#set document(title: "9.7 Exercises", author: "Jiří Lebl") #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")) == 9.7#h(0.6em)Exercises === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/Appendix_A%3A_Linear_Algebra/A.1%3A_Vectors_Mappings_and_Matrices")[A.1: Vectors, Mappings, and Matrices] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ On a piece of graph paper draw the vectors: + #math.equation(block: false, alt: "[ 2; 5 ]")[$[ 2 \ 5 ]$] + #math.equation(block: false, alt: "[ minus 2; minus 4 ]")[$[ − 2 \ − 4 ]$] + #math.equation(block: false, alt: "open parenthesis 3 , minus 4 close parenthesis")[$( 3 , − 4 )$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ On a piece of graph paper draw the vector #math.equation(block: false, alt: "open parenthesis 1 , 2 close parenthesis")[$( 1 , 2 )$] starting at (based at) the given point: + based at #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] + based at #math.equation(block: false, alt: "open parenthesis 1 , 2 close parenthesis")[$( 1 , 2 )$] + based at #math.equation(block: false, alt: "open parenthesis 0 , minus 1 close parenthesis")[$( 0 , − 1 )$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ On a piece of graph paper draw the following operations. Draw and label the vectors involved in the operations as well as the result: + #math.equation(block: false, alt: "[ 1; minus 4 ] plus [ 2; 3 ]")[$[ 1 \ − 4 ] + [ 2 \ 3 ]$] + #math.equation(block: false, alt: "[ minus 3; 2 ] minus [ 1; 3 ]")[$[ − 3 \ 2 ] − [ 1 \ 3 ]$] + #math.equation(block: false, alt: "3 [ 2; 1 ]")[$3 [ 2 \ 1 ]$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute the magnitude of + #math.equation(block: false, alt: "[ 7; 2 ]")[$[ 7 \ 2 ]$] + #math.equation(block: false, alt: "[ minus 2; 3; 1 ]")[$[ − 2 \ 3 \ 1 ]$] + #math.equation(block: false, alt: "open parenthesis 1 , 3 , minus 4 close parenthesis")[$( 1 , 3 , − 4 )$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute + #math.equation(block: false, alt: "[ 2; 3 ] plus [ 7; minus 8 ]")[$[ 2 \ 3 ] + [ 7 \ − 8 ]$] + #math.equation(block: false, alt: "[ minus 2; 3 ] minus [ 6; minus 4 ]")[$[ − 2 \ 3 ] − [ 6 \ − 4 ]$] + #math.equation(block: false, alt: "minus [ minus 3; 2 ]")[$− [ − 3 \ 2 ]$] + #math.equation(block: false, alt: "4 [ minus 1; 5 ]")[$4 [ − 1 \ 5 ]$] + #math.equation(block: false, alt: "5 [ 1; 0 ] plus 9 [ 0; 1 ]")[$5 [ 1 \ 0 ] + 9 [ 0 \ 1 ]$] + #math.equation(block: false, alt: "3 [ 1; minus 8 ] minus 2 [ 3; minus 1 ]")[$3 [ 1 \ − 8 ] − 2 [ 3 \ − 1 ]$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the unit vector in the direction of the given vector + #math.equation(block: false, alt: "[ 1; minus 3 ]")[$[ 1 \ − 3 ]$] + #math.equation(block: false, alt: "[ 2; 1; minus 1 ]")[$[ 2 \ 1 \ − 1 ]$] + #math.equation(block: false, alt: "open parenthesis 3 , 1 , minus 2 close parenthesis")[$( 3 , 1 , − 2 )$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ If #math.equation(block: false, alt: "x → equals open parenthesis 1 , 2 close parenthesis")[$arrow(x) = ( 1 , 2 )$] and #math.equation(block: false, alt: "y →")[$arrow(y)$] are added together, we find #math.equation(block: false, alt: "x → plus y → equals open parenthesis 0 , 2 close parenthesis")[$arrow(x) + arrow(y) = ( 0 , 2 )$]. What is #math.equation(block: false, alt: "y →")[$arrow(y)$]? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Write #math.equation(block: false, alt: "open parenthesis 1 , 2 , 3 close parenthesis")[$( 1 , 2 , 3 )$] as a linear combination of the standard basis vectors #math.equation(block: false, alt: "e → sub 1")[$arrow(e)_(1)$], #math.equation(block: false, alt: "e → sub 2")[$arrow(e)_(2)$], and #math.equation(block: false, alt: "e → sub 3")[$arrow(e)_(3)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ If the magnitude of #math.equation(block: false, alt: "x →")[$arrow(x)$] is 4, what is the magnitude of + #math.equation(block: false, alt: "0 x →")[$0 arrow(x)$] + #math.equation(block: false, alt: "3 x →")[$3 arrow(x)$] + #math.equation(block: false, alt: "minus x →")[$− arrow(x)$] + #math.equation(block: false, alt: "minus 4 x →")[$− 4 arrow(x)$] + #math.equation(block: false, alt: "x → plus x →")[$arrow(x) + arrow(x)$] + #math.equation(block: false, alt: "x → minus x →")[$arrow(x) − arrow(x)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose a linear mapping #math.equation(block: false, alt: "F : R squared → R squared")[$F : R^(2) → R^(2)$] takes #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$] to #math.equation(block: false, alt: "open parenthesis 2 , minus 1 close parenthesis")[$( 2 , − 1 )$] and it takes #math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$] to #math.equation(block: false, alt: "open parenthesis 3 , 3 close parenthesis")[$( 3 , 3 )$]. Where does it take + #math.equation(block: false, alt: "open parenthesis 1 , 1 close parenthesis")[$( 1 , 1 )$] + #math.equation(block: false, alt: "open parenthesis 2 , 0 close parenthesis")[$( 2 , 0 )$] + #math.equation(block: false, alt: "open parenthesis 2 , minus 1 close parenthesis")[$( 2 , − 1 )$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose a linear mapping #math.equation(block: false, alt: "F : R cubed → R squared")[$F : R^(3) → R^(2)$] takes #math.equation(block: false, alt: "open parenthesis 1 , 0 , 0 close parenthesis")[$( 1 , 0 , 0 )$] to #math.equation(block: false, alt: "open parenthesis 2 , 1 close parenthesis")[$( 2 , 1 )$], it takes #math.equation(block: false, alt: "open parenthesis 0 , 1 , 0 close parenthesis")[$( 0 , 1 , 0 )$] to #math.equation(block: false, alt: "open parenthesis 3 , 4 close parenthesis")[$( 3 , 4 )$], and it takes #math.equation(block: false, alt: "open parenthesis 0 , 0 , 1 close parenthesis")[$( 0 , 0 , 1 )$] to #math.equation(block: false, alt: "open parenthesis 5 , 6 close parenthesis")[$( 5 , 6 )$]. Write down the matrix representing the mapping #math.equation(block: false, alt: "F")[$F$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose that a mapping #math.equation(block: false, alt: "F : R squared → ℝ squared")[$F : R^(2) → ℝ^(2)$] takes #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$] to #math.equation(block: false, alt: "open parenthesis 1 , 2 close parenthesis")[$( 1 , 2 )$], #math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$] to #math.equation(block: false, alt: "open parenthesis 3 , 4 close parenthesis")[$( 3 , 4 )$], and #math.equation(block: false, alt: "open parenthesis 1 , 1 close parenthesis")[$( 1 , 1 )$] to #math.equation(block: false, alt: "open parenthesis 0 , minus 1 close parenthesis")[$( 0 , − 1 )$]. Explain why #math.equation(block: false, alt: "F")[$F$] is not linear. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[(challenging)] Let #math.equation(block: false, alt: "R cubed")[$R^(3)$] represent the space of quadratic polynomials in #math.equation(block: false, alt: "t")[$t$]: a point #math.equation(block: false, alt: "open parenthesis a sub 0 , a sub 1 , a sub 2 close parenthesis")[$( a_(0) , a_(1) , a_(2) )$] in #math.equation(block: false, alt: "R cubed")[$R^(3)$] represents the polynomial #math.equation(block: false, alt: "a sub 0 plus a sub 1 t plus a sub 2 t squared")[$a_(0) + a_(1) t + a_(2) t^(2)$]. Consider the derivative #math.equation(block: false, alt: "the fraction d over d t")[$frac(d, d t)$] as a mapping of #math.equation(block: false, alt: "R cubed")[$R^(3)$] to #math.equation(block: false, alt: "R cubed")[$R^(3)$], and note that #math.equation(block: false, alt: "the fraction d over d t")[$frac(d, d t)$] is linear. Write down #math.equation(block: false, alt: "the fraction d over d t")[$frac(d, d t)$] as a #math.equation(block: false, alt: "3 times 3")[$3 × 3$] matrix. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute the magnitude of + #math.equation(block: false, alt: "[ 1; 3 ]")[$[ 1 \ 3 ]$] + #math.equation(block: false, alt: "[ 2; 3; minus 1 ]")[$[ 2 \ 3 \ − 1 ]$] + #math.equation(block: false, alt: "open parenthesis minus 2 , 1 , minus 2 close parenthesis")[$( − 2 , 1 , − 2 )$] Answer + #math.equation(block: false, alt: "the square root of 10")[$sqrt(10)$] + #math.equation(block: false, alt: "the square root of 14")[$sqrt(14)$] + #math.equation(block: false, alt: "3")[$3$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the unit vector in the direction of the given vector + #math.equation(block: false, alt: "[ minus 1; 1 ]")[$[ − 1 \ 1 ]$] + #math.equation(block: false, alt: "[ 1; minus 1; 2 ]")[$[ 1 \ − 1 \ 2 ]$] + #math.equation(block: false, alt: "open parenthesis 2 , minus 5 , 2 close parenthesis")[$( 2 , − 5 , 2 )$] Answer + #math.equation(block: false, alt: "the fraction minus 1 over the square root of 2; the fraction 1 over the square root of 2")[$frac(− 1, sqrt(2)) \ frac(1, sqrt(2))$] + #math.equation(block: false, alt: "the fraction 1 over the square root of 6; the fraction minus 1 over the square root of 6; the fraction 2 over the square root of 6")[$frac(1, sqrt(6)) \ frac(− 1, sqrt(6)) \ frac(2, sqrt(6))$] + #math.equation(block: false, alt: "open parenthesis the fraction 2 over the square root of 33 , the fraction minus 5 over the square root of 33 , the fraction 2 over the square root of 33 close parenthesis")[$( frac(2, sqrt(33)) , frac(− 5, sqrt(33)) , frac(2, sqrt(33)) )$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute + #math.equation(block: false, alt: "[ 3; 1 ] plus [ 6; minus 3 ]")[$[ 3 \ 1 ] + [ 6 \ − 3 ]$] + #math.equation(block: false, alt: "[ minus 1; 2 ] minus [ 2; minus 1 ]")[$[ − 1 \ 2 ] − [ 2 \ − 1 ]$] + #math.equation(block: false, alt: "minus [ minus 5; 3 ]")[$− [ − 5 \ 3 ]$] + #math.equation(block: false, alt: "2 [ minus 2; 4 ]")[$2 [ − 2 \ 4 ]$] + #math.equation(block: false, alt: "3 [ 1; 0 ] plus 7 [ 0; 1 ]")[$3 [ 1 \ 0 ] + 7 [ 0 \ 1 ]$] + #math.equation(block: false, alt: "2 [ 2; minus 3 ] minus 6 [ 2; minus 1 ]")[$2 [ 2 \ − 3 ] − 6 [ 2 \ − 1 ]$] Answer + #math.equation(block: false, alt: "9; minus 2")[$9 \ − 2$] + #math.equation(block: false, alt: "minus 3; 3")[$− 3 \ 3$] + #math.equation(block: false, alt: "5; minus 3")[$5 \ − 3$] + #math.equation(block: false, alt: "minus 4; 8")[$− 4 \ 8$] + #math.equation(block: false, alt: "3; 7")[$3 \ 7$] + #math.equation(block: false, alt: "minus 8; 3")[$− 8 \ 3$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ If the magnitude of #math.equation(block: false, alt: "x →")[$arrow(x)$] is 5, what is the magnitude of + #math.equation(block: false, alt: "4 x →")[$4 arrow(x)$] + #math.equation(block: false, alt: "minus 2 x →")[$− 2 arrow(x)$] + #math.equation(block: false, alt: "minus 4 x →")[$− 4 arrow(x)$] Answer + #math.equation(block: false, alt: "20")[$20$] + #math.equation(block: false, alt: "10")[$10$] + #math.equation(block: false, alt: "20")[$20$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose a linear mapping #math.equation(block: false, alt: "F : R squared → R squared")[$F : R^(2) → R^(2)$] takes #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$] to #math.equation(block: false, alt: "open parenthesis 1 , minus 1 close parenthesis")[$( 1 , − 1 )$] and it takes #math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$] to #math.equation(block: false, alt: "open parenthesis 2 , 0 close parenthesis")[$( 2 , 0 )$]. Where does it take + #math.equation(block: false, alt: "open parenthesis 1 , 1 close parenthesis")[$( 1 , 1 )$] + #math.equation(block: false, alt: "open parenthesis 0 , 2 close parenthesis")[$( 0 , 2 )$] + #math.equation(block: false, alt: "open parenthesis 1 , minus 1 close parenthesis")[$( 1 , − 1 )$] Answer + #math.equation(block: false, alt: "open parenthesis 3 , minus 1 close parenthesis")[$( 3 , − 1 )$] + #math.equation(block: false, alt: "open parenthesis 4 , 0 close parenthesis")[$( 4 , 0 )$] + #math.equation(block: false, alt: "open parenthesis minus 1 , minus 1 close parenthesis")[$( − 1 , − 1 )$] ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/Appendix_A%3A_Linear_Algebra/A.2%3A_Matrix_Algebra")[A.2: Matrix Algebra] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Add the following matrices + #math.equation(block: false, alt: "[ minus 1, 2, 2; 5, 8, minus 1 ] plus [ 3, 2, 3; 8, 3, 5 ]")[$[ − 1 & 2 & 2 \ 5 & 8 & − 1 ] + [ 3 & 2 & 3 \ 8 & 3 & 5 ]$] + #math.equation(block: false, alt: "[ 1, 2, 4; 2, 3, 1; 0, 5, 1 ] plus [ 2, minus 8, minus 3; 3, 1, 0; 6, minus 4, 1 ]")[$[ 1 & 2 & 4 \ 2 & 3 & 1 \ 0 & 5 & 1 ] + [ 2 & − 8 & − 3 \ 3 & 1 & 0 \ 6 & − 4 & 1 ]$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute + #math.equation(block: false, alt: "3 [ 0, 3; minus 2, 2 ] plus 6 [ 1, 5; minus 1, 5 ]")[$3 [ 0 & 3 \ − 2 & 2 ] + 6 [ 1 & 5 \ − 1 & 5 ]$] + #math.equation(block: false, alt: "2 [ minus 3, 1; 2, 2 ] minus 3 [ 2, minus 1; 3, 2 ]")[$2 [ − 3 & 1 \ 2 & 2 ] − 3 [ 2 & − 1 \ 3 & 2 ]$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Multiply the following matrices + #math.equation(block: false, alt: "[ minus 1, 2; 3, 1; 5, 8 ] [ 3, minus 1, 3, 1; 8, 3, 2, minus 3 ]")[$[ − 1 & 2 \ 3 & 1 \ 5 & 8 ] [ 3 & − 1 & 3 & 1 \ 8 & 3 & 2 & − 3 ]$] + #math.equation(block: false, alt: "[ 1, 2, 3; 3, 1, 1; 1, 0, 3 ] [ 2, 3, 1, 7; 1, 2, 3, minus 1; 1, minus 1, 3, 0 ]")[$[ 1 & 2 & 3 \ 3 & 1 & 1 \ 1 & 0 & 3 ] [ 2 & 3 & 1 & 7 \ 1 & 2 & 3 & − 1 \ 1 & − 1 & 3 & 0 ]$] + #math.equation(block: false, alt: "[ 4, 1, 6, 3; 5, 6, 5, 0; 4, 6, 6, 0 ] [ 2, 5; 1, 2; 3, 5; 5, 6 ]")[$[ 4 & 1 & 6 & 3 \ 5 & 6 & 5 & 0 \ 4 & 6 & 6 & 0 ] [ 2 & 5 \ 1 & 2 \ 3 & 5 \ 5 & 6 ]$] + #math.equation(block: false, alt: "[ 1, 1, 4; 0, 5, 1 ] [ 2, 2; 1, 0; 6, 4 ]")[$[ 1 & 1 & 4 \ 0 & 5 & 1 ] [ 2 & 2 \ 1 & 0 \ 6 & 4 ]$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute the inverse of the given matrices + #math.equation(block: false, alt: "[ minus 3 ]")[$[ − 3 ]$] + #math.equation(block: false, alt: "[ 0, minus 1; 1, 0 ]")[$[ 0 & − 1 \ 1 & 0 ]$] + #math.equation(block: false, alt: "[ 1, 4; 1, 3 ]")[$[ 1 & 4 \ 1 & 3 ]$] + #math.equation(block: false, alt: "[ 2, 2; 1, 4 ]")[$[ 2 & 2 \ 1 & 4 ]$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute the inverse of the given matrices + #math.equation(block: false, alt: "[ minus 2, 0; 0, 1 ]")[$[ − 2 & 0 \ 0 & 1 ]$] + #math.equation(block: false, alt: "[ 3, 0, 0; 0, minus 2, 0; 0, 0, 1 ]")[$[ 3 & 0 & 0 \ 0 & − 2 & 0 \ 0 & 0 & 1 ]$] + #math.equation(block: false, alt: "[ 1, 0, 0, 0; 0, minus 1, 0, 0; 0, 0, 0.01, 0; 0, 0, 0, minus 5 ]")[$[ 1 & 0 & 0 & 0 \ 0 & − 1 & 0 & 0 \ 0 & 0 & 0.01 & 0 \ 0 & 0 & 0 & − 5 ]$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Add the following matrices + #math.equation(block: false, alt: "[ 2, 1, 0; 1, 1, minus 1 ] plus [ 5, 3, 4; 1, 2, 5 ]")[$[ 2 & 1 & 0 \ 1 & 1 & − 1 ] + [ 5 & 3 & 4 \ 1 & 2 & 5 ]$] + #math.equation(block: false, alt: "[ 6, minus 2, 3; 7, 3, 3; 8, minus 1, 2 ] plus [ minus 1, minus 1, minus 3; 6, 7, 3; minus 9, 4, minus 1 ]")[$[ 6 & − 2 & 3 \ 7 & 3 & 3 \ 8 & − 1 & 2 ] + [ − 1 & − 1 & − 3 \ 6 & 7 & 3 \ − 9 & 4 & − 1 ]$] Answer Add texts here. Do not delete this text first. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute + #math.equation(block: false, alt: "2 [ 1, 2; 3, 4 ] plus 3 [ minus 1, 3; 1, 2 ]")[$2 [ 1 & 2 \ 3 & 4 ] + 3 [ − 1 & 3 \ 1 & 2 ]$] + #math.equation(block: false, alt: "3 [ 2, minus 1; 1, 3 ] minus 2 [ 2, 1; minus 1, 2 ]")[$3 [ 2 & − 1 \ 1 & 3 ] − 2 [ 2 & 1 \ − 1 & 2 ]$] Answer Add texts here. Do not delete this text first. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Multiply the following matrices + #math.equation(block: false, alt: "[ 2, 1, 4; 3, 4, 4 ] [ 2, 4; 6, 3; 3, 5 ]")[$[ 2 & 1 & 4 \ 3 & 4 & 4 ] [ 2 & 4 \ 6 & 3 \ 3 & 5 ]$] + #math.equation(block: false, alt: "[ 0, 3, 3; 2, minus 2, 1; 3, 5, minus 2 ] [ 6, 6, 2; 4, 6, 0; 2, 0, 4 ]")[$[ 0 & 3 & 3 \ 2 & − 2 & 1 \ 3 & 5 & − 2 ] [ 6 & 6 & 2 \ 4 & 6 & 0 \ 2 & 0 & 4 ]$] + #math.equation(block: false, alt: "[ 3, 4, 1; 2, minus 1, 0; 4, minus 1, 5 ] [ 0, 2, 5, 0; 2, 0, 5, 2; 3, 6, 1, 6 ]")[$[ 3 & 4 & 1 \ 2 & − 1 & 0 \ 4 & − 1 & 5 ] [ 0 & 2 & 5 & 0 \ 2 & 0 & 5 & 2 \ 3 & 6 & 1 & 6 ]$] + #math.equation(block: false, alt: "[ minus 2, minus 2; 5, 3; 2, 1 ] [ 0, 3; 1, 3 ]")[$[ − 2 & − 2 \ 5 & 3 \ 2 & 1 ] [ 0 & 3 \ 1 & 3 ]$] Answer Add texts here. Do not delete this text first. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute the inverse of the given matrices + #math.equation(block: false, alt: "[ 2 ]")[$[ 2 ]$] + #math.equation(block: false, alt: "[ 0, 1; 1, 0 ]")[$[ 0 & 1 \ 1 & 0 ]$] + #math.equation(block: false, alt: "[ 1, 2; 3, 5 ]")[$[ 1 & 2 \ 3 & 5 ]$] + #math.equation(block: false, alt: "[ 4, 2; 4, 4 ]")[$[ 4 & 2 \ 4 & 4 ]$] Answer Add texts here. Do not delete this text first. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute the inverse of the given matrices + #math.equation(block: false, alt: "[ 2, 0; 0, 3 ]")[$[ 2 & 0 \ 0 & 3 ]$] + #math.equation(block: false, alt: "[ 4, 0, 0; 0, 5, 0; 0, 0, minus 1 ]")[$[ 4 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & − 1 ]$] + #math.equation(block: false, alt: "[ minus 1, 0, 0, 0; 0, 2, 0, 0; 0, 0, 3, 0; 0, 0, 0, 0.1 ]")[$[ − 1 & 0 & 0 & 0 \ 0 & 2 & 0 & 0 \ 0 & 0 & 3 & 0 \ 0 & 0 & 0 & 0.1 ]$] Answer Add texts here. Do not delete this text first. ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/Appendix_A%3A_Linear_Algebra/A.3%3A_Elimination")[A.3: Elimination] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute the reduced row echelon form for the following matrices: + #math.equation(block: false, alt: "[ 1, 3, 1; 0, 1, 1 ]")[$[ 1 & 3 & 1 \ 0 & 1 & 1 ]$] + #math.equation(block: false, alt: "[ 3, 3; 6, minus 3 ]")[$[ 3 & 3 \ 6 & − 3 ]$] + #math.equation(block: false, alt: "[ 3, 6; minus 2, minus 3 ]")[$[ 3 & 6 \ − 2 & − 3 ]$] + #math.equation(block: false, alt: "[ 6, 6, 7, 7; 1, 1, 0, 1 ]")[$[ 6 & 6 & 7 & 7 \ 1 & 1 & 0 & 1 ]$] + #math.equation(block: false, alt: "[ 9, 3, 0, 2; 8, 6, 3, 6; 7, 9, 7, 9 ]")[$[ 9 & 3 & 0 & 2 \ 8 & 6 & 3 & 6 \ 7 & 9 & 7 & 9 ]$] + #math.equation(block: false, alt: "[ 2, 1, 3, minus 3; 6, 0, 0, minus 1; minus 2, 4, 4, 3 ]")[$[ 2 & 1 & 3 & − 3 \ 6 & 0 & 0 & − 1 \ − 2 & 4 & 4 & 3 ]$] + #math.equation(block: false, alt: "[ 6, 6, 5; 0, minus 2, 2; 6, 5, 6 ]")[$[ 6 & 6 & 5 \ 0 & − 2 & 2 \ 6 & 5 & 6 ]$] + #math.equation(block: false, alt: "[ 0, 2, 0, minus 1; 6, 6, minus 3, 3; 6, 2, minus 3, 5 ]")[$[ 0 & 2 & 0 & − 1 \ 6 & 6 & − 3 & 3 \ 6 & 2 & − 3 & 5 ]$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute the inverse of the given matrices + #math.equation(block: false, alt: "[ 1, 0, 0; 0, 0, 1; 0, 1, 0 ]")[$[ 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 ]$] + #math.equation(block: false, alt: "[ 1, 1, 1; 0, 2, 1; 0, 0, 1 ]")[$[ 1 & 1 & 1 \ 0 & 2 & 1 \ 0 & 0 & 1 ]$] + #math.equation(block: false, alt: "[ 1, 2, 3; 2, 0, 1; 0, 2, 1 ]")[$[ 1 & 2 & 3 \ 2 & 0 & 1 \ 0 & 2 & 1 ]$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve (find all solutions), or show no solution exists + #math.equation(block: false, alt: "4 x sub 1 plus 3 x sub 2 equals minus 2; minus x sub 1 plus x sub 2 equals 4")[$4 x_(1) + 3 x_(2) = − 2 \ − x_(1) + x_(2) = 4$] + #math.equation(block: false, alt: "x sub 1 plus 5 x sub 2 plus 3 x sub 3 equals 7; 8 x sub 1 plus 7 x sub 2 plus 8 x sub 3 equals 8; 4 x sub 1 plus 8 x sub 2 plus 6 x sub 3 equals 4")[$x_(1) + 5 x_(2) + 3 x_(3) = 7 \ 8 x_(1) + 7 x_(2) + 8 x_(3) = 8 \ 4 x_(1) + 8 x_(2) + 6 x_(3) = 4$] + #math.equation(block: false, alt: "4 x sub 1 plus 8 x sub 2 plus 2 x sub 3 equals 3; minus x sub 1 minus 2 x sub 2 plus 3 x sub 3 equals 1; 4 x sub 1 plus 8 x sub 2 plus 3 x sub 3 } equals 2")[$4 x_(1) + 8 x_(2) + 2 x_(3) = 3 \ − x_(1) − 2 x_(2) + 3 x_(3) = 1 \ 4 x_(1) + 8 x_(2) + 3 x_(3) \} = 2$] + #math.equation(block: false, alt: "x plus 2 y plus 3 z equals 4; 2 x minus y plus 3 z equals 1; 3 x plus y plus 6 z equals 6")[$x + 2 y + 3 z = 4 \ 2 x − y + 3 z = 1 \ 3 x + y + 6 z = 6$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ By computing the inverse, solve the following systems for #math.equation(block: false, alt: "x →")[$arrow(x)$]. + #math.equation(block: false, alt: "[ 4, 1; minus 1, 3 ] x → equals [ 13; 26 ]")[$[ 4 & 1 \ − 1 & 3 ] arrow(x) = [ 13 \ 26 ]$] + #math.equation(block: false, alt: "[ 3, 3; 3, 4 ] x → equals [ 2; minus 1 ]")[$[ 3 & 3 \ 3 & 4 ] arrow(x) = [ 2 \ − 1 ]$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute the rank of the given matrices + #math.equation(block: false, alt: "[ 6, 3, 5; 1, 4, 1; 7, 7, 6 ]")[$[ 6 & 3 & 5 \ 1 & 4 & 1 \ 7 & 7 & 6 ]$] + #math.equation(block: false, alt: "[ 5, minus 2, minus 1; 3, 0, 6; 2, 4, 5 ]")[$[ 5 & − 2 & − 1 \ 3 & 0 & 6 \ 2 & 4 & 5 ]$] + #math.equation(block: false, alt: "[ 1, 2, 3; minus 1, minus 2, minus 3; 2, 4, 6 ]")[$[ 1 & 2 & 3 \ − 1 & − 2 & − 3 \ 2 & 4 & 6 ]$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For the matrices in Exercise #math.equation(block: false, alt: "A .3 .5")[$A ".3" ".5"$], find a linearly independent set of row vectors that span the row space (they don’t need to be rows of the matrix). ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For the matrices in Exercise #math.equation(block: false, alt: "A .3 .5")[$A ".3" ".5"$], find a linearly independent set of columns that span the column space. That is, find the pivot columns of the matrices. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a linearly independent subset of the following vectors that has the same span. #math.equation(block: true, alt: "[ minus 1; 1; 2 ] , [ 2; minus 2; minus 4 ] , [ minus 2; 4; 1 ] , [ minus 1; 3; minus 2 ]")[$[ − 1 \ 1 \ 2 ] , #h(1em) [ 2 \ − 2 \ − 4 ] , #h(1em) [ − 2 \ 4 \ 1 ] , #h(1em) [ − 1 \ 3 \ − 2 ]$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute the reduced row echelon form for the following matrices: + #math.equation(block: false, alt: "[ 1, 0, 1; 0, 1, 0 ]")[$[ 1 & 0 & 1 \ 0 & 1 & 0 ]$] + #math.equation(block: false, alt: "[ 1, 2; 3, 4 ]")[$[ 1 & 2 \ 3 & 4 ]$] + #math.equation(block: false, alt: "[ 1, 1; minus 2, minus 2 ]")[$[ 1 & 1 \ − 2 & − 2 ]$] + #math.equation(block: false, alt: "[ 1, minus 3, 1; 4, 6, minus 2; minus 2, 6, minus 2 ]")[$[ 1 & − 3 & 1 \ 4 & 6 & − 2 \ − 2 & 6 & − 2 ]$] + #math.equation(block: false, alt: "[ 2, 2, 5, 2; 1, minus 2, 4, minus 1; 0, 3, 1, minus 2 ]")[$[ 2 & 2 & 5 & 2 \ 1 & − 2 & 4 & − 1 \ 0 & 3 & 1 & − 2 ]$] + #math.equation(block: false, alt: "[ minus 2, 6, 4, 3; 6, 0, minus 3, 0; 4, 2, minus 1, 1 ]")[$[ − 2 & 6 & 4 & 3 \ 6 & 0 & − 3 & 0 \ 4 & 2 & − 1 & 1 ]$] + #math.equation(block: false, alt: "[ 0, 0, 0, 0; 0, 0, 0, 0 ]")[$[ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 ]$] + #math.equation(block: false, alt: "[ 1, 2, 3, 3; 1, 2, 3, 5 ]")[$[ 1 & 2 & 3 & 3 \ 1 & 2 & 3 & 5 ]$] Answer + #math.equation(block: false, alt: "1 0 1; 0 1 0")[$1 0 1 \ 0 1 0$] + #math.equation(block: false, alt: "1 0; 0 1")[$1 0 \ 0 1$] + #math.equation(block: false, alt: "1 1; 0 0")[$1 1 \ 0 0$] + #math.equation(block: false, alt: "1 0 0; 0 1 minus the fraction 1 over 3; 0 0 0")[$1 0 0 \ 0 1 − frac(1, 3) \ 0 0 0$] + #math.equation(block: false, alt: "1 0 0 the fraction 77 over 15; 0 1 0 minus the fraction 2 over 15; 0 0 1 minus the fraction 8 over 5")[$1 0 0 frac(77, 15) \ 0 1 0 − frac(2, 15) \ 0 0 1 − frac(8, 5)$] + #math.equation(block: false, alt: "1 0 minus the fraction 1 over 2 0; 0 1 the fraction 1 over 2 the fraction 1 over 2; 0 0 0 0")[$1 0 − frac(1, 2) 0 \ 0 1 frac(1, 2) frac(1, 2) \ 0 0 0 0$] + #math.equation(block: false, alt: "0 0 0 0; 0 0 0 0")[$0 0 0 0 \ 0 0 0 0$] + #math.equation(block: false, alt: "1 2 3 0; 0 0 0 1")[$1 2 3 0 \ 0 0 0 1$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute the inverse of the given matrices + #math.equation(block: false, alt: "[ 0, 1, 0; minus 1, 0, 0; 0, 0, 1 ]")[$[ 0 & 1 & 0 \ − 1 & 0 & 0 \ 0 & 0 & 1 ]$] + #math.equation(block: false, alt: "[ 1, 1, 1; 1, 1, 0; 1, 0, 0 ]")[$[ 1 & 1 & 1 \ 1 & 1 & 0 \ 1 & 0 & 0 ]$] + #math.equation(block: false, alt: "[ 2, 4, 0; 2, 2, 3; 2, 4, 1 ]")[$[ 2 & 4 & 0 \ 2 & 2 & 3 \ 2 & 4 & 1 ]$] Answer + #math.equation(block: false, alt: "0 minus 1 0; 1 0 0; 0 0 1")[$0 − 1 0 \ 1 0 0 \ 0 0 1$] + #math.equation(block: false, alt: "0 0 1; 0 1 minus 1; 1 minus 1 0")[$0 0 1 \ 0 1 − 1 \ 1 − 1 0$] + #math.equation(block: false, alt: "the fraction 5 over 2 1 minus 3; minus 1 minus the fraction 1 over 2 the fraction 3 over 2; minus 1 0 1")[$frac(5, 2) 1 − 3 \ − 1 − frac(1, 2) frac(3, 2) \ − 1 0 1$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve (find all solutions), or show no solution exists + #math.equation(block: false, alt: "4 x sub 1 plus 3 x sub 2 equals minus 1; 5 x sub 1 plus 6 x sub 2 equals 4")[$4 x_(1) + 3 x_(2) = − 1 \ 5 x_(1) + 6 x_(2) = 4$] + #math.equation(block: false, alt: "5 x plus 6 y plus 5 z equals 7; 6 x plus 8 y plus 6 z equals minus 1; 5 x plus 2 y plus 5 z equals 2")[$5 x + 6 y + 5 z = 7 \ 6 x + 8 y + 6 z = − 1 \ 5 x + 2 y + 5 z = 2$] + #math.equation(block: false, alt: "a plus b plus c equals minus 1; a plus 5 b plus 6 c equals minus 1; minus 2 a plus 5 b plus 6 c equals 8")[$a + b + c = − 1 \ a + 5 b + 6 c = − 1 \ − 2 a + 5 b + 6 c = 8$] + #math.equation(block: false, alt: "minus 2 x sub 1 plus 2 x sub 2 plus 8 x sub 3 equals 6; x sub 2 plus x sub 3 equals 2; x sub 1 plus 4 x sub 2 plus x sub 3 equals 7")[$− 2 x_(1) + 2 x_(2) + 8 x_(3) = 6 \ x_(2) + x_(3) = 2 \ x_(1) + 4 x_(2) + x_(3) = 7$] Answer + #math.equation(block: false, alt: "x sub 1 equals minus 2 , x sub 2 equals the fraction 7 over 3")[$x_(1) = − 2 , #h(0.222em) x_(2) = frac(7, 3)$] + no solution + #math.equation(block: false, alt: "a equals minus 3 , b equals 10 , c equals minus 8")[$a = − 3 , #h(0.222em) b = 10 , #h(0.222em) c = − 8$] + #math.equation(block: false, alt: "x sub 3")[$x_(3)$] is free, #math.equation(block: false, alt: "x sub 1 equals minus 1 plus 3 x sub 3")[$x_(1) = − 1 + 3 x_(3)$], #math.equation(block: false, alt: "x sub 2 equals 2 minus x sub 3")[$x_(2) = 2 − x_(3)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ By computing the inverse, solve the following systems for #math.equation(block: false, alt: "x →")[$arrow(x)$]. + #math.equation(block: false, alt: "[ minus 1, 1; 3, 3 ] x → equals [ 4; 6 ]")[$[ − 1 & 1 \ 3 & 3 ] arrow(x) = [ 4 \ 6 ]$] + #math.equation(block: false, alt: "[ 2, 7; 1, 6 ] x → equals [ 1; 3 ]")[$[ 2 & 7 \ 1 & 6 ] arrow(x) = [ 1 \ 3 ]$] Answer + #math.equation(block: false, alt: "minus 1; 3")[$− 1 \ 3$] + #math.equation(block: false, alt: "minus 3; 1")[$− 3 \ 1$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute the rank of the given matrices + #math.equation(block: false, alt: "[ 7, minus 1, 6; 7, 7, 7; 7, 6, 2 ]")[$[ 7 & − 1 & 6 \ 7 & 7 & 7 \ 7 & 6 & 2 ]$] + #math.equation(block: false, alt: "[ 1, 1, 1; 1, 1, 1; 2, 2, 2 ]")[$[ 1 & 1 & 1 \ 1 & 1 & 1 \ 2 & 2 & 2 ]$] + #math.equation(block: false, alt: "[ 0, 3, minus 1; 6, 3, 1; 4, 7, minus 1 ]")[$[ 0 & 3 & − 1 \ 6 & 3 & 1 \ 4 & 7 & − 1 ]$] Answer + #math.equation(block: false, alt: "3")[$3$] + #math.equation(block: false, alt: "1")[$1$] + #math.equation(block: false, alt: "2")[$2$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For the matrices in Exercise #math.equation(block: false, alt: "A .3 .13")[$A ".3" ".13"$], find a linearly independent set of row vectors that span the row space (they don’t need to be rows of the matrix). Answer + #math.equation(block: false, alt: "1 0 0")[$1 0 0$], #math.equation(block: false, alt: "0 1 0")[$0 1 0$], #math.equation(block: false, alt: "0 0 1")[$0 0 1$] + #math.equation(block: false, alt: "1 1 1")[$1 1 1$] + #math.equation(block: false, alt: "1 0 the fraction 1 over 3")[$1 0 frac(1, 3)$], #math.equation(block: false, alt: "0 1 minus the fraction 1 over 3")[$0 1 − frac(1, 3)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For the matrices in Exercise #math.equation(block: false, alt: "A .3 .13")[$A ".3" ".13"$], find a linearly independent set of columns that span the column space. That is, find the pivot columns of the matrices. Answer + #math.equation(block: false, alt: "7; 7; 7")[$7 \ 7 \ 7$], #math.equation(block: false, alt: "minus 1; 7; 6")[$− 1 \ 7 \ 6$], #math.equation(block: false, alt: "7; 6; 2")[$7 \ 6 \ 2$] + #math.equation(block: false, alt: "1; 1; 2")[$1 \ 1 \ 2$] + #math.equation(block: false, alt: "0; 6; 4")[$0 \ 6 \ 4$], #math.equation(block: false, alt: "3; 3; 7")[$3 \ 3 \ 7$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a linearly independent subset of the following vectors that has the same span. #math.equation(block: true, alt: "[ 0; 0; 0 ] , [ 3; 1; minus 5 ] , [ 0; 3; minus 1 ] , [ minus 3; 2; 4 ]")[$[ 0 \ 0 \ 0 ] , #h(1em) [ 3 \ 1 \ − 5 ] , #h(1em) [ 0 \ 3 \ − 1 ] , #h(1em) [ − 3 \ 2 \ 4 ]$] Answer #math.equation(block: true, alt: "3; 1; minus 5")[$3 \ 1 \ − 5$], #math.equation(block: true, alt: "0; 3; minus 1")[$0 \ 3 \ − 1$] ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/Appendix_A%3A_Linear_Algebra/A.4%3A_Subspaces_Dimension_and_The_Kernel")[A.4: Subspaces, Dimension, and The Kernel] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For the following sets of vectors, find a basis for the subspace spanned by the vectors, and find the dimension of the subspace. + #math.equation(block: false, alt: "[ 1; 1; 1 ] , [ minus 1; minus 1; minus 1 ]")[$[ 1 \ 1 \ 1 ] , #h(1em) [ − 1 \ − 1 \ − 1 ]$] + #math.equation(block: false, alt: "[ 1; 0; 5 ] , [ 0; 1; 0 ] , [ 0; minus 1; 0 ]")[$[ 1 \ 0 \ 5 ] , #h(1em) [ 0 \ 1 \ 0 ] , #h(1em) [ 0 \ − 1 \ 0 ]$] + #math.equation(block: false, alt: "[ minus 4; minus 3; 5 ] , [ 2; 3; 3 ] , [ 2; 0; 2 ]")[$[ − 4 \ − 3 \ 5 ] , #h(1em) [ 2 \ 3 \ 3 ] , #h(1em) [ 2 \ 0 \ 2 ]$] + #math.equation(block: false, alt: "[ 1; 3; 0 ] , [ 0; 2; 2 ] , [ minus 1; minus 1; 2 ]")[$[ 1 \ 3 \ 0 ] , #h(1em) [ 0 \ 2 \ 2 ] , #h(1em) [ − 1 \ − 1 \ 2 ]$] + #math.equation(block: false, alt: "[ 1; 3 ] , [ 0; 2 ] , [ minus 1; minus 1 ]")[$[ 1 \ 3 ] , #h(1em) [ 0 \ 2 ] , #h(1em) [ − 1 \ − 1 ]$] + #math.equation(block: false, alt: "[ 3; 1; 3 ] , [ 2; 4; minus 4 ] , [ minus 5; minus 5; minus 2 ]")[$[ 3 \ 1 \ 3 ] , #h(1em) [ 2 \ 4 \ − 4 ] , #h(1em) [ − 5 \ − 5 \ − 2 ]$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For the following matrices, find a basis for the kernel (nullspace). + #math.equation(block: false, alt: "[ 1, 1, 1; 1, 1, 5; 1, 1, minus 4 ]")[$[ 1 & 1 & 1 \ 1 & 1 & 5 \ 1 & 1 & − 4 ]$] + #math.equation(block: false, alt: "[ 2, minus 1, minus 3; 4, 0, minus 4; minus 1, 1, 2 ]")[$[ 2 & − 1 & − 3 \ 4 & 0 & − 4 \ − 1 & 1 & 2 ]$] + #math.equation(block: false, alt: "[ minus 4, 4, 4; minus 1, 1, 1; minus 5, 5, 5 ]")[$[ − 4 & 4 & 4 \ − 1 & 1 & 1 \ − 5 & 5 & 5 ]$] + #math.equation(block: false, alt: "[ minus 2, 1, 1, 1; minus 4, 2, 2, 2; 1, 0, 4, 3 ]")[$[ − 2 & 1 & 1 & 1 \ − 4 & 2 & 2 & 2 \ 1 & 0 & 4 & 3 ]$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose a #math.equation(block: false, alt: "5 times 5")[$5 × 5$] matrix #math.equation(block: false, alt: "A")[$A$] has rank 3. What is the nullity? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose that #math.equation(block: false, alt: "X")[$X$] is the set of all the vectors of #math.equation(block: false, alt: "ℝ cubed")[$ℝ^(3)$] whose third component is zero. Is #math.equation(block: false, alt: "X")[$X$] a subspace? And if so, find a basis and the dimension. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Consider a square matrix #math.equation(block: false, alt: "A")[$A$], and suppose that #math.equation(block: false, alt: "x →")[$arrow(x)$] is a nonzero vector such that #math.equation(block: false, alt: "A x → equals 0 →")[$A arrow(x) = arrow(0)$]. What does the Fredholm alternative say about invertibility of #math.equation(block: false, alt: "A")[$A$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Consider #math.equation(block: true, alt: "M equals [ 1, 2, 3; 2, ?, ?; minus 1, ?, ? ] .")[$M = [ 1 & 2 & 3 \ 2 & ? & ? \ − 1 & ? & ? ] .$] If the nullity of this matrix is 2, fill in the question marks. Hint: What is the rank? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For the following sets of vectors, find a basis for the subspace spanned by the vectors, and find the dimension of the subspace. + #math.equation(block: false, alt: "[ 1; 2 ] , [ 1; 1 ]")[$[ 1 \ 2 ] , #h(1em) [ 1 \ 1 ]$] + #math.equation(block: false, alt: "[ 1; 1; 1 ] , [ 2; 2; 2 ] , [ 1; 1; 2 ]")[$[ 1 \ 1 \ 1 ] , #h(1em) [ 2 \ 2 \ 2 ] , #h(1em) [ 1 \ 1 \ 2 ]$] + #math.equation(block: false, alt: "[ 5; 3; 1 ] , [ 5; minus 1; 5 ] , [ minus 1; 3; minus 4 ]")[$[ 5 \ 3 \ 1 ] , #h(1em) [ 5 \ − 1 \ 5 ] , #h(1em) [ − 1 \ 3 \ − 4 ]$] + #math.equation(block: false, alt: "[ 2; 2; 4 ] , [ 2; 2; 3 ] , [ 4; 4; minus 3 ]")[$[ 2 \ 2 \ 4 ] , #h(1em) [ 2 \ 2 \ 3 ] , #h(1em) [ 4 \ 4 \ − 3 ]$] + #math.equation(block: false, alt: "[ 1; 0 ] , [ 2; 0 ] , [ 3; 0 ]")[$[ 1 \ 0 ] , #h(1em) [ 2 \ 0 ] , #h(1em) [ 3 \ 0 ]$] + #math.equation(block: false, alt: "[ 1; 0; 0 ] , [ 2; 0; 0 ] , [ 0; 1; 2 ]")[$[ 1 \ 0 \ 0 ] , #h(1em) [ 2 \ 0 \ 0 ] , #h(1em) [ 0 \ 1 \ 2 ]$] Answer + #math.equation(block: false, alt: "1; 2")[$1 \ 2$], #math.equation(block: false, alt: "1; 1")[$1 \ 1$] dimension #math.equation(block: false, alt: "2")[$2$], + #math.equation(block: false, alt: "1; 1; 1")[$1 \ 1 \ 1$], #math.equation(block: false, alt: "1; 1; 2")[$1 \ 1 \ 2$] dimension #math.equation(block: false, alt: "2")[$2$], + #math.equation(block: false, alt: "5; 3; 1")[$5 \ 3 \ 1$], #math.equation(block: false, alt: "5; minus 1; 5")[$5 \ − 1 \ 5$], #math.equation(block: false, alt: "minus 1; 3; minus 4")[$− 1 \ 3 \ − 4$] dimension #math.equation(block: false, alt: "3")[$3$], + #math.equation(block: false, alt: "2; 2; 4")[$2 \ 2 \ 4$], #math.equation(block: false, alt: "2; 2; 3")[$2 \ 2 \ 3$] dimension #math.equation(block: false, alt: "2")[$2$], + #math.equation(block: false, alt: "1; 1")[$1 \ 1$] dimension #math.equation(block: false, alt: "1")[$1$], + #math.equation(block: false, alt: "1; 0; 0")[$1 \ 0 \ 0$], #math.equation(block: false, alt: "0; 1; 2")[$0 \ 1 \ 2$] dimension #math.equation(block: false, alt: "2")[$2$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For the following matrices, find a basis for the kernel (nullspace). + #math.equation(block: false, alt: "[ 2, 6, 1, 9; 1, 3, 2, 9; 3, 9, 0, 9 ]")[$[ 2 & 6 & 1 & 9 \ 1 & 3 & 2 & 9 \ 3 & 9 & 0 & 9 ]$] + #math.equation(block: false, alt: "[ 2, minus 2, minus 5; minus 1, 1, 5; minus 5, 5, minus 3 ]")[$[ 2 & − 2 & − 5 \ − 1 & 1 & 5 \ − 5 & 5 & − 3 ]$] + #math.equation(block: false, alt: "[ 1, minus 5, minus 4; 2, 3, 5; minus 3, 5, 2 ]")[$[ 1 & − 5 & − 4 \ 2 & 3 & 5 \ − 3 & 5 & 2 ]$] + #math.equation(block: false, alt: "[ 0, 4, 4; 0, 1, 1; 0, 5, 5 ]")[$[ 0 & 4 & 4 \ 0 & 1 & 1 \ 0 & 5 & 5 ]$] Answer + #math.equation(block: false, alt: "3; minus 1; 0; 0")[$3 \ − 1 \ 0 \ 0$], #math.equation(block: false, alt: "3; 0; 3; minus 1")[$3 \ 0 \ 3 \ − 1$] + #math.equation(block: false, alt: "minus 1; minus 1; 0")[$− 1 \ − 1 \ 0$] + #math.equation(block: false, alt: "1; 1; minus 1")[$1 \ 1 \ − 1$] + #math.equation(block: false, alt: "minus 1; 0; 0")[$− 1 \ 0 \ 0$], #math.equation(block: false, alt: "0; 1; minus 1")[$0 \ 1 \ − 1$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose the column space of a #math.equation(block: false, alt: "9 times 5")[$9 × 5$] matrix #math.equation(block: false, alt: "A")[$A$] of dimension 3. Find + Rank of #math.equation(block: false, alt: "A")[$A$]. + Nullity of #math.equation(block: false, alt: "A")[$A$]. + Dimension of the row space of #math.equation(block: false, alt: "A")[$A$]. + Dimension of the nullspace of #math.equation(block: false, alt: "A")[$A$]. + Size of the maximum subset of linearly independent rows of #math.equation(block: false, alt: "A")[$A$]. Answer + #math.equation(block: false, alt: "3")[$3$] + #math.equation(block: false, alt: "2")[$2$] + #math.equation(block: false, alt: "3")[$3$] + #math.equation(block: false, alt: "2")[$2$] + #math.equation(block: false, alt: "3")[$3$] ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/Appendix_A%3A_Linear_Algebra/A.5%3A_Inner_Product_and_Projections")[A.5: Inner Product and Projections] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the #math.equation(block: false, alt: "s")[$s$] that makes the following vectors orthogonal: #math.equation(block: false, alt: "open parenthesis 1 , 2 , 3 close parenthesis")[$( 1 , 2 , 3 )$], #math.equation(block: false, alt: "open parenthesis 1 , 1 , s close parenthesis")[$( 1 , 1 , s )$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the angle #math.equation(block: false, alt: "θ")[$θ$] between #math.equation(block: false, alt: "open parenthesis 1 , 3 , 1 close parenthesis")[$( 1 , 3 , 1 )$], #math.equation(block: false, alt: "open parenthesis 2 , 1 , minus 1 close parenthesis")[$( 2 , 1 , − 1 )$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Given that #math.equation(block: false, alt: "⟨ v → , w → ⟩ equals 3")[$⟨ arrow(v) , arrow(w) ⟩ = 3$] and #math.equation(block: false, alt: "⟨ v → , u → ⟩ equals minus 1")[$⟨ arrow(v) , arrow(u) ⟩ = − 1$] compute + #math.equation(block: false, alt: "⟨ u → , 2 v → ⟩")[$⟨ arrow(u) , 2 arrow(v) ⟩$] + #math.equation(block: false, alt: "⟨ v → , 2 w → plus 3 u → ⟩")[$⟨ arrow(v) , 2 arrow(w) + 3 arrow(u) ⟩$] + #math.equation(block: false, alt: "⟨ w → plus 3 u → , v → ⟩")[$⟨ arrow(w) + 3 arrow(u) , arrow(v) ⟩$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose #math.equation(block: false, alt: "v → equals open parenthesis 1 , 1 , minus 1 close parenthesis")[$arrow(v) = ( 1 , 1 , − 1 )$]. Find + #math.equation(block: false, alt: "proj sub v → open parenthesis open parenthesis 1 , 0 , 0 close parenthesis close parenthesis")[$op("proj")_(arrow(v)) ( ( 1 , 0 , 0 ) )$] + #math.equation(block: false, alt: "proj sub v → open parenthesis open parenthesis 1 , 2 , 3 close parenthesis close parenthesis")[$op("proj")_(arrow(v)) ( ( 1 , 2 , 3 ) )$] + #math.equation(block: false, alt: "proj sub v → open parenthesis open parenthesis 1 , minus 1 , 0 close parenthesis close parenthesis")[$op("proj")_(arrow(v)) ( ( 1 , − 1 , 0 ) )$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Consider the vectors #math.equation(block: false, alt: "open parenthesis 1 , 2 , 3 close parenthesis")[$( 1 , 2 , 3 )$], #math.equation(block: false, alt: "open parenthesis minus 3 , 0 , 1 close parenthesis")[$( − 3 , 0 , 1 )$], #math.equation(block: false, alt: "open parenthesis 1 , minus 5 , 3 close parenthesis")[$( 1 , − 5 , 3 )$]. + Check that the vectors are linearly independent and so form a basis. + Check that the vectors are mutually orthogonal, and are therefore an orthogonal basis. + Represent #math.equation(block: false, alt: "open parenthesis 1 , 1 , 1 close parenthesis")[$( 1 , 1 , 1 )$] as a linear combination of this basis. + Make the basis orthonormal. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "S")[$S$] be the subspace spanned by #math.equation(block: false, alt: "open parenthesis 1 , 3 , minus 1 close parenthesis")[$( 1 , 3 , − 1 )$], #math.equation(block: false, alt: "open parenthesis 1 , 1 , 1 close parenthesis")[$( 1 , 1 , 1 )$]. Find an orthogonal basis of #math.equation(block: false, alt: "S")[$S$] by the Gram-Schmidt process. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Starting with #math.equation(block: false, alt: "open parenthesis 1 , 2 , 3 close parenthesis")[$( 1 , 2 , 3 )$], #math.equation(block: false, alt: "open parenthesis 1 , 1 , 1 close parenthesis")[$( 1 , 1 , 1 )$], #math.equation(block: false, alt: "open parenthesis 2 , 2 , 0 close parenthesis")[$( 2 , 2 , 0 )$], follow the Gram-Schmidt process to find an orthogonal basis of #math.equation(block: false, alt: "ℝ cubed")[$ℝ^(3)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find an orthogonal basis of #math.equation(block: false, alt: "ℝ cubed")[$ℝ^(3)$] such that #math.equation(block: false, alt: "open parenthesis 3 , 1 , minus 2 close parenthesis")[$( 3 , 1 , − 2 )$] is one of the vectors. Hint: First find two extra vectors to make a linearly independent set. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Using cosines and sines of #math.equation(block: false, alt: "θ")[$θ$], find a unit vector #math.equation(block: false, alt: "u →")[$arrow(u)$] in #math.equation(block: false, alt: "ℝ squared")[$ℝ^(2)$] that makes angle #math.equation(block: false, alt: "θ")[$θ$] with #math.equation(block: false, alt: "ı → equals open parenthesis 1 , 0 close parenthesis")[$arrow(ı) = ( 1 , 0 )$]. What is #math.equation(block: false, alt: "⟨ ı → , u → ⟩")[$⟨ arrow(ı) , arrow(u) ⟩$]? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the #math.equation(block: false, alt: "s")[$s$] that makes the following vectors orthogonal: #math.equation(block: false, alt: "open parenthesis 1 , 1 , 1 close parenthesis")[$( 1 , 1 , 1 )$], #math.equation(block: false, alt: "open parenthesis 1 , s , 1 close parenthesis")[$( 1 , s , 1 )$]. Answer #math.equation(block: true, alt: "s equals minus 2")[$s = − 2$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the angle #math.equation(block: false, alt: "θ")[$θ$] between #math.equation(block: false, alt: "open parenthesis 1 , 2 , 3 close parenthesis")[$( 1 , 2 , 3 )$], #math.equation(block: false, alt: "open parenthesis 1 , 1 , 1 close parenthesis")[$( 1 , 1 , 1 )$]. Answer #math.equation(block: true, alt: "θ approximately equals 0.3876")[$θ ≈ 0.3876$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Given that #math.equation(block: false, alt: "⟨ v → , w → ⟩ equals 1")[$⟨ arrow(v) , arrow(w) ⟩ = 1$] and #math.equation(block: false, alt: "⟨ v → , u → ⟩ equals minus 1")[$⟨ arrow(v) , arrow(u) ⟩ = − 1$] and #math.equation(block: false, alt: "‖ v → ‖ equals 3")[$‖ arrow(v) ‖ = 3$] and + #math.equation(block: false, alt: "⟨ 3 u → , 5 v → ⟩")[$⟨ 3 arrow(u) , 5 arrow(v) ⟩$] + #math.equation(block: false, alt: "⟨ v → , 2 w → plus 3 u → ⟩")[$⟨ arrow(v) , 2 arrow(w) + 3 arrow(u) ⟩$] + #math.equation(block: false, alt: "⟨ w → plus 3 v → , v → ⟩")[$⟨ arrow(w) + 3 arrow(v) , arrow(v) ⟩$] Answer + #math.equation(block: false, alt: "minus 15")[$− 15$] + #math.equation(block: false, alt: "minus 1")[$− 1$] + #math.equation(block: false, alt: "28")[$28$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose #math.equation(block: false, alt: "v → equals open parenthesis 1 , 0 , minus 1 close parenthesis")[$arrow(v) = ( 1 , 0 , − 1 )$]. Find + #math.equation(block: false, alt: "proj sub v → open parenthesis open parenthesis 0 , 2 , 1 close parenthesis close parenthesis")[$op("proj")_(arrow(v)) ( ( 0 , 2 , 1 ) )$] + #math.equation(block: false, alt: "proj sub v → open parenthesis open parenthesis 1 , 0 , 1 close parenthesis close parenthesis")[$op("proj")_(arrow(v)) ( ( 1 , 0 , 1 ) )$] + #math.equation(block: false, alt: "proj sub v → open parenthesis open parenthesis 4 , minus 1 , 0 close parenthesis close parenthesis")[$op("proj")_(arrow(v)) ( ( 4 , − 1 , 0 ) )$] Answer + #math.equation(block: false, alt: "open parenthesis minus the fraction 1 over 2 , 0 , the fraction 1 over 2 close parenthesis")[$( − frac(1, 2) , 0 , frac(1, 2) )$] + #math.equation(block: false, alt: "open parenthesis 0 , 0 , 0 close parenthesis")[$( 0 , 0 , 0 )$] + #math.equation(block: false, alt: "open parenthesis 2 , 0 , minus 2 close parenthesis")[$( 2 , 0 , − 2 )$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The vectors #math.equation(block: false, alt: "open parenthesis 1 , 1 , minus 1 close parenthesis")[$( 1 , 1 , − 1 )$], #math.equation(block: false, alt: "open parenthesis 2 , minus 1 , 1 close parenthesis")[$( 2 , − 1 , 1 )$], #math.equation(block: false, alt: "open parenthesis 1 , minus 5 , 3 close parenthesis")[$( 1 , − 5 , 3 )$] form an orthogonal basis. Represent the following vectors in terms of this basis: + #math.equation(block: false, alt: "open parenthesis 1 , minus 8 , 4 close parenthesis")[$( 1 , − 8 , 4 )$] + #math.equation(block: false, alt: "open parenthesis 5 , minus 7 , 5 close parenthesis")[$( 5 , − 7 , 5 )$] + #math.equation(block: false, alt: "open parenthesis 0 , minus 6 , 2 close parenthesis")[$( 0 , − 6 , 2 )$] Answer + #math.equation(block: false, alt: "open parenthesis 1 , 1 , minus 1 close parenthesis minus open parenthesis 2 , minus 1 , 1 close parenthesis plus 2 open parenthesis 1 , minus 5 , 3 close parenthesis")[$( 1 , 1 , − 1 ) − ( 2 , − 1 , 1 ) + 2 ( 1 , − 5 , 3 )$] + #math.equation(block: false, alt: "2 open parenthesis 2 , minus 1 , 1 close parenthesis plus open parenthesis 1 , minus 5 , 3 close parenthesis")[$2 ( 2 , − 1 , 1 ) + ( 1 , − 5 , 3 )$] + #math.equation(block: false, alt: "2 open parenthesis 1 , 1 , minus 1 close parenthesis minus 2 open parenthesis 2 , minus 1 , 1 close parenthesis plus 2 open parenthesis 1 , minus 5 , 3 close parenthesis")[$2 ( 1 , 1 , − 1 ) − 2 ( 2 , − 1 , 1 ) + 2 ( 1 , − 5 , 3 )$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "S")[$S$] be the subspace spanned by #math.equation(block: false, alt: "open parenthesis 2 , minus 1 , 1 close parenthesis")[$( 2 , − 1 , 1 )$], #math.equation(block: false, alt: "open parenthesis 2 , 2 , 2 close parenthesis")[$( 2 , 2 , 2 )$]. Find an orthogonal basis of #math.equation(block: false, alt: "S")[$S$] by the Gram-Schmidt process. Answer #math.equation(block: true, alt: "open parenthesis 2 , minus 1 , 1 close parenthesis")[$( 2 , − 1 , 1 )$], #math.equation(block: true, alt: "open parenthesis the fraction 2 over 3 , the fraction 8 over 3 , the fraction 4 over 3 close parenthesis")[$( frac(2, 3) , frac(8, 3) , frac(4, 3) )$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Starting with #math.equation(block: false, alt: "open parenthesis 1 , 1 , minus 1 close parenthesis")[$( 1 , 1 , − 1 )$], #math.equation(block: false, alt: "open parenthesis 2 , 3 , minus 1 close parenthesis")[$( 2 , 3 , − 1 )$], #math.equation(block: false, alt: "open parenthesis 1 , minus 1 , 1 close parenthesis")[$( 1 , − 1 , 1 )$], follow the Gram-Schmidt process to find an orthogonal basis of #math.equation(block: false, alt: "ℝ cubed")[$ℝ^(3)$]. Answer #math.equation(block: true, alt: "open parenthesis 1 , 1 , minus 1 close parenthesis")[$( 1 , 1 , − 1 )$], #math.equation(block: true, alt: "open parenthesis 0 , 1 , 1 close parenthesis")[$( 0 , 1 , 1 )$], #math.equation(block: true, alt: "open parenthesis the fraction 4 over 3 , minus the fraction 2 over 3 , the fraction 2 over 3 close parenthesis")[$( frac(4, 3) , − frac(2, 3) , frac(2, 3) )$] ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/Appendix_A%3A_Linear_Algebra/A.6%3A_Determinant")[A.6: Determinant] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute the determinant of the following matrices: + #math.equation(block: false, alt: "[ 3 ]")[$[ 3 ]$] + #math.equation(block: false, alt: "[ 1, 3; 2, 1 ]")[$[ 1 & 3 \ 2 & 1 ]$] + #math.equation(block: false, alt: "[ 2, 1; 4, 2 ]")[$[ 2 & 1 \ 4 & 2 ]$] + #math.equation(block: false, alt: "[ 1, 2, 3; 0, 4, 5; 0, 0, 6 ]")[$[ 1 & 2 & 3 \ 0 & 4 & 5 \ 0 & 0 & 6 ]$] + #math.equation(block: false, alt: "[ 2, 1, 0; minus 2, 7, minus 3; 0, 2, 0 ]")[$[ 2 & 1 & 0 \ − 2 & 7 & − 3 \ 0 & 2 & 0 ]$] + #math.equation(block: false, alt: "[ 2, 1, 3; 8, 6, 3; 7, 9, 7 ]")[$[ 2 & 1 & 3 \ 8 & 6 & 3 \ 7 & 9 & 7 ]$] + #math.equation(block: false, alt: "[ 0, 2, 5, 7; 0, 0, 2, minus 3; 3, 4, 5, 7; 0, 0, 2, 4 ]")[$[ 0 & 2 & 5 & 7 \ 0 & 0 & 2 & − 3 \ 3 & 4 & 5 & 7 \ 0 & 0 & 2 & 4 ]$] + #math.equation(block: false, alt: "[ 0, 1, 2, 0; 1, 1, minus 1, 2; 1, 1, 2, 1; 2, minus 1, minus 2, 3 ]")[$[ 0 & 1 & 2 & 0 \ 1 & 1 & − 1 & 2 \ 1 & 1 & 2 & 1 \ 2 & − 1 & − 2 & 3 ]$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For which #math.equation(block: false, alt: "x")[$x$] are the following matrices singular (not invertible). + #math.equation(block: false, alt: "[ 2, 3; 2, x ]")[$[ 2 & 3 \ 2 & x ]$] + #math.equation(block: false, alt: "[ 2, x; 1, 2 ]")[$[ 2 & x \ 1 & 2 ]$] + #math.equation(block: false, alt: "[ x, 1; 4, x ]")[$[ x & 1 \ 4 & x ]$] + #math.equation(block: false, alt: "[ x, 0, 1; 1, 4, 2; 1, 6, 2 ]")[$[ x & 0 & 1 \ 1 & 4 & 2 \ 1 & 6 & 2 ]$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute #math.equation(block: true, alt: "det open parenthesis [ to the power 2, 1, 2, 3; 0, 8, 6, 5; 0, 0, 3, 9; 0, 0, 0, 1 close parenthesis")[$det \( \[^(2 & 1 & 2 & 3 \ 0 & 8 & 6 & 5 \ 0 & 0 & 3 & 9 \ 0 & 0 & 0 & 1) \)$] without computing the inverse. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose #math.equation(block: true, alt: "L equals [ 1, 0, 0, 0; 2, 1, 0, 0; 7, π, 1, 0; 2 to the power 8, 5, minus 99, 1 ] and U equals [ 5, 9, 1, minus sin open parenthesis 1 close parenthesis; 0, 1, 88, minus 1; 0, 0, 1, 3; 0, 0, 0, 1 ] .")[$L = [ 1 & 0 & 0 & 0 \ 2 & 1 & 0 & 0 \ 7 & π & 1 & 0 \ 2^(8) & 5 & − 99 & 1 ] #h(2em) "and" #h(2em) U = [ 5 & 9 & 1 & − sin ( 1 ) \ 0 & 1 & 88 & − 1 \ 0 & 0 & 1 & 3 \ 0 & 0 & 0 & 1 ] .$] Let #math.equation(block: false, alt: "A equals L U")[$A = L U$]. Compute #math.equation(block: false, alt: "det open parenthesis A close parenthesis")[$det ( A )$] in a simple way, without computing what is #math.equation(block: false, alt: "A")[$A$]. Hint: First read off #math.equation(block: false, alt: "det open parenthesis L close parenthesis")[$det ( L )$] and #math.equation(block: false, alt: "det open parenthesis U close parenthesis")[$det ( U )$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Consider the linear mapping from #math.equation(block: false, alt: "R squared")[$R^(2)$] to #math.equation(block: false, alt: "R squared")[$R^(2)$] given by the matrix #math.equation(block: false, alt: "A equals [ 1, x; 2, 1 ]")[$A = [ 1 & x \ 2 & 1 ]$] for some number #math.equation(block: false, alt: "x")[$x$]. You wish to make #math.equation(block: false, alt: "A")[$A$] such that it doubles the area of every geometric figure. What are the possibilities for #math.equation(block: false, alt: "x")[$x$] (there are two answers). ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "S")[$S$] are #math.equation(block: false, alt: "n times n")[$n × n$] matrices, and #math.equation(block: false, alt: "S")[$S$] is invertible. Suppose that #math.equation(block: false, alt: "det open parenthesis A close parenthesis equals 3")[$det ( A ) = 3$]. Compute #math.equation(block: false, alt: "det open parenthesis S to the power minus 1 A S close parenthesis")[$det ( S^(− 1) A S )$] and #math.equation(block: false, alt: "det open parenthesis S A S to the power minus 1 close parenthesis")[$det ( S A S^(− 1) )$]. Justify your answer using the theorems in this section. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "A")[$A$] be an #math.equation(block: false, alt: "n times n")[$n × n$] matrix such that #math.equation(block: false, alt: "det open parenthesis A close parenthesis equals 1")[$det ( A ) = 1$]. Compute #math.equation(block: false, alt: "det open parenthesis x A close parenthesis")[$det ( x A )$] given a number #math.equation(block: false, alt: "x")[$x$]. Hint: First try computing #math.equation(block: false, alt: "det open parenthesis x I close parenthesis")[$det ( x I )$], then note that #math.equation(block: false, alt: "x A equals open parenthesis x I close parenthesis A")[$x A = ( x I ) A$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute the determinant of the following matrices: + #math.equation(block: false, alt: "[ minus 2 ]")[$[ − 2 ]$] + #math.equation(block: false, alt: "[ 2, minus 2; 1, 3 ]")[$[ 2 & − 2 \ 1 & 3 ]$] + #math.equation(block: false, alt: "[ 2, 2; 2, 2 ]")[$[ 2 & 2 \ 2 & 2 ]$] + #math.equation(block: false, alt: "[ 2, 9, minus 11; 0, minus 1, 5; 0, 0, 3 ]")[$[ 2 & 9 & − 11 \ 0 & − 1 & 5 \ 0 & 0 & 3 ]$] + #math.equation(block: false, alt: "[ 2, 1, 0; minus 2, 7, 3; 1, 1, 0 ]")[$[ 2 & 1 & 0 \ − 2 & 7 & 3 \ 1 & 1 & 0 ]$] + #math.equation(block: false, alt: "[ 5, 1, 3; 4, 1, 1; 4, 5, 1 ]")[$[ 5 & 1 & 3 \ 4 & 1 & 1 \ 4 & 5 & 1 ]$] + #math.equation(block: false, alt: "[ 3, 2, 5, 7; 0, 0, 2, 0; 0, 4, 5, 0; 2, 1, 2, 4 ]")[$[ 3 & 2 & 5 & 7 \ 0 & 0 & 2 & 0 \ 0 & 4 & 5 & 0 \ 2 & 1 & 2 & 4 ]$] + #math.equation(block: false, alt: "[ 0, 2, 1, 0; 1, 2, minus 3, 4; 5, 6, minus 7, 8; 1, 2, 3, minus 2 ]")[$[ 0 & 2 & 1 & 0 \ 1 & 2 & − 3 & 4 \ 5 & 6 & − 7 & 8 \ 1 & 2 & 3 & − 2 ]$] Answer + #math.equation(block: false, alt: "minus 2")[$− 2$] + #math.equation(block: false, alt: "8")[$8$] + #math.equation(block: false, alt: "0")[$0$] + #math.equation(block: false, alt: "minus 6")[$− 6$] + #math.equation(block: false, alt: "minus 3")[$− 3$] + #math.equation(block: false, alt: "28")[$28$] + #math.equation(block: false, alt: "16")[$16$] + #math.equation(block: false, alt: "minus 24")[$− 24$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For which #math.equation(block: false, alt: "x")[$x$] are the following matrices singular (not invertible). + #math.equation(block: false, alt: "[ 1, 3; 1, x ]")[$[ 1 & 3 \ 1 & x ]$] + #math.equation(block: false, alt: "[ 3, x; 1, 3 ]")[$[ 3 & x \ 1 & 3 ]$] + #math.equation(block: false, alt: "[ x, 3; 3, x ]")[$[ x & 3 \ 3 & x ]$] + #math.equation(block: false, alt: "[ x, 1, 0; 1, 4, 0; 1, 6, 2 ]")[$[ x & 1 & 0 \ 1 & 4 & 0 \ 1 & 6 & 2 ]$] Answer + #math.equation(block: false, alt: "3")[$3$] + #math.equation(block: false, alt: "9")[$9$] + #math.equation(block: false, alt: "3")[$3$] + #math.equation(block: false, alt: "the fraction 1 over 4")[$frac(1, 4)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute #math.equation(block: true, alt: "det open parenthesis [ to the power 3, 4, 7, 12; 0, minus 1, 9, minus 8; 0, 0, minus 2, 4; 0, 0, 0, 2 close parenthesis")[$det \( \[^(3 & 4 & 7 & 12 \ 0 & − 1 & 9 & − 8 \ 0 & 0 & − 2 & 4 \ 0 & 0 & 0 & 2) \)$] without computing the inverse. Answer #math.equation(block: true, alt: "12")[$12$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[(challenging)] Find all the #math.equation(block: false, alt: "x")[$x$] that make the matrix inverse #math.equation(block: true, alt: "[ to the power 1, 2; 1, x")[$\[^(1 & 2 \ 1 & x)$] have only integer entries (no fractions). Note that there are two answers. Answer #math.equation(block: false, alt: "1")[$1$] and #math.equation(block: false, alt: "3")[$3$] ]