(a) Answer The value of A + B is [ − 4 9 1 5 − 6 − 5 15 − 5 − 2 3 10 − 10 − 4 − 2 − 2 ] . Explanation Given: The matrices A = [ − 1 4 0 3 − 2 2 5 4 − 1 0 8 − 6 − 4 − 1 0 ] and B = [ − 3 5 1 2 − 4 − 7 10 − 9 − 1 3 2 − 4 0 − 1 − 2 ] . Concept used: Operations like addition and subtraction of matrices are only possible when they are of same dimension. Dimension of matrix with a rows and b columns is given by a × b . If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar. Calculation: Add the matrix A and Bas follows: A + B = [ − 1 4 0 3 − 2 2 5 4 − 1 0 8 − 6 − 4 − 1 0 ] + [ − 3 5 1 2 − 4 − 7 10 − 9 − 1 3 2 − 4 0 − 1 − 2 ] = [ − 1 − 3 4 + 5 1 + 0 3 + 2 − 2 − 4 2 − 7 5 + 10 4 − 9 − 1 − 1 0 + 3 8 + 2 − 6 − 4 − 4 + 0 − 1 − 1 0 − 2 ] = [ − 4 9 1 5 − 6 − 5 15 − 5 − 2 3 10 − 10 − 4 − 2 − 2 ] Thus, the value of A + B is [ − 4 9 1 5 − 6 − 5 15 − 5 − 2 3 10 − 10 − 4 − 2 − 2 ] . Now, by graphing utility of matrix capability, Hence, the result is verified. (b) To determine To find A − B and check your result by graphing utility. Answer The value of A − B is [ − 2 − 1 − 1 1 2 9 − 5 13 0 3 6 − 2 − 4 0 2 ] . Explanation Given: The matrices A = [ − 1 4 0 3 − 2 2 5 4 − 1 0 8 − 6 − 4 − 1 0 ] and B = [ − 3 5 1 2 − 4 − 7 10 − 9 − 1 3 2 − 4 0 − 1 − 2 ] . Concept used: Operations like addition and subtraction of matrices are only possible when they are of same dimension. Dimension of matrix with a rows and b columns is given by a × b . If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar. Calculation: Subtract the matrices A and Bas follows: A − B = [ − 1 4 0 3 − 2 2 5 4 − 1 0 8 − 6 − 4 − 1 0 ] − [ − 3 5 1 2 − 4 − 7 10 − 9 − 1 3 2 − 4 0 − 1 − 2 ] = [ − 1 + 3 4 − 5 0 − 1 3 − 2 − 2 + 4 2 + 7 5 − 10 4 + 9 − 1 + 1 0 + 3 8 − 2 − 6 + 4 − 4 − 0 − 1 + 1 0 + 2 ] = [ − 2 − 1 − 1 1 2 9 − 5 13 0 3 6 − 2 − 4 0 2 ] Thus, the value of A − B is [ − 2 − 1 − 1 1 2 9 − 5 13 0 3 6 − 2 − 4 0 2 ] . Now, by graphing utility of matrix capability, Hence, the result is verified. (c) To determine To find 3 A and check your result by graphing utility. Answer The value of 3 A is [ − 3 12 0 9 − 6 6 15 12 − 3 0 24 − 18 − 12 − 3 0 ] . Explanation Given: The matrix is A = [ − 1 4 0 3 − 2 2 5 4 − 1 0 8 − 6 − 4 − 1 0 ] . Concept used: Operations like addition and subtraction of matrices are only possible when they are of same dimension. Dimension of matrix with a rows and b columns is given by a × b . If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar. Calculation: Multiply all the elements of matrix A by 3 as follows: 3 A = 3 [ − 1 4 0 3 − 2 2 5 4 − 1 0 8 − 6 − 4 − 1 0 ] = [ − 3 12 0 9 − 6 6 15 12 − 3 0 24 − 18 − 12 − 3 0 ] Thus, the value of 3 A is [ − 3 12 0 9 − 6 6 15 12 − 3 0 24 − 18 − 12 − 3 0 ] . Now, by graphing utility of matrix capability, Hence, the result is verified. (d) To determine To find 3 A − 2 B and check your result by graphing utility. Answer The value of 3 A − 2 B is [ 3 2 − 2 5 2 20 − 5 30 − 1 − 6 20 − 10 − 12 − 1 4 ] . Explanation Given: The matrices A = [ − 1 4 0 3 − 2 2 5 4 − 1 0 8 − 6 − 4 − 1 0 ] and B = [ − 3 5 1 2 − 4 − 7 10 − 9 − 1 3 2 − 4 0 − 1 − 2 ] . Concept used: Operations like addition and subtraction of matrices are only possible when they are of same dimension. Dimension of matrix with a rows and b columns is given by a × b . If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar. Calculation: The value of 3 A − 2 B is calculated as follows: Thus, the value of 3 A − 2 B is [ 3 2 − 2 5 2 20 − 5 30 − 1 − 6 20 − 10 − 12 − 1 4 ] . Now, by graphing utility of matrix capability, Hence, the result is verified.

Precalculus with Limits: A Graphing Approach

7th Edition

ISBN: 9781305071711

Author: Ron Larson

Publisher: Brooks/Cole

See similar textbooks

Videos

Question

Chapter 7.5, Problem 18E

(a)

To determine

To find A+B and check your result by graphingutility.

(a)

Expert Solution

Answer to Problem 18E

The value of A+B is [−4915−6−515−5−2310−10−4−2−2] .

Explanation of Solution

Given:

The matrices A=[−1403−2254−108−6−4−10] and B=[−3512−4−710−9−132−40−1−2] .

Concept used:

Operations like addition and subtraction of matrices are only possible when they are of same dimension.

Dimension of matrix with a rows and b columns is given by a×b .

If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar.

Calculation:

Add the matrix A and Bas follows:

A+B=[−1403−2254−108−6−4−10]+[−3512−4−710−9−132−40−1−2]=[−1−34+51+03+2−2−42−75+104−9−1−10+38+2−6−4−4+0−1−10−2]=[−4915−6−515−5−2310−10−4−2−2]

Thus, the value of A+B is [−4915−6−515−5−2310−10−4−2−2] .

Now, by graphing utility of matrix capability,

Precalculus with Limits: A Graphing Approach, Chapter 7.5, Problem 18E , additional homework tip 1

Hence, the result is verified.

(b)

To determine

To find A−B and check your result by graphing utility.

(b)

Expert Solution

Answer to Problem 18E

The value of A−B is [−2−1−1129−513036−2−402] .

Explanation of Solution

Given:

The matrices A=[−1403−2254−108−6−4−10] and B=[−3512−4−710−9−132−40−1−2] .

Concept used:

Operations like addition and subtraction of matrices are only possible when they are of same dimension.

Dimension of matrix with a rows and b columns is given by a×b .

If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar.

Calculation:

Subtract the matrices A and Bas follows:

A−B=[−1403−2254−108−6−4−10]−[−3512−4−710−9−132−40−1−2]=[−1+34−50−13−2−2+42+75−104+9−1+10+38−2−6+4−4−0−1+10+2]=[−2−1−1129−513036−2−402]

Thus, the value of A−B is [−2−1−1129−513036−2−402] .

Now, by graphing utility of matrix capability,

Precalculus with Limits: A Graphing Approach, Chapter 7.5, Problem 18E , additional homework tip 2

Hence, the result is verified.

(c)

To determine

To find 3A and check your result by graphing utility.

(c)

Expert Solution

Answer to Problem 18E

The value of 3A is [−31209−661512−3024−18−12−30] .

Explanation of Solution

Given:

The matrix is A=[−1403−2254−108−6−4−10] .

Concept used:

Operations like addition and subtraction of matrices are only possible when they are of same dimension.

Dimension of matrix with a rows and b columns is given by a×b .

If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar.

Calculation:

Multiply all the elements of matrix A by 3 as follows:

3A=3[−1403−2254−108−6−4−10]=[−31209−661512−3024−18−12−30]

Thus, the value of 3A is [−31209−661512−3024−18−12−30] .

Now, by graphing utility of matrix capability,

Precalculus with Limits: A Graphing Approach, Chapter 7.5, Problem 18E , additional homework tip 3

Hence, the result is verified.

(d)

To determine

To find 3A−2B and check your result by graphing utility.

(d)

Expert Solution

Answer to Problem 18E

The value of 3A−2B is [32−25220−530−1−620−10−12−14] .

Explanation of Solution

Given:

The matrices A=[−1403−2254−108−6−4−10] and B=[−3512−4−710−9−132−40−1−2] .

Concept used:

Operations like addition and subtraction of matrices are only possible when they are of same dimension.

Dimension of matrix with a rows and b columns is given by a×b .

If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar.

Calculation:

The value of 3A−2B is calculated as follows:

Thus, the value of 3A−2B is [32−25220−530−1−620−10−12−14] .

Now, by graphing utility of matrix capability,

Precalculus with Limits: A Graphing Approach, Chapter 7.5, Problem 18E , additional homework tip 4

Hence, the result is verified.

Chapter 7.5, Problem 17E

Chapter 7.5, Problem 19E

Chapter 7 Solutions

Precalculus with Limits: A Graphing Approach

Ch. 7.1 - Prob. 1E Ch. 7.1 - Prob. 2E Ch. 7.1 - Prob. 3E Ch. 7.1 - Prob. 4E Ch. 7.1 - Prob. 5E Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - Prob. 8E Ch. 7.1 - Prob. 9E Ch. 7.1 - Prob. 10E

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.