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Math  /  Algebra

Question23. Let A=[−612−36]A=\left[\begin{array}{rr}-6 & 12 \\ -3 & 6\end{array}\right] and w=[21]w=\left[\begin{array}{l}2 \\ 1\end{array}\right]. Determine if ww is in Col Al Is in in Nol A?
24. Let A=[−8−2−9648404]A=\left[\begin{array}{rrr}-8 & -2 & -9 \\ 6 & 4 & 8 \\ 4 & 0 & 4\end{array}\right] and w=[21−2]w=\left[\begin{array}{r}2 \\ 1 \\ -2\end{array}\right]. Determine if w is in ColA\mathrm{Col} A. Is w in Nul A?
In Exercises 15 and 16 , find AA ouch that the given set is Col⁡A\operatorname{Col} A
15. {[2s+3tr+s−2t4r+s3r−s−t]:r,s,t\left\{\left[\begin{array}{c}2 s+3 t \\ r+s-2 t \\ 4 r+s \\ 3 r-s-t\end{array}\right]: r, s, t\right. real }\}
16. {[b−c2b+c+d5c−4dd]:b,c,d\left\{\left[\begin{array}{c}b-c \\ 2 b+c+d \\ 5 c-4 d \\ d\end{array}\right]: b, c, d\right. real }\}

Studdy Solution

STEP 1

1. A vector w w is in the column space of a matrix A A if there exists a vector x x such that Ax=w Ax = w .
2. A vector w w is in the null space of a matrix A A if Aw=0 Aw = 0 .
3. For problems involving column space, we need to check if the vector w w can be expressed as a linear combination of the columns of A A .
4. For problems involving null space, we need to check if multiplying A A by w w results in the zero vector.
5. To find a matrix A A such that a given set is Col⁡A\operatorname{Col} A, we need to express the given vectors as columns of A A .

STEP 2

1. Problem 23: Determine if w w is in Col⁡A\operatorname{Col} A.
2. Problem 23: Determine if w w is in Nul⁡A\operatorname{Nul} A.
3. Problem 24: Determine if w w is in Col⁡A\operatorname{Col} A.
4. Problem 24: Determine if w w is in Nul⁡A\operatorname{Nul} A.
5. Problem 15: Find matrix A A such that the given set is Col⁡A\operatorname{Col} A.
6. Problem 16: Find matrix A A such that the given set is Col⁡A\operatorname{Col} A.

STEP 3

For Problem 23, check if w=[21] w = \begin{bmatrix} 2 \\ 1 \end{bmatrix} is in Col⁡A\operatorname{Col} A where A=[−612−36] A = \begin{bmatrix} -6 & 12 \\ -3 & 6 \end{bmatrix} .
To determine this, solve the matrix equation Ax=w Ax = w :
[−612−36][x1x2]=[21]\begin{bmatrix} -6 & 12 \\ -3 & 6 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}
This results in the system of equations:
−6x1+12x2=2-6x_1 + 12x_2 = 2 −3x1+6x2=1-3x_1 + 6x_2 = 1
Check for consistency by solving this system.

STEP 4

Solve the system using row reduction or substitution. Notice that the second equation is a multiple of the first:
Divide the first equation by 2:
−3x1+6x2=1-3x_1 + 6x_2 = 1
This matches the second equation, confirming consistency. Therefore, w w is in Col⁡A\operatorname{Col} A.

STEP 5

For Problem 23, check if w=[21] w = \begin{bmatrix} 2 \\ 1 \end{bmatrix} is in Nul⁡A\operatorname{Nul} A.
Calculate Aw Aw :
A[21]=[−612−36][21]=[−6(2)+12(1)−3(2)+6(1)]=[00]A \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} -6 & 12 \\ -3 & 6 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} -6(2) + 12(1) \\ -3(2) + 6(1) \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}
Since Aw=0 Aw = 0 , w w is in Nul⁡A\operatorname{Nul} A.

STEP 6

For Problem 24, check if w=[21−2] w = \begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix} is in Col⁡A\operatorname{Col} A where A=[−8−2−9648404] A = \begin{bmatrix} -8 & -2 & -9 \\ 6 & 4 & 8 \\ 4 & 0 & 4 \end{bmatrix} .
Solve the matrix equation Ax=w Ax = w :
[−8−2−9648404][x1x2x3]=[21−2]\begin{bmatrix} -8 & -2 & -9 \\ 6 & 4 & 8 \\ 4 & 0 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix}
Check for consistency by solving this system using row reduction.

STEP 7

Perform row reduction on the augmented matrix:
[−8−2−926481404−2]\left[\begin{array}{ccc|c} -8 & -2 & -9 & 2 \\ 6 & 4 & 8 & 1 \\ 4 & 0 & 4 & -2 \end{array}\right]
After row reduction, if the system is consistent, w w is in Col⁡A\operatorname{Col} A. If not, it is not.

STEP 8

For Problem 24, check if w=[21−2] w = \begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix} is in Nul⁡A\operatorname{Nul} A.
Calculate Aw Aw :
A[21−2]=[−8−2−9648404][21−2]A \begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix} = \begin{bmatrix} -8 & -2 & -9 \\ 6 & 4 & 8 \\ 4 & 0 & 4 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix}
Compute the result and check if it equals the zero vector.

STEP 9

For Problem 15, find matrix A A such that the given set is Col⁡A\operatorname{Col} A:
The set is:
{[2s+3tr+s−2t4r+s3r−s−t]:r,s,t real}\left\{\begin{bmatrix} 2s + 3t \\ r + s - 2t \\ 4r + s \\ 3r - s - t \end{bmatrix} : r, s, t \text{ real} \right\}
Express each vector in terms of r,s,t r, s, t and form columns of A A .

STEP 10

The columns of A A are the coefficients of r,s,t r, s, t :
A=[02311−24103−1−1]A = \begin{bmatrix} 0 & 2 & 3 \\ 1 & 1 & -2 \\ 4 & 1 & 0 \\ 3 & -1 & -1 \end{bmatrix}

STEP 11

For Problem 16, find matrix A A such that the given set is Col⁡A\operatorname{Col} A:
The set is:
{[b−c2b+c+d5c−4dd]:b,c,d real}\left\{\begin{bmatrix} b - c \\ 2b + c + d \\ 5c - 4d \\ d \end{bmatrix} : b, c, d \text{ real} \right\}
Express each vector in terms of b,c,d b, c, d and form columns of A A .

STEP 12

The columns of A A are the coefficients of b,c,d b, c, d :
A=[1−1021105−4001]A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 1 & 1 \\ 0 & 5 & -4 \\ 0 & 0 & 1 \end{bmatrix}
The solutions to the problems are as follows: - Problem 23: w w is in both Col⁡A\operatorname{Col} A and Nul⁡A\operatorname{Nul} A. - Problem 24: Check row reduction for Col⁡A\operatorname{Col} A and compute Aw Aw for Nul⁡A\operatorname{Nul} A. - Problem 15: A=[02311−24103−1−1] A = \begin{bmatrix} 0 & 2 & 3 \\ 1 & 1 & -2 \\ 4 & 1 & 0 \\ 3 & -1 & -1 \end{bmatrix} . - Problem 16: A=[1−1021105−4001] A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 1 & 1 \\ 0 & 5 & -4 \\ 0 & 0 & 1 \end{bmatrix} .

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