 determinant is equal to a1 times minor of a1 minus b1 times minor of b1 plus c1 times minor of c1.], alt: "A 3 by 3 determinant is equal to a1 times minor of a1 minus b1 times minor of b1 plus c1 times minor of c1.", caption: none)
- #strong[Sign Pattern:] When expanding by minors using a row or column, the sign of the terms in the expansion follow the following pattern. #linebreak() 
#math.equation(block: true, alt: "| plus minus plus; minus plus minus; plus minus plus |")[$| + #h(1.5em) − #h(1.5em) + \ − #h(1.5em) + #h(1.5em) − \ + #h(1.5em) − #h(1.5em) + |$]
- #strong[Cramer’s Rule:] For the system of equations #math.equation(block: false, alt: "{ a sub 1 x plus b sub 1 y equals k sub 1; a sub 2 x plus b sub 2 y equals k sub 2 ,")[$\{ a_(1) x + b_(1) y = k_(1) \ a_(2) x + b_(2) y = k_(2) ,$] the solution #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] can be determined by #linebreak() 
#figure(figph[x is Dx upon D and y is Dy upon D where D is determinant with row 1: a1, b1 and row 2 a2, b2, use coefficients of the variables; Dx is determinant with row 1: k1, b1 and row 2: k2, b2, replace the x coefficients with the consonants; Dy is determinant with row 1: a1, k1 and row 2: a2, k2, replace the y coefficients with constants.], alt: "x is Dx upon D and y is Dy upon D where D is determinant with row 1: a1, b1 and row 2 a2, b2, use coefficients of the variables; Dx is determinant with row 1: k1, b1 and row 2: k2, b2, replace the x coefficients with the consonants; Dy is determinant with row 1: a1, k1 and row 2: a2, k2, replace the y coefficients with constants.", caption: none)

 #linebreak() 
Notice that to form the determinant #emph[D], we use take the coefficients of the variables.
- #strong[How to solve a system of two equations using Cramer’s rule.]
+ Evaluate the determinant #emph[D], using the coefficients of the variables.
+ Evaluate the determinant #math.equation(block: false, alt: "D sub x .")[$D_(x) .$] Use the constants in place of the #emph[x] coefficients.
+ Evaluate the determinant #math.equation(block: false, alt: "D sub y .")[$D_(y) .$] Use the constants in place of the #emph[y] coefficients.
+ Find #emph[x] and #emph[y]. #math.equation(block: false, alt: "x equals the fraction D sub x over D ,")[$x = frac(D_(x), D) ,$] #math.equation(block: false, alt: "y equals the fraction D sub y over D .")[$y = frac(D_(y), D) .$]
+ Write the solution as an ordered pair.
+ Check that the ordered pair is a solution to #strong[both] original equations.
+ #strong[Dependent and Inconsistent Systems of Equations:] For any system of equations, where the #strong[value of the determinant] #math.equation(block: false, alt: "D equals 0 ,")[$D = 0 ,$] #linebreak() 
#math.equation(block: false, alt: "Value of determinants, Type of system, Solution; D equals 0 and D sub x , D sub y and D sub z are all zero, consistent and dependent, infinitely many solutions; D equals 0 and D sub x , D sub y and D sub z are not all zero, inconsistent, no solution")[$"Value of determinants" &  &  & "Type of system" &  &  & "Solution" \ D = 0 #h(0.2em) "and" #h(0.2em) D_(x) , D_(y) #h(0.2em) "and" #h(0.2em) D_(z) #h(0.2em) "are all zero" &  &  & "consistent and dependent" &  &  & "infinitely many solutions" \ D = 0 #h(0.2em) "and" #h(0.2em) D_(x) , D_(y) #h(0.2em) "and" #h(0.2em) D_(z) #h(0.2em) "are not all zero" &  &  & "inconsistent" &  &  & "no solution"$]
+ #strong[Test for Collinear Points:] Three points #math.equation(block: false, alt: "open parenthesis x sub 1 , y sub 1 close parenthesis ,")[$( x_(1) , y_(1) ) ,$] #math.equation(block: false, alt: "open parenthesis x sub 2 , y sub 2 close parenthesis ,")[$( x_(2) , y_(2) ) ,$] and #math.equation(block: false, alt: "open parenthesis x sub 3 , y sub 3 close parenthesis")[$( x_(3) , y_(3) )$] are collinear if and only if #linebreak() 
#math.equation(block: true, alt: "| x sub 1 y sub 1 1; x sub 2 y sub 2 1; x sub 3 y sub 3 1 | equals 0")[$| x_(1) #h(1.5em) y_(1) #h(1.5em) 1 \ x_(2) #h(1.5em) y_(2) #h(1.5em) 1 \ x_(3) #h(1.5em) y_(3) #h(1.5em) 1 | = 0$]


==== Practice Makes Perfect

#strong[Evaluate the Determinant of a 2 × 2 Matrix]

In the following exercises, evaluate the determinant of each square matrix.

#math.equation(block: true, alt: "[ 6 −2; 3 −1 ]")[$[ 6 #h(1.5em) −2 \ 3 #h(1.5em) −1 ]$]

#math.equation(block: true, alt: "[ −4 8; −3 5 ]")[$[ −4 #h(1.5em) 8 \ −3 #h(1.5em) 5 ]$]

#solutionbox[
4
]

#math.equation(block: true, alt: "[ −3, 5; 0, −4 ]")[$[ −3 &  &  & 5 \ 0 &  &  & −4 ]$]

#math.equation(block: true, alt: "[ −2, 0; 7, −5 ]")[$[ −2 &  &  & 0 \ 7 &  &  & −5 ]$]

#solutionbox[
10
]

#strong[Evaluate the Determinant of a 3 × 3 Matrix]

In the following exercises, find and then evaluate the indicated minors.

#math.equation(block: false, alt: "| 3, −1, 4; −1, 0, −2; −4, 1, 5 |")[$| 3 &  &  & −1 &  &  & 4 \ −1 &  &  & 0 &  &  & −2 \ −4 &  &  & 1 &  &  & 5 |$] #linebreak() Find the minor ⓐ #math.equation(block: false, alt: "a sub 1")[$a_(1)$] ⓑ #math.equation(block: false, alt: "b sub 2")[$b_(2)$] ⓒ #math.equation(block: false, alt: "c sub 3")[$c_(3)$]

#math.equation(block: false, alt: "| −1, −3, 2; 4, −2, −1; −2, 0, −3 |")[$| −1 &  &  & −3 &  &  & 2 \ 4 &  &  & −2 &  &  & −1 \ −2 &  &  & 0 &  &  & −3 |$] #linebreak() Find the minor ⓐ #math.equation(block: false, alt: "a sub 1")[$a_(1)$] ⓑ #math.equation(block: false, alt: "b sub 1")[$b_(1)$] ⓒ #math.equation(block: false, alt: "c sub 2")[$c_(2)$]

#solutionbox[
ⓐ 6  ⓑ #math.equation(block: false, alt: "−14")[$−14$] ⓒ #math.equation(block: false, alt: "−6")[$−6$]
]

#math.equation(block: false, alt: "| 2, −3, −4; −1, 2, −3; 0, −1, −2 |")[$| 2 &  &  & −3 &  &  & −4 \ −1 &  &  & 2 &  &  & −3 \ 0 &  &  & −1 &  &  & −2 |$] #linebreak() Find the minor ⓐ #math.equation(block: false, alt: "a sub 2")[$a_(2)$] ⓑ #math.equation(block: false, alt: "b sub 2")[$b_(2)$] ⓒ #math.equation(block: false, alt: "c sub 2")[$c_(2)$]

#math.equation(block: false, alt: "| −2, −2, 3; 1, −3, 0; −2, 3, −2 |")[$| −2 &  &  & −2 &  &  & 3 \ 1 &  &  & −3 &  &  & 0 \ −2 &  &  & 3 &  &  & −2 |$] #linebreak() Find the minor ⓐ #math.equation(block: false, alt: "a sub 3")[$a_(3)$] ⓑ #math.equation(block: false, alt: "b sub 3")[$b_(3)$] ⓒ #math.equation(block: false, alt: "c sub 3")[$c_(3)$]

#solutionbox[
ⓐ 9  ⓑ #math.equation(block: false, alt: "−3")[$−3$] ⓒ 8
]

In the following exercises, evaluate each determinant by expanding by minors along the first row.

#math.equation(block: true, alt: "| −2, 3, −1; −1, 2, −2; 3, 1, −3 |")[$| −2 &  &  & 3 &  &  & −1 \ −1 &  &  & 2 &  &  & −2 \ 3 &  &  & 1 &  &  & −3 |$]

#math.equation(block: true, alt: "| 4, −1, −2; −3, −2, 1; −2, −5, 7 |")[$| 4 &  &  & −1 &  &  & −2 \ −3 &  &  & −2 &  &  & 1 \ −2 &  &  & −5 &  &  & 7 |$]

#solutionbox[
#math.equation(block: true, alt: "−77")[$−77$]
]

#math.equation(block: true, alt: "| −2, −3, −4; 5, −6, 7; −1, 2, 0 |")[$| −2 &  &  & −3 &  &  & −4 \ 5 &  &  & −6 &  &  & 7 \ −1 &  &  & 2 &  &  & 0 |$]

#math.equation(block: true, alt: "| 1, 3, −2; 5, −6, 4; 0, −2, −1 |")[$| 1 &  &  & 3 &  &  & −2 \ 5 &  &  & −6 &  &  & 4 \ 0 &  &  & −2 &  &  & −1 |$]

#solutionbox[
49
]

In the following exercises, evaluate each determinant by expanding by minors.

#math.equation(block: true, alt: "| −5, −1, −4; 4, 0, −3; 2, −2, 6 |")[$| −5 &  &  & −1 &  &  & −4 \ 4 &  &  & 0 &  &  & −3 \ 2 &  &  & −2 &  &  & 6 |$]

#math.equation(block: true, alt: "| 4, −1, 3; 3, −2, 2; −1, 0, 4 |")[$| 4 &  &  & −1 &  &  & 3 \ 3 &  &  & −2 &  &  & 2 \ −1 &  &  & 0 &  &  & 4 |$]

#solutionbox[
#math.equation(block: true, alt: "−24")[$−24$]
]

#math.equation(block: true, alt: "| 3, 5, 4; −1, 3, 0; −2, 6, 1 |")[$| 3 &  &  & 5 &  &  & 4 \ −1 &  &  & 3 &  &  & 0 \ −2 &  &  & 6 &  &  & 1 |$]

#math.equation(block: true, alt: "| 2, −4, −3; 5, −1, −4; 3, 2, 0 |")[$| 2 &  &  & −4 &  &  & −3 \ 5 &  &  & −1 &  &  & −4 \ 3 &  &  & 2 &  &  & 0 |$]

#solutionbox[
25
]

#strong[Use Cramer’s Rule to Solve Systems of Equations]

In the following exercises, solve each system of equations using Cramer’s Rule.

#math.equation(block: true, alt: "{ −2 x plus 3 y equals 3; x plus 3 y equals 12")[$\{ −2 x + 3 y = 3 \ x + 3 y = 12$]

#math.equation(block: true, alt: "{ x minus 2 y equals −5; 2 x minus 3 y equals −4")[$\{ x − 2 y = −5 \ 2 x − 3 y = −4$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis 7 , 6 close parenthesis")[$( 7 , 6 )$]
]

#math.equation(block: true, alt: "{ x minus 3 y equals −9; 2 x plus 5 y equals 4")[$\{ x − 3 y = −9 \ 2 x + 5 y = 4$]

#math.equation(block: true, alt: "{ 2 x plus y equals −4; 3 x minus 2 y equals −6")[$\{ 2 x + y = −4 \ 3 x − 2 y = −6$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis −2 , 0 close parenthesis")[$( −2 , 0 )$]
]

#math.equation(block: true, alt: "{ x minus 2 y equals −5; 2 x minus 3 y equals −4")[$\{ x − 2 y = −5 \ 2 x − 3 y = −4$]

#math.equation(block: true, alt: "{ x minus 3 y equals −9; 2 x plus 5 y equals 4")[$\{ x − 3 y = −9 \ 2 x + 5 y = 4$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis −3 , 2 close parenthesis")[$( −3 , 2 )$]
]

#math.equation(block: true, alt: "{ 5 x minus 3 y equals −1; 2 x minus y equals 2")[$\{ 5 x − 3 y = −1 \ 2 x − y = 2$]

#math.equation(block: true, alt: "{ 3 x plus 8 y equals −3; 2 x plus 5 y equals −3")[$\{ 3 x + 8 y = −3 \ 2 x + 5 y = −3$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis −9 , 3 close parenthesis")[$( −9 , 3 )$]
]

#math.equation(block: true, alt: "{ 6 x minus 5 y plus 2 z equals 3; 2 x plus y minus 4 z equals 5; 3 x minus 3 y plus z equals −1")[$\{ 6 x − 5 y + 2 z = 3 \ 2 x + y − 4 z = 5 \ 3 x − 3 y + z = −1$]

#math.equation(block: true, alt: "{ 4 x minus 3 y plus z equals 7; 2 x minus 5 y minus 4 z equals 3; 3 x minus 2 y minus 2 z equals −7")[$\{ 4 x − 3 y + z = 7 \ 2 x − 5 y − 4 z = 3 \ 3 x − 2 y − 2 z = −7$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis −3 , −5 , 4 close parenthesis")[$( −3 , −5 , 4 )$]
]

#math.equation(block: true, alt: "{ 2 x minus 5 y plus 3 z equals 8; 3 x minus y plus 4 z equals 7; x plus 3 y plus 2 z equals −3")[$\{ 2 x − 5 y + 3 z = 8 \ 3 x − y + 4 z = 7 \ x + 3 y + 2 z = −3$]

#math.equation(block: true, alt: "{ 11 x plus 9 y plus 2 z equals −9; 7 x plus 5 y plus 3 z equals −7; 4 x plus 3 y plus z equals −3")[$\{ 11 x + 9 y + 2 z = −9 \ 7 x + 5 y + 3 z = −7 \ 4 x + 3 y + z = −3$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis 2 , −3 , −2 close parenthesis")[$( 2 , −3 , −2 )$]
]

#math.equation(block: true, alt: "{ x plus 2 z equals 0; 4 y plus 3 z equals −2; 2 x minus 5 y equals 3")[$\{ x + 2 z = 0 \ 4 y + 3 z = −2 \ 2 x − 5 y = 3$]

#math.equation(block: true, alt: "{ 2 x plus 5 y equals 4; 3 y minus z equals 3; 4 x plus 3 z equals −3")[$\{ 2 x + 5 y = 4 \ 3 y − z = 3 \ 4 x + 3 z = −3$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis −3 , 2 , 3 close parenthesis")[$( −3 , 2 , 3 )$]
]

#math.equation(block: true, alt: "{ 2 y plus 3 z equals −1; 5 x plus 3 y equals −6; 7 x plus z equals 1")[$\{ 2 y + 3 z = −1 \ 5 x + 3 y = −6 \ 7 x + z = 1$]

#math.equation(block: true, alt: "{ 3 x minus z equals −3; 5 y plus 2 z equals −6; 4 x plus 3 y equals −8")[$\{ 3 x − z = −3 \ 5 y + 2 z = −6 \ 4 x + 3 y = −8$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis −2 , 0 , −3 close parenthesis")[$( −2 , 0 , −3 )$]
]

#math.equation(block: true, alt: "{ 2 x plus y equals 3; 6 x plus 3 y equals 9")[$\{ 2 x + y = 3 \ 6 x + 3 y = 9$]

#math.equation(block: true, alt: "{ x minus 4 y equals −1; −3 x plus 12 y equals 3")[$\{ x − 4 y = −1 \ −3 x + 12 y = 3$]

#solutionbox[
infinitely many solutions
]

#math.equation(block: true, alt: "{ −3 x minus y equals 4; 6 x plus 2 y equals −16")[$\{ −3 x − y = 4 \ 6 x + 2 y = −16$]

#math.equation(block: true, alt: "{ 4 x plus 3 y equals 2; 20 x plus 15 y equals 5")[$\{ 4 x + 3 y = 2 \ 20 x + 15 y = 5$]

#solutionbox[
inconsistent
]

#math.equation(block: true, alt: "{ x plus y minus 3 z equals −1; y minus z equals 0; − x plus 2 y equals 1")[$\{ x + y − 3 z = −1 \ y − z = 0 \ "−" x + 2 y = 1$]

#math.equation(block: true, alt: "{ 2 x plus 3 y plus z equals 12; x plus y plus z equals 9; 3 x plus 4 y plus 2 z equals 20")[$\{ 2 x + 3 y + z = 12 \ x + y + z = 9 \ 3 x + 4 y + 2 z = 20$]

#solutionbox[
inconsistent
]

#math.equation(block: true, alt: "{ 3 x plus 4 y minus 3 z equals −2; 2 x plus 3 y minus z equals −12; x plus y minus 2 z equals 6")[$\{ 3 x + 4 y − 3 z = −2 \ 2 x + 3 y − z = −12 \ x + y − 2 z = 6$]

#math.equation(block: true, alt: "{ x minus 2 y plus 3 z equals 1; x plus y minus 3 z equals 7; 3 x minus 4 y plus 5 z equals 7")[$\{ x − 2 y + 3 z = 1 \ x + y − 3 z = 7 \ 3 x − 4 y + 5 z = 7$]

#solutionbox[
infinitely many solutions
]

#strong[Solve Applications Using Determinants]

In the following exercises, determine whether the given points are collinear.

#math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis ,")[$( 0 , 1 ) ,$] #math.equation(block: false, alt: "open parenthesis 2 , 0 close parenthesis ,")[$( 2 , 0 ) ,$] and #math.equation(block: false, alt: "open parenthesis −2 , 2 close parenthesis .")[$( −2 , 2 ) .$]

#math.equation(block: false, alt: "open parenthesis 0 , −5 close parenthesis ,")[$( 0 , −5 ) ,$] #math.equation(block: false, alt: "open parenthesis −2 , −2 close parenthesis ,")[$( −2 , −2 ) ,$] and #math.equation(block: false, alt: "open parenthesis 2 , −8 close parenthesis .")[$( 2 , −8 ) .$]

#solutionbox[
yes
]

#math.equation(block: false, alt: "open parenthesis 4 , −3 close parenthesis ,")[$( 4 , −3 ) ,$] #math.equation(block: false, alt: "open parenthesis 6 , −4 close parenthesis ,")[$( 6 , −4 ) ,$] and #math.equation(block: false, alt: "open parenthesis 2 , −2 close parenthesis .")[$( 2 , −2 ) .$]

#math.equation(block: false, alt: "open parenthesis −2 , 1 close parenthesis ,")[$( −2 , 1 ) ,$] #math.equation(block: false, alt: "open parenthesis −4 , 4 close parenthesis ,")[$( −4 , 4 ) ,$] and #math.equation(block: false, alt: "open parenthesis 0 , −2 close parenthesis .")[$( 0 , −2 ) .$]

#solutionbox[
yes
]


==== Writing Exercises

Explain the difference between a square matrix and its determinant. Give an example of each.

Explain what is meant by the minor of an entry in a square matrix.

#solutionbox[
Answers will vary.
]

Explain how to decide which row or column you will use to expand a #math.equation(block: false, alt: "3 times 3")[$3 #h(0.2em) × #h(0.2em) 3$] determinant.

Explain the steps for solving a system of equations using Cramer’s rule.

#solutionbox[
Answers will vary.
]


==== Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

#figure(figph[This table has 4 columns, 4 rows and a header row. The header row labels each column: I ca, confidently, with some help and no, I don’t get it. The first column has the following statements: Evaluate the Determinant of a 2 by 2 Matrix, Evaluate the Determinant of a 3 by 3 Matrix, Use Cramer’s Rule to Solve Systems of Equations, Solve Applications Using Determinants. The remaining columns are blank.], alt: "This table has 4 columns, 4 rows and a header row. The header row labels each column: I ca, confidently, with some help and no, I don’t get it. The first column has the following statements: Evaluate the Determinant of a 2 by 2 Matrix, Evaluate the Determinant of a 3 by 3 Matrix, Use Cramer’s Rule to Solve Systems of Equations, Solve Applications Using Determinants. The remaining columns are blank.", caption: none)

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?