(a) Answer The value of A + B is [ − 1 − 1 − 2 − 3 10 2 ] . Explanation Given: The matrices A = [ 1 − 1 3 0 6 9 ] and B = [ − 2 0 − 5 − 3 4 − 7 ] . 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 − 1 3 0 6 9 ] + [ − 2 0 − 5 − 3 4 − 7 ] = [ 1 − 2 − 1 + 0 3 − 5 0 − 3 6 + 4 9 − 7 ] = [ − 1 − 1 − 2 − 3 10 2 ] Thus, the value of A + B is [ − 1 − 1 − 2 − 3 10 2 ] . Now, by graphing utility of matrix capability, Hence, the result is verified. (b) To determine The value of A − B , if A = [ 1 − 1 3 0 6 9 ] and B = [ − 2 0 − 5 − 3 4 − 7 ] . Answer The value of A − B is [ 3 − 1 8 3 2 16 ] . Explanation Given: The matrices A = [ 1 − 1 3 0 6 9 ] and B = [ − 2 0 − 5 − 3 4 − 7 ] . 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 − 1 3 0 6 9 ] − [ − 2 0 − 5 − 3 4 − 7 ] = [ 1 + 2 − 1 − 0 3 + 5 0 + 3 6 − 4 9 + 7 ] = [ 3 − 1 8 3 2 16 ] Thus, the value of A − B is [ 3 − 1 8 3 2 16 ] . Now, by graphing utility of matrix capability, Hence, the result is verified. (c) To determine The value of 3 A , if A = [ 1 − 1 3 0 6 9 ] . Answer The value of 3 A is [ 3 − 3 9 0 18 27 ] . Explanation Given: The matrix is A = [ 1 − 1 3 0 6 9 ] . 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 − 1 3 0 6 9 ] = [ 3 ⋅ 1 3 ⋅ − 1 3 ⋅ 3 3 ⋅ 0 3 ⋅ 6 3 ⋅ 9 ] = [ 3 − 3 9 0 18 27 ] Thus, the value of 3 A is [ 3 − 3 9 0 18 27 ] . Now, by graphing utility of matrix capability, Hence, the result is verified. (d) To determine The value of 3 A − 2 B , if A = [ 1 − 1 3 0 6 9 ] and B = [ − 2 0 − 5 − 3 4 − 7 ] . Answer The value of 3 A − 2 B is [ 7 − 3 19 6 10 41 ] . Explanation Given: The matrices A = [ 1 − 1 3 0 6 9 ] and B = [ − 2 0 − 5 − 3 4 − 7 ] . 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: 3 A − 2 B = 3 [ 1 − 1 3 0 6 9 ] − 2 [ − 2 0 − 5 − 3 4 − 7 ] = [ 3 − 3 9 0 18 27 ] + [ 4 0 10 6 − 8 14 ] = [ 3 + 4 − 3 + 0 9 + 10 0 + 6 18 − 8 27 + 14 ] = [ 7 − 3 19 6 10 41 ] Thus, the value of 3 A − 2 B is [ 7 − 3 19 6 10 41 ] . 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

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Question

Chapter 7.5, Problem 16E

(a)

To determine

The value of A+B , if A=[1−13069] and B=[−20−5−34−7] .

(a)

Expert Solution

Answer to Problem 16E

The value of A+B is [−1−1−2−3102] .

Explanation of Solution

Given:

The matrices A=[1−13069] and B=[−20−5−34−7] .

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=[1−13069]+[−20−5−34−7]=[1−2−1+03−50−36+49−7]=[−1−1−2−3102]

Thus, the value of A+B is [−1−1−2−3102] .

Now, by graphing utility of matrix capability,

Precalculus with Limits: A Graphing Approach, Chapter 7.5, Problem 16E , additional homework tip 1

Hence, the result is verified.

(b)

To determine

The value of A−B , if A=[1−13069] and B=[−20−5−34−7] .

(b)

Expert Solution

Answer to Problem 16E

The value of A−B is [3−183216] .

Explanation of Solution

Given:

The matrices A=[1−13069] and B=[−20−5−34−7] .

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=[1−13069]−[−20−5−34−7]=[1+2−1−03+50+36−49+7]=[3−183216]

Thus, the value of A−B is [3−183216] .

Now, by graphing utility of matrix capability,

Precalculus with Limits: A Graphing Approach, Chapter 7.5, Problem 16E , additional homework tip 2

Hence, the result is verified.

(c)

To determine

The value of 3A , if A=[1−13069] .

(c)

Expert Solution

Answer to Problem 16E

The value of 3A is [3−3901827] .

Explanation of Solution

Given:

The matrix is A=[1−13069] .

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[1−13069]=[3⋅13⋅−13⋅33⋅03⋅63⋅9]=[3−3901827]

Thus, the value of 3A is [3−3901827] .

Now, by graphing utility of matrix capability,

Precalculus with Limits: A Graphing Approach, Chapter 7.5, Problem 16E , additional homework tip 3

Hence, the result is verified.

(d)

To determine

The value of 3A−2B , if A=[1−13069] and B=[−20−5−34−7] .

(d)

Expert Solution

Answer to Problem 16E

The value of 3A−2B is [7−31961041] .

Explanation of Solution

Given:

The matrices A=[1−13069] and B=[−20−5−34−7] .

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:

3A−2B=3[1−13069]−2[−20−5−34−7]=[3−3901827]+[40106−814]=[3+4−3+09+100+618−827+14]=[7−31961041]

Thus, the value of 3A−2B is [7−31961041] .

Now, by graphing utility of matrix capability,

Precalculus with Limits: A Graphing Approach, Chapter 7.5, Problem 16E , additional homework tip 4

Hence, the result is verified.

Chapter 7.5, Problem 15E

Chapter 7.5, Problem 17E

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

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