A.7 Exercises
Your Turn
On a piece of graph paper draw the vector ( 1 , 2 ) (1,2)
based at ( 0 , 0 ) (0,0) based at ( 1 , 2 ) (1,2) based at ( 0 , − 1 ) (0,-1) Your Turn
On a piece of graph paper draw the following operations. Draw and label the vectors involved in the operations as well as the result:
[
1
−
4
]
+
[
2
3
]
\begin{bmatrix} 1 \\ -4 \end{bmatrix} + \begin{bmatrix} 2 \\ 3 \end{bmatrix}
[
−
3
2
]
−
[
1
3
]
\begin{bmatrix} -3 \\ 2 \end{bmatrix} - \begin{bmatrix} 1 \\ 3 \end{bmatrix}
3
[
2
1
]
3\begin{bmatrix} 2 \\ 1 \end{bmatrix}
Your Turn
Compute the magnitude of
[
7
2
]
\begin{bmatrix} 7 \\ 2 \end{bmatrix}
[
−
2
3
1
]
\begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix}
(
1
,
3
,
−
4
)
(1,3,-4)
Your Turn
Compute
[
2
3
]
+
[
7
−
8
]
\begin{bmatrix} 2 \\ 3 \end{bmatrix} + \begin{bmatrix} 7 \\ -8 \end{bmatrix}
[
−
2
3
]
−
[
6
−
4
]
\begin{bmatrix} -2 \\ 3 \end{bmatrix} - \begin{bmatrix} 6 \\ -4 \end{bmatrix}
−
[
−
3
2
]
-\begin{bmatrix} -3 \\ 2 \end{bmatrix}
4
[
−
1
5
]
4\begin{bmatrix} -1 \\ 5 \end{bmatrix}
5
[
1
0
]
+
9
[
0
1
]
5\begin{bmatrix} 1 \\ 0 \end{bmatrix} + 9 \begin{bmatrix} 0 \\ 1 \end{bmatrix}
3
[
1
−
8
]
−
2
[
3
−
1
]
3\begin{bmatrix} 1 \\ -8 \end{bmatrix} - 2 \begin{bmatrix} 3 \\ -1 \end{bmatrix}
Your Turn
If x → = ( 1 , 2 ) \vec{x} = (1,2) y → \vec{y} x → + y → = ( 0 , 2 ) \vec{x}+\vec{y} = (0,2) y → \vec{y}
Your Turn
Write ( 1 , 2 , 3 ) (1,2,3) e → 1 \vec{e}_1 e → 2 \vec{e}_2 e → 3 \vec{e}_3
Your Turn
If the magnitude of x → \vec{x}
0
x
→
0\vec{x}
3
x
→
3\vec{x}
−
x
→
-\vec{x}
−
4
x
→
-4\vec{x}
x
→
+
x
→
\vec{x}+\vec{x}
x
→
−
x
→
\vec{x}-\vec{x}
Your Turn
Suppose a linear mapping F : R 2 → R 2 F \colon {\mathbb R}^2 \to {\mathbb R}^2 ( 1 , 0 ) (1,0) ( 2 , − 1 ) (2,-1) ( 0 , 1 ) (0,1) ( 3 , 3 ) (3,3)
(
1
,
1
)
(1,1)
(
2
,
0
)
(2,0)
(
2
,
−
1
)
(2,-1)
Your Turn
Suppose a linear mapping F : R 3 → R 2 F \colon {\mathbb R}^3 \to {\mathbb R}^2 ( 1 , 0 , 0 ) (1,0,0) ( 2 , 1 ) (2,1) ( 0 , 1 , 0 ) (0,1,0) ( 3 , 4 ) (3,4) ( 0 , 0 , 1 ) (0,0,1) ( 5 , 6 ) (5,6) F F
Your Turn
Suppose that a mapping F : R 2 → ℝ 2 F \colon {\mathbb R}^2 \to \mathbb{R}^2 ( 1 , 0 ) (1,0) ( 1 , 2 ) (1,2) ( 0 , 1 ) (0,1) ( 3 , 4 ) (3,4) ( 1 , 1 ) (1,1) ( 0 , − 1 ) (0,-1) F F
Your Turn
(challenging)
Let R 3 {\mathbb R}^3 t t ( a 0 , a 1 , a 2 ) (a_0,a_1,a_2) R 3 {\mathbb R}^3 a 0 + a 1 t + a 2 t 2 a_0 + a_1 t + a_2 t^2 d d t \frac{d}{dt} R 3 {\mathbb R}^3 R 3 {\mathbb R}^3 d d t \frac{d}{dt} d d t \frac{d}{dt} 3 × 3 3 \times 3
Your Turn
Compute the magnitude of
[
1
3
]
\begin{bmatrix} 1 \\ 3 \end{bmatrix}
[
2
3
−
1
]
\begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix}
(
−
2
,
1
,
−
2
)
(-2,1,-2)
Answer
10
\sqrt{10}
14
\sqrt{14}
3
3
Your Turn
Compute
[
3
1
]
+
[
6
−
3
]
\begin{bmatrix} 3 \\ 1 \end{bmatrix} + \begin{bmatrix} 6 \\ -3 \end{bmatrix}
[
−
1
2
]
−
[
2
−
1
]
\begin{bmatrix} -1 \\ 2 \end{bmatrix} - \begin{bmatrix} 2 \\ -1 \end{bmatrix}
−
[
−
5
3
]
-\begin{bmatrix} -5 \\ 3 \end{bmatrix}
2
[
−
2
4
]
2\begin{bmatrix} -2 \\ 4 \end{bmatrix}
3
[
1
0
]
+
7
[
0
1
]
3\begin{bmatrix} 1 \\ 0 \end{bmatrix} + 7 \begin{bmatrix} 0 \\ 1 \end{bmatrix}
2
[
2
−
3
]
−
6
[
2
−
1
]
2\begin{bmatrix} 2 \\ -3 \end{bmatrix} - 6 \begin{bmatrix} 2 \\ -1 \end{bmatrix}
Answer
[
9
−
2
]
\left[\begin{array}{c}{9}\\{-2}\end{array}\right]
[
−
3
3
]
\left[\begin{array}{c}{-3}\\{3}\end{array}\right]
[
5
−
3
]
\left[\begin{array}{c}{5}\\{-3}\end{array}\right]
[
−
4
8
]
\left[\begin{array}{c}{-4}\\{8}\end{array}\right]
[
3
7
]
\left[\begin{array}{c}{3}\\{7}\end{array}\right]
[
−
8
3
]
\left[\begin{array}{c}{-8}\\{3}\end{array}\right]
Your Turn
If the magnitude of x → \vec{x}
4
x
→
4\vec{x}
−
2
x
→
-2\vec{x}
−
4
x
→
-4\vec{x}
Answer
20
20
10
10
20
20
Your Turn
Suppose a linear mapping F : R 2 → R 2 F \colon {\mathbb R}^2 \to {\mathbb R}^2 ( 1 , 0 ) (1,0) ( 1 , − 1 ) (1,-1) ( 0 , 1 ) (0,1) ( 2 , 0 ) (2,0)
(
1
,
1
)
(1,1)
(
0
,
2
)
(0,2)
(
1
,
−
1
)
(1,-1)
Answer
(
3
,
−
1
)
(3,-1)
(
4
,
0
)
(4,0)
(
−
1
,
−
1
)
(-1,-1)
Your Turn
Compute
3
[
0
3
−
2
2
]
+
6
[
1
5
−
1
5
]
3\begin{bmatrix} 0 & 3 \\ -2 & 2 \end{bmatrix} + 6 \begin{bmatrix} 1 & 5 \\ -1 & 5 \end{bmatrix}
2
[
−
3
1
2
2
]
−
3
[
2
−
1
3
2
]
2\begin{bmatrix} -3 & 1 \\ 2 & 2 \end{bmatrix} - 3 \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix}
Your Turn
Add the following matrices
[
2
1
0
1
1
−
1
]
+
[
5
3
4
1
2
5
]
\begin{bmatrix} 2 & 1 & 0 \\ 1 & 1 & -1 \end{bmatrix} + \begin{bmatrix} 5 & 3 & 4 \\ 1 & 2 & 5 \end{bmatrix}
[
6
−
2
3
7
3
3
8
−
1
2
]
+
[
−
1
−
1
−
3
6
7
3
−
9
4
−
1
]
\begin{bmatrix} 6 & -2 & 3 \\ 7 & 3 & 3 \\ 8 & -1 & 2 \end{bmatrix} + \begin{bmatrix} -1 & -1 & -3 \\ 6 & 7 & 3 \\ -9 & 4 & -1 \end{bmatrix}
Answer
Add texts here. Do not delete this text first.
Your Turn
Multiply the following matrices
[
2
1
4
3
4
4
]
[
2
4
6
3
3
5
]
\begin{bmatrix} 2 & 1 & 4 \\ 3 & 4 & 4 \end{bmatrix} \begin{bmatrix} 2 & 4 \\ 6 & 3 \\ 3 & 5 \end{bmatrix}
[
0
3
3
2
−
2
1
3
5
−
2
]
[
6
6
2
4
6
0
2
0
4
]
\begin{bmatrix} 0 & 3 & 3 \\ 2 & -2 & 1 \\ 3 & 5 & -2 \end{bmatrix} \begin{bmatrix} 6 & 6 & 2 \\ 4 & 6 & 0 \\ 2 & 0 & 4 \end{bmatrix}
[
3
4
1
2
−
1
0
4
−
1
5
]
[
0
2
5
0
2
0
5
2
3
6
1
6
]
\begin{bmatrix} 3 & 4 & 1 \\ 2 & -1 & 0 \\ 4 & -1 & 5 \end{bmatrix} \begin{bmatrix} 0 & 2 & 5 & 0 \\ 2 & 0 & 5 & 2 \\ 3 & 6 & 1 & 6 \end{bmatrix}
[
−
2
−
2
5
3
2
1
]
[
0
3
1
3
]
\begin{bmatrix} -2 & -2 \\ 5 & 3 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 0 & 3 \\ 1 & 3 \end{bmatrix}
Answer
Add texts here. Do not delete this text first.
Your Turn
Compute the inverse of the given matrices
[
2
]
\begin{bmatrix} 2 \end{bmatrix}
[
0
1
1
0
]
\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}
[
1
2
3
5
]
\begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}
[
4
2
4
4
]
\begin{bmatrix} 4 & 2 \\ 4 & 4 \end{bmatrix}
Answer
Add texts here. Do not delete this text first.
Your Turn
Compute the inverse of the given matrices
[
2
0
0
3
]
\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}
[
4
0
0
0
5
0
0
0
−
1
]
\begin{bmatrix} 4 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & -1 \end{bmatrix}
[
−
1
0
0
0
0
2
0
0
0
0
3
0
0
0
0
0.1
]
\begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0.1 \end{bmatrix}
Answer
Add texts here. Do not delete this text first.
Your Turn
Solve (find all solutions), or show no solution exists
4
x
1
+
3
x
2
=
−
2
−
x
1
+
x
2
=
4
\begin{aligned} 4x_1+3x_2 & = -2 \\ -x_1+\phantom{3} x_2 & = 4 \end{aligned}
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
\begin{aligned} x_1+5x_2+3x_3 & = 7 \\ 8x_1+7x_2+8x_3 & = 8 \\ 4x_1+8x_2+6x_3 & = 4 \end{aligned}
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
=
2
\begin{aligned} 4x_1+8x_2+2x_3 & = 3 \\ -x_1-2x_2+3x_3 & = 1 \\ 4x_1+8x_2 \phantom{{}+3x_3} & = 2 \end{aligned}
x
+
2
y
+
3
z
=
4
2
x
−
y
+
3
z
=
1
3
x
+
y
+
6
z
=
6
\begin{aligned} x+2y+3z & = 4 \\ 2 x-\phantom{2} y+3z & = 1 \\ 3 x+\phantom{2} y+6z & = 6 \end{aligned}
Your Turn
By computing the inverse, solve the following systems for x → \vec{x}
[
4
1
−
1
3
]
x
→
=
[
13
26
]
\begin{bmatrix} 4 & 1 \\ -1 & 3 \end{bmatrix} \vec{x} = \begin{bmatrix} 13 \\ 26 \end{bmatrix}
[
3
3
3
4
]
x
→
=
[
2
−
1
]
\begin{bmatrix} 3 & 3 \\ 3 & 4 \end{bmatrix} \vec{x} = \begin{bmatrix} 2 \\ -1 \end{bmatrix}
Your Turn
For the matrices in Exercise 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).
Your Turn
For the matrices in Exercise A.3.5, find a linearly independent set of columns that span the column space. That is, find the pivot columns of the matrices.
Your Turn
Find a linearly independent subset of the following vectors that has the same span.
[ − 1 1 2 ] , [ 2 − 2 − 4 ] , [ − 2 4 1 ] , [ − 1 3 − 2 ] \begin{bmatrix} -1 \\ 1 \\ 2 \end{bmatrix}, \quad \begin{bmatrix} 2 \\ -2 \\ -4 \end{bmatrix}, \quad \begin{bmatrix} -2 \\ 4 \\ 1 \end{bmatrix}, \quad \begin{bmatrix} -1 \\ 3 \\ -2 \end{bmatrix} \nonumber
Your Turn
Solve (find all solutions), or show no solution exists
4
x
1
+
3
x
2
=
−
1
5
x
1
+
6
x
2
=
4
\begin{aligned} 4x_1+3x_2 & = -1 \\ 5x_1+6x_2 & = 4 \end{aligned}
5
x
+
6
y
+
5
z
=
7
6
x
+
8
y
+
6
z
=
−
1
5
x
+
2
y
+
5
z
=
2
\begin{aligned} 5x+6y+5z & = 7 \\ 6x+8y+6z & = -1 \\ 5x+2y+5z & = 2 \end{aligned}
a
+
b
+
c
=
−
1
a
+
5
b
+
6
c
=
−
1
−
2
a
+
5
b
+
6
c
=
8
\begin{aligned} a+\phantom{5}b+\phantom{6}c & = -1 \\ a+5b+6c & = -1 \\ -2a+5b+6c & = 8 \end{aligned}
−
2
x
1
+
2
x
2
+
8
x
3
=
6
x
2
+
x
3
=
2
x
1
+
4
x
2
+
x
3
=
7
\begin{aligned} -2 x_1+2x_2+8x_3 & = 6 \\ x_2+\phantom{8}x_3 & = 2 \\ x_1+4x_2+\phantom{8}x_3 & = 7 \end{aligned}
Answer
x
1
=
−
2
,
x
2
=
7
3
x_{1}=-2,\: x_{2}=\frac{7}{3}
no solution
a
=
−
3
,
b
=
10
,
c
=
−
8
a=-3,\: b=10,\: c=-8
x 3 x_{3} x 1 = − 1 + 3 x 3 x_{1}=-1+3x_{3} x 2 = 2 − x 3 x_{2}=2-x_{3} Your Turn
By computing the inverse, solve the following systems for x → \vec{x}
[
−
1
1
3
3
]
x
→
=
[
4
6
]
\begin{bmatrix} -1 & 1 \\ 3 & 3 \end{bmatrix} \vec{x} = \begin{bmatrix} 4 \\ 6 \end{bmatrix}
[
2
7
1
6
]
x
→
=
[
1
3
]
\begin{bmatrix} 2 & 7 \\ 1 & 6 \end{bmatrix} \vec{x} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}
Answer
[
−
1
3
]
\left[\begin{array}{c}{-1}\\{3}\end{array}\right]
[
−
3
1
]
\left[\begin{array}{c}{-3}\\{1}\end{array}\right]
Your Turn
For the matrices in Exercise 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
[ 1 0 0 ] \left[\begin{array}{ccc}{1}&{0}&{0}\end{array}\right] [ 0 1 0 ] \left[\begin{array}{ccc}{0}&{1}&{0}\end{array}\right] [ 0 0 1 ] \left[\begin{array}{ccc}{0}&{0}&{1}\end{array}\right]
[
1
1
1
]
\left[\begin{array}{ccc}{1}&{1}&{1}\end{array}\right]
[ 1 0 1 3 ] \left[\begin{array}{ccc}{1}&{0}&{\frac{1}{3}}\end{array}\right] [ 0 1 − 1 3 ] \left[\begin{array}{ccc}{0}&{1}&{-\frac{1}{3}}\end{array}\right] Your Turn
For the matrices in Exercise 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
[ 7 7 7 ] \left[\begin{array}{c}{7}\\{7}\\{7}\end{array}\right] [ − 1 7 6 ] \left[\begin{array}{c}{-1}\\{7}\\{6}\end{array}\right] [ 7 6 2 ] \left[\begin{array}{c}{7}\\{6}\\{2}\end{array}\right]
[
1
1
2
]
\left[\begin{array}{c}{1}\\{1}\\{2}\end{array}\right]
[ 0 6 4 ] \left[\begin{array}{c}{0}\\{6}\\{4}\end{array}\right] [ 3 3 7 ] \left[\begin{array}{c}{3}\\{3}\\{7}\end{array}\right] Your Turn
Find a linearly independent subset of the following vectors that has the same span.
[ 0 0 0 ] , [ 3 1 − 5 ] , [ 0 3 − 1 ] , [ − 3 2 4 ] \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} 3 \\ 1 \\ -5 \end{bmatrix}, \quad \begin{bmatrix} 0 \\ 3 \\ -1 \end{bmatrix}, \quad \begin{bmatrix} -3 \\ 2 \\ 4 \end{bmatrix} \nonumber
Answer
[ 3 1 − 5 ] \left[\begin{array}{c}{3}\\{1}\\{-5}\end{array}\right] [ 0 3 − 1 ] \left[\begin{array}{c}{0}\\{3}\\{-1}\end{array}\right]
Your Turn
For the following sets of vectors, find a basis for the subspace spanned by the vectors, and find the dimension of the subspace.
[
1
1
1
]
,
[
−
1
−
1
−
1
]
\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \quad \begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix}
[
1
0
5
]
,
[
0
1
0
]
,
[
0
−
1
0
]
\begin{bmatrix} 1 \\ 0 \\ 5 \end{bmatrix}, \quad \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} 0 \\ -1 \\ 0 \end{bmatrix}
[
−
4
−
3
5
]
,
[
2
3
3
]
,
[
2
0
2
]
\begin{bmatrix} -4 \\ -3 \\ 5 \end{bmatrix}, \quad \begin{bmatrix} 2 \\ 3 \\ 3 \end{bmatrix}, \quad \begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix}
[
1
3
0
]
,
[
0
2
2
]
,
[
−
1
−
1
2
]
\begin{bmatrix} 1 \\ 3 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} 0 \\ 2 \\ 2 \end{bmatrix}, \quad \begin{bmatrix} -1 \\ -1 \\ 2 \end{bmatrix}
[
1
3
]
,
[
0
2
]
,
[
−
1
−
1
]
\begin{bmatrix} 1 \\ 3 \end{bmatrix}, \quad \begin{bmatrix} 0 \\ 2 \end{bmatrix}, \quad \begin{bmatrix} -1 \\ -1 \end{bmatrix}
[
3
1
3
]
,
[
2
4
−
4
]
,
[
−
5
−
5
−
2
]
\begin{bmatrix} 3 \\ 1 \\ 3 \end{bmatrix}, \quad \begin{bmatrix} 2 \\ 4 \\ -4 \end{bmatrix}, \quad \begin{bmatrix} -5 \\ -5 \\ -2 \end{bmatrix}
Your Turn
For the following matrices, find a basis for the kernel (nullspace).
[
1
1
1
1
1
5
1
1
−
4
]
\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 5 \\ 1 & 1 & -4 \end{bmatrix}
[
2
−
1
−
3
4
0
−
4
−
1
1
2
]
\begin{bmatrix} 2 & -1 & -3 \\ 4 & 0 & -4 \\ -1 & 1 & 2 \end{bmatrix}
[
−
4
4
4
−
1
1
1
−
5
5
5
]
\begin{bmatrix} -4 & 4 & 4 \\ -1 & 1 & 1 \\ -5 & 5 & 5 \end{bmatrix}
[
−
2
1
1
1
−
4
2
2
2
1
0
4
3
]
\begin{bmatrix} -2 & 1 & 1 & 1 \\ -4 & 2 & 2 & 2 \\ 1 & 0 & 4 & 3 \end{bmatrix}
Your Turn
Suppose that X X ℝ 3 {\mathbb{R}}^3 X X
Your Turn
Consider a square matrix A A x → \vec{x} A x → = 0 → A \vec{x} = \vec{0} A A
Your Turn
For the following sets of vectors, find a basis for the subspace spanned by the vectors, and find the dimension of the subspace.
[
1
2
]
,
[
1
1
]
\begin{bmatrix} 1 \\ 2 \end{bmatrix}, \quad \begin{bmatrix} 1 \\ 1 \end{bmatrix}
[
1
1
1
]
,
[
2
2
2
]
,
[
1
1
2
]
\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \quad \begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix}, \quad \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}
[
5
3
1
]
,
[
5
−
1
5
]
,
[
−
1
3
−
4
]
\begin{bmatrix} 5 \\ 3 \\ 1 \end{bmatrix}, \quad \begin{bmatrix} 5 \\ -1 \\ 5 \end{bmatrix}, \quad \begin{bmatrix} -1 \\ 3 \\ -4 \end{bmatrix}
[
2
2
4
]
,
[
2
2
3
]
,
[
4
4
−
3
]
\begin{bmatrix} 2 \\ 2 \\ 4 \end{bmatrix}, \quad \begin{bmatrix} 2 \\ 2 \\ 3 \end{bmatrix}, \quad \begin{bmatrix} 4 \\ 4 \\ -3 \end{bmatrix}
[
1
0
]
,
[
2
0
]
,
[
3
0
]
\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} 2 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} 3 \\ 0 \end{bmatrix}
[
1
0
0
]
,
[
2
0
0
]
,
[
0
1
2
]
\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} 2 \\ 0 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}
Answer
[ 1 2 ] \left[\begin{array}{c}{1}\\{2}\end{array}\right] [ 1 1 ] \left[\begin{array}{c}{1}\\{1}\end{array}\right] 2 2 [ 1 1 1 ] \left[\begin{array}{c}{1}\\{1}\\{1}\end{array}\right] [ 1 1 2 ] \left[\begin{array}{c}{1}\\{1}\\{2}\end{array}\right] 2 2 [ 5 3 1 ] \left[\begin{array}{c}{5}\\{3}\\{1}\end{array}\right] [ 5 − 1 5 ] \left[\begin{array}{c}{5}\\{-1}\\{5}\end{array}\right] [ − 1 3 − 4 ] \left[\begin{array}{c}{-1}\\{3}\\{-4}\end{array}\right] 3 3 [ 2 2 4 ] \left[\begin{array}{c}{2}\\{2}\\{4}\end{array}\right] [ 2 2 3 ] \left[\begin{array}{c}{2}\\{2}\\{3}\end{array}\right] 2 2 [ 1 1 ] \left[\begin{array}{c}{1}\\{1}\end{array}\right] 1 1 [ 1 0 0 ] \left[\begin{array}{c}{1}\\{0}\\{0}\end{array}\right] [ 0 1 2 ] \left[\begin{array}{c}{0}\\{1}\\{2}\end{array}\right] 2 2 Your Turn
For the following matrices, find a basis for the kernel (nullspace).
[
2
6
1
9
1
3
2
9
3
9
0
9
]
\begin{bmatrix} 2 & 6 & 1 & 9 \\ 1 & 3 & 2 & 9 \\ 3 & 9 & 0 & 9 \end{bmatrix}
[
2
−
2
−
5
−
1
1
5
−
5
5
−
3
]
\begin{bmatrix} 2 & -2 & -5 \\ -1 & 1 & 5 \\ -5 & 5 & -3 \end{bmatrix}
[
1
−
5
−
4
2
3
5
−
3
5
2
]
\begin{bmatrix} 1 & -5 & -4 \\ 2 & 3 & 5 \\ -3 & 5 & 2 \end{bmatrix}
[
0
4
4
0
1
1
0
5
5
]
\begin{bmatrix} 0 & 4 & 4 \\ 0 & 1 & 1 \\ 0 & 5 & 5 \end{bmatrix}
Answer
[ 3 − 1 0 0 ] \left[\begin{array}{c}{3}\\{-1}\\{0}\\{0}\end{array}\right] [ 3 0 3 − 1 ] \left[\begin{array}{c}{3}\\{0}\\{3}\\{-1}\end{array}\right]
[
−
1
−
1
0
]
\left[\begin{array}{c}{-1}\\{-1}\\{0}\end{array}\right]
[
1
1
−
1
]
\left[\begin{array}{c}{1}\\{1}\\{-1}\end{array}\right]
[ − 1 0 0 ] \left[\begin{array}{c}{-1}\\{0}\\{0}\end{array}\right] [ 0 1 − 1 ] \left[\begin{array}{c}{0}\\{1}\\{-1}\end{array}\right] Your Turn
Find the s s ( 1 , 2 , 3 ) (1,2,3) ( 1 , 1 , s ) (1,1,s)
Your Turn
Find the angle θ \theta ( 1 , 3 , 1 ) (1,3,1) ( 2 , 1 , − 1 ) (2,1,-1)
Your Turn
Given that ⟨ v → , w → ⟩ = 3 \langle \vec{v}, \vec{w} \rangle = 3 ⟨ v → , u → ⟩ = − 1 \langle \vec{v}, \vec{u} \rangle = -1
⟨
u
→
,
2
v
→
⟩
\langle \vec{u}, 2 \vec{v} \rangle
⟨
v
→
,
2
w
→
+
3
u
→
⟩
\langle \vec{v}, 2 \vec{w} + 3 \vec{u} \rangle
⟨
w
→
+
3
u
→
,
v
→
⟩
\langle \vec{w} + 3 \vec{u}, \vec{v} \rangle
Your Turn
Suppose v → = ( 1 , 1 , − 1 ) \vec{v} = (1,1,-1)
proj
v
→
(
(
1
,
0
,
0
)
)
\operatorname{proj}_{\vec{v}}\bigl( (1,0,0) \bigr)
proj
v
→
(
(
1
,
2
,
3
)
)
\operatorname{proj}_{\vec{v}}\bigl( (1,2,3) \bigr)
proj
v
→
(
(
1
,
−
1
,
0
)
)
\operatorname{proj}_{\vec{v}}\bigl( (1,-1,0) \bigr)
Your Turn
Consider the vectors ( 1 , 2 , 3 ) (1,2,3) ( − 3 , 0 , 1 ) (-3,0,1) ( 1 , − 5 , 3 ) (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 ( 1 , 1 , 1 ) (1,1,1) Make the basis orthonormal. Your Turn
Let S S ( 1 , 3 , − 1 ) (1,3,-1) ( 1 , 1 , 1 ) (1,1,1) S S
Your Turn
Starting with ( 1 , 2 , 3 ) (1,2,3) ( 1 , 1 , 1 ) (1,1,1) ( 2 , 2 , 0 ) (2,2,0) ℝ 3 {\mathbb{R}}^3
Your Turn
Find an orthogonal basis of ℝ 3 {\mathbb{R}}^3 ( 3 , 1 , − 2 ) (3,1,-2)
Your Turn
Using cosines and sines of θ \theta u → \vec{u} ℝ 2 {\mathbb{R}}^2 θ \theta ı → = ( 1 , 0 ) \vec{\imath} = (1,0) ⟨ ı → , u → ⟩ \langle \vec{\imath}, \vec{u} \rangle
Your Turn
Find the s s ( 1 , 1 , 1 ) (1,1,1) ( 1 , s , 1 ) (1,s,1)
Answer
s
=
−
2
s=-2
Your Turn
Find the angle θ \theta ( 1 , 2 , 3 ) (1,2,3) ( 1 , 1 , 1 ) (1,1,1)
Answer
θ
≈
0.3876
\theta\approx 0.3876
Your Turn
Given that ⟨ v → , w → ⟩ = 1 \langle \vec{v}, \vec{w} \rangle = 1 ⟨ v → , u → ⟩ = − 1 \langle \vec{v}, \vec{u} \rangle = -1 ‖ v → ‖ = 3 \lVert \vec{v} \rVert = 3
⟨
3
u
→
,
5
v
→
⟩
\langle 3 \vec{u}, 5 \vec{v} \rangle
⟨
v
→
,
2
w
→
+
3
u
→
⟩
\langle \vec{v}, 2 \vec{w} + 3 \vec{u} \rangle
⟨
w
→
+
3
v
→
,
v
→
⟩
\langle \vec{w} + 3 \vec{v}, \vec{v} \rangle
Answer
−
15
-15
−
1
-1
28
28
Your Turn
Suppose v → = ( 1 , 0 , − 1 ) \vec{v} = (1,0,-1)
proj
v
→
(
(
0
,
2
,
1
)
)
\operatorname{proj}_{\vec{v}}\bigl( (0,2,1) \bigr)
proj
v
→
(
(
1
,
0
,
1
)
)
\operatorname{proj}_{\vec{v}}\bigl( (1,0,1) \bigr)
proj
v
→
(
(
4
,
−
1
,
0
)
)
\operatorname{proj}_{\vec{v}}\bigl( (4,-1,0) \bigr)
Answer
(
−
1
2
,
0
,
1
2
)
\left(-\frac{1}{2},0,\frac{1}{2}\right)
(
0
,
0
,
0
)
(0,0,0)
(
2
,
0
,
−
2
)
(2,0,-2)
Your Turn
The vectors ( 1 , 1 , − 1 ) (1,1,-1) ( 2 , − 1 , 1 ) (2,-1,1) ( 1 , − 5 , 3 ) (1,-5,3)
(
1
,
−
8
,
4
)
(1,-8,4)
(
5
,
−
7
,
5
)
(5,-7,5)
(
0
,
−
6
,
2
)
(0,-6,2)
Answer
(
1
,
1
,
−
1
)
−
(
2
,
−
1
,
1
)
+
2
(
1
,
−
5
,
3
)
(1,1,-1)-(2,-1,1)+2(1,-5,3)
2
(
2
,
−
1
,
1
)
+
(
1
,
−
5
,
3
)
2(2,-1,1)+(1,-5,3)
2
(
1
,
1
,
−
1
)
−
2
(
2
,
−
1
,
1
)
+
2
(
1
,
−
5
,
3
)
2(1,1,-1)-2(2,-1,1)+2(1,-5,3)
Your Turn
Let S S ( 2 , − 1 , 1 ) (2,-1,1) ( 2 , 2 , 2 ) (2,2,2) S S
Answer
( 2 , − 1 , 1 ) (2,-1,1) ( 2 3 , 8 3 , 4 3 ) \left(\frac{2}{3},\frac{8}{3},\frac{4}{3}\right)
Your Turn
Starting with ( 1 , 1 , − 1 ) (1,1,-1) ( 2 , 3 , − 1 ) (2,3,-1) ( 1 , − 1 , 1 ) (1,-1,1) ℝ 3 {\mathbb{R}}^3
Answer
( 1 , 1 , − 1 ) (1,1,-1) ( 0 , 1 , 1 ) (0,1,1) ( 4 3 , − 2 3 , 2 3 ) \left(\frac{4}{3},-\frac{2}{3},\frac{2}{3}\right)
Your Turn
Compute
det ( [ 2 1 2 3 0 8 6 5 0 0 3 9 0 0 0 1 ] − 1 ) \det \left( \begin{bmatrix} 2 & 1 & 2 & 3 \\ 0 & 8 & 6 & 5 \\ 0 & 0 & 3 & 9 \\ 0 & 0 & 0 & 1 \end{bmatrix}^{-1} \right) \nonumber
without computing the inverse.
Your Turn
Suppose
L = [ 1 0 0 0 2 1 0 0 7 π 1 0 2 8 5 − 99 1 ] and U = [ 5 9 1 − sin ( 1 ) 0 1 88 − 1 0 0 1 3 0 0 0 1 ] . L = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 \\ 7 & \pi & 1 & 0 \\ 2^8 & 5 & -99 & 1 \end{bmatrix} \qquad \text{and} \qquad U = \begin{bmatrix} 5 & 9 & 1 & -\sin(1) \\ 0 & 1 & 88 & -1 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \end{bmatrix}. \nonumber
Let A = L U A = LU det ( A ) \det(A) A A det ( L ) \det(L) det ( U ) \det(U) Your Turn
Consider the linear mapping from R 2 {\mathbb R}^2 R 2 {\mathbb R}^2 A = [ 1 x 2 1 ] A = \left[ \begin{smallmatrix} 1 & x \\ 2 & 1 \end{smallmatrix} \right] x x A A x x
Your Turn
Suppose A A S S n × n n \times n S S det ( A ) = 3 \det(A) = 3 det ( S − 1 A S ) \det(S^{-1}AS) det ( S A S − 1 ) \det(SAS^{-1})
Your Turn
Let A A n × n n \times n det ( A ) = 1 \det(A)=1 det ( x A ) \det(x A) x x det ( x I ) \det(xI) x A = ( x I ) A xA = (xI)A
Your Turn
Compute
det ( [ 3 4 7 12 0 − 1 9 − 8 0 0 − 2 4 0 0 0 2 ] − 1 ) \det \left( \begin{bmatrix} 3 & 4 & 7 & 12 \\ 0 & -1 & 9 & -8 \\ 0 & 0 & -2 & 4 \\ 0 & 0 & 0 & 2 \end{bmatrix}^{-1} \right) \nonumber
without computing the inverse.
Answer
12
12
Your Turn
(challenging)
Find all the x x
[ 1 2 1 x ] − 1 \begin{bmatrix} 1 & 2 \\ 1 & x \end{bmatrix}^{-1} \nonumber
have only integer entries (no fractions). Note that there are two answers.
Answer
1 1 3 3
Adapted from Differential Equations for Engineers by Jiří Lebl (https://www.jirka.org/diffyqs/), © Jiří Lebl, licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0 .
