6-2 Matrix Multiplication, Inverses and Determinants

Find AB and BA, if possible.

8. A =

B =

SOLUTION:

A = ; B =

A is a 2 × 3 matrix and B is a 4 × 2 matrix. Because the number of columns of A is not equal to the number of rows of B, AB is undefined.

B is a 4 × 2 matrix and A is a 2 × 3 matrix. Because the number of columns of B is equal to the number of rows of A, BA exists.

To find the first entry of BA, find the sum of the products of the entries in row 1 of B and column 1 of A. Follow the same procedure for row 2 column 1 of BA and the remaining entries.

ANSWER:

AB is undefined; BA = .

Write each system of equations as a matrix equation, AX = B. Then use Gauss-Jordan elimination on the augmented matrix to solve they system. eSolutions Manual - Powered by Cognero Page 1 11.

SOLUTION: Write the system matrix in form AX = B. Make sure you align the variables. For the first matrix, the first column should include x1, the second column x2, and the third column x3. The column matrix and the matrix of constant terms should be listed in order.

· =

Write the augmented matrix . Attach the matrix of constant terms to the end of the 3 × 3 matrix.

Use Gauss-Jordan elimination to solve the system. First, use elementary row operations to transform A into I.

The solution of the equation is given by X.

Therefore, the solution is (1, −5, −6). ANSWER:

· = ; (1, −5, −6)

Determine whether A and B are inverse matrices.

20. A =

B =

SOLUTION:

A = ; B =

If A and B are inverse matrices, then AB = BA = I.

Because AB = BA = I, B = A−1 and A = B−1. ANSWER: yes

24. A =

B =

SOLUTION:

A = ;B =

If A and B are inverse matrices, then AB = BA = I.

AB ≠ I, so A and B are not inverses. ANSWER: no

Find A -1, if it exists. If A -1 does not exist, write singular.

28. A =

SOLUTION: Create the doubly augmented matrix .

Apply elementary row operations to write the matrix in reduced row-echelon form.

A row of 0s has been formed, so the first 2 columns cannot become the identity matrix. Therefore, A is singular. ANSWER: singular

Find the determinant of each matrix. Then find the inverse of the matrix, if it exists.

36.

SOLUTION: Find the determinant.

det(A) = |A| = = ad − bc

Because the determinant is not 0, the matrix is invertible. Find the inverse matrix.

Confirm that AA−1 = A−1A = I.

ANSWER:

−23;

39.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

−11;

42.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

12; Find AB and BA, if possible.

8. A =

B =

SOLUTION:

A = ; B =

A is a 2 × 3 matrix and B is a 4 × 2 matrix. Because the number of columns of A is not equal to the number of rows of B, AB is undefined.

B is a 4 × 2 matrix and A is a 2 × 3 matrix. Because the number of columns of B is equal to the number of rows of A, BA exists.

To find the first entry of BA, find the sum of the products of the entries in row 1 of B and column 1 of A. Follow the same procedure for row 2 column 1 of BA and the remaining entries.

ANSWER:

AB is undefined; BA = .

6-2 Matrix Multiplication, Inverses and Determinants Write each system of equations as a matrix equation, AX = B. Then use Gauss-Jordan elimination on the augmented matrix to solve they system.

11.

· =

Write the augmented matrix . Attach the matrix of constant terms to the end of the 3 × 3 matrix.

Use Gauss-Jordan elimination to solve the system. First, use elementary row operations to transform A into I.

eSolutions Manual - Powered by Cognero Page 2

The solution of the equation is given by X.

Therefore, the solution is (1, −5, −6). ANSWER:

· = ; (1, −5, −6)

Determine whether A and B are inverse matrices.

20. A =

B =

SOLUTION:

A = ; B =

If A and B are inverse matrices, then AB = BA = I.

Because AB = BA = I, B = A−1 and A = B−1. ANSWER: yes

24. A =

B =

SOLUTION:

A = ;B =

If A and B are inverse matrices, then AB = BA = I.

AB ≠ I, so A and B are not inverses. ANSWER: no

Find A -1, if it exists. If A -1 does not exist, write singular.

28. A =

SOLUTION: Create the doubly augmented matrix .

Apply elementary row operations to write the matrix in reduced row-echelon form.

A row of 0s has been formed, so the first 2 columns cannot become the identity matrix. Therefore, A is singular. ANSWER: singular

Find the determinant of each matrix. Then find the inverse of the matrix, if it exists.

36.

SOLUTION: Find the determinant.

det(A) = |A| = = ad − bc

Because the determinant is not 0, the matrix is invertible. Find the inverse matrix.

Confirm that AA−1 = A−1A = I.

ANSWER:

−23;

39.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

−11;

42.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

12; Find AB and BA, if possible.

8. A =

B =

SOLUTION:

A = ; B =

A is a 2 × 3 matrix and B is a 4 × 2 matrix. Because the number of columns of A is not equal to the number of rows of B, AB is undefined.

B is a 4 × 2 matrix and A is a 2 × 3 matrix. Because the number of columns of B is equal to the number of rows of A, BA exists.

To find the first entry of BA, find the sum of the products of the entries in row 1 of B and column 1 of A. Follow the same procedure for row 2 column 1 of BA and the remaining entries.

ANSWER:

AB is undefined; BA = .

Write each system of equations as a matrix equation, AX = B. Then use Gauss-Jordan elimination on the augmented matrix to solve they system.

11.

· =

Write the augmented matrix . Attach the matrix of constant terms to the end of the 3 × 3 matrix.

6-2 Matrix Multiplication, Inverses and Determinants Use Gauss-Jordan elimination to solve the system. First, use elementary row operations to transform A into I.

The solution of the equation is given by X.

Therefore, the solution is (1, −5, −6). ANSWER:

· = ; (1, −5, −6) eSolutions Manual - Powered by Cognero Page 3

Determine whether A and B are inverse matrices.

20. A =

B =

SOLUTION:

A = ; B =

If A and B are inverse matrices, then AB = BA = I.

Because AB = BA = I, B = A−1 and A = B−1. ANSWER: yes

24. A =

B =

SOLUTION:

A = ;B =

If A and B are inverse matrices, then AB = BA = I.

AB ≠ I, so A and B are not inverses. ANSWER: no

Find A -1, if it exists. If A -1 does not exist, write singular.

28. A =

SOLUTION: Create the doubly augmented matrix .

Apply elementary row operations to write the matrix in reduced row-echelon form.

A row of 0s has been formed, so the first 2 columns cannot become the identity matrix. Therefore, A is singular. ANSWER: singular

Find the determinant of each matrix. Then find the inverse of the matrix, if it exists.

36.

SOLUTION: Find the determinant.

det(A) = |A| = = ad − bc

Because the determinant is not 0, the matrix is invertible. Find the inverse matrix.

Confirm that AA−1 = A−1A = I.

ANSWER:

−23;

39.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

−11;

42.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

12; Find AB and BA, if possible.

8. A =

B =

SOLUTION:

A = ; B =

A is a 2 × 3 matrix and B is a 4 × 2 matrix. Because the number of columns of A is not equal to the number of rows of B, AB is undefined.

B is a 4 × 2 matrix and A is a 2 × 3 matrix. Because the number of columns of B is equal to the number of rows of A, BA exists.

To find the first entry of BA, find the sum of the products of the entries in row 1 of B and column 1 of A. Follow the same procedure for row 2 column 1 of BA and the remaining entries.

ANSWER:

AB is undefined; BA = .

Write each system of equations as a matrix equation, AX = B. Then use Gauss-Jordan elimination on the augmented matrix to solve they system.

11.

· =

Write the augmented matrix . Attach the matrix of constant terms to the end of the 3 × 3 matrix.

Use Gauss-Jordan elimination to solve the system. First, use elementary row operations to transform A into I.

The solution of the equation is given by X.

6-2 Therefore,Matrix Multiplication,the solution is (1, −5, Inverses−6). and Determinants ANSWER:

· = ; (1, −5, −6)

Determine whether A and B are inverse matrices.

20. A =

B =

SOLUTION:

A = ; B =

If A and B are inverse matrices, then AB = BA = I.

Because AB = BA = I, B = A−1 and A = B−1. ANSWER: yes

24. A =

B =

SOLUTION:

A = ;B =

If A and B are inverse matrices, then AB = BA = I.

eSolutions Manual - Powered by Cognero Page 4

AB ≠ I, so A and B are not inverses. ANSWER: no

Find A -1, if it exists. If A -1 does not exist, write singular.

28. A =

SOLUTION: Create the doubly augmented matrix .

Apply elementary row operations to write the matrix in reduced row-echelon form.

A row of 0s has been formed, so the first 2 columns cannot become the identity matrix. Therefore, A is singular. ANSWER: singular

Find the determinant of each matrix. Then find the inverse of the matrix, if it exists.

36.

SOLUTION: Find the determinant.

det(A) = |A| = = ad − bc

Because the determinant is not 0, the matrix is invertible. Find the inverse matrix.

Confirm that AA−1 = A−1A = I.

ANSWER:

−23;

39.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

−11;

42.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

12; Find AB and BA, if possible.

8. A =

B =

SOLUTION:

A = ; B =

A is a 2 × 3 matrix and B is a 4 × 2 matrix. Because the number of columns of A is not equal to the number of rows of B, AB is undefined.

B is a 4 × 2 matrix and A is a 2 × 3 matrix. Because the number of columns of B is equal to the number of rows of A, BA exists.

To find the first entry of BA, find the sum of the products of the entries in row 1 of B and column 1 of A. Follow the same procedure for row 2 column 1 of BA and the remaining entries.

ANSWER:

AB is undefined; BA = .

Write each system of equations as a matrix equation, AX = B. Then use Gauss-Jordan elimination on the augmented matrix to solve they system.

11.

· =

Write the augmented matrix . Attach the matrix of constant terms to the end of the 3 × 3 matrix.

Use Gauss-Jordan elimination to solve the system. First, use elementary row operations to transform A into I.

The solution of the equation is given by X.

Therefore, the solution is (1, −5, −6). ANSWER:

· = ; (1, −5, −6)

Determine whether A and B are inverse matrices.

20. A =

B =

SOLUTION:

A = ; B =

If A and B are inverse matrices, then AB = BA = I.

Because AB = BA = I, B = A−1 and A = B−1. ANSWER: 6-2 yesMatrix Multiplication, Inverses and Determinants

24. A =

B =

SOLUTION:

A = ;B =

If A and B are inverse matrices, then AB = BA = I.

AB ≠ I, so A and B are not inverses. ANSWER: no

Find A -1, if it exists. If A -1 does not exist, write singular.

28. A =

SOLUTION: Create the doubly augmented matrix .

Apply elementary row operations to write the matrix in reduced row-echelon form.

A row of 0s has been formed, so the first 2 columns cannot become the identity matrix. Therefore, A is singular. ANSWER: singular

Find the determinant of each matrix. Then find the inverse of the matrix, if it exists.

36.

SOLUTION: Find the determinant. eSolutionsdet(AManual) = |A-|Powered = by=Cognero ad − bc Page 5

Because the determinant is not 0, the matrix is invertible. Find the inverse matrix.

Confirm that AA−1 = A−1A = I.

ANSWER:

−23;

39.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

−11;

42.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

12; Find AB and BA, if possible.

8. A =

B =

SOLUTION:

A = ; B =

A is a 2 × 3 matrix and B is a 4 × 2 matrix. Because the number of columns of A is not equal to the number of rows of B, AB is undefined.

B is a 4 × 2 matrix and A is a 2 × 3 matrix. Because the number of columns of B is equal to the number of rows of A, BA exists.

To find the first entry of BA, find the sum of the products of the entries in row 1 of B and column 1 of A. Follow the same procedure for row 2 column 1 of BA and the remaining entries.

ANSWER:

AB is undefined; BA = .

Write each system of equations as a matrix equation, AX = B. Then use Gauss-Jordan elimination on the augmented matrix to solve they system.

11.

· =

Write the augmented matrix . Attach the matrix of constant terms to the end of the 3 × 3 matrix.

Use Gauss-Jordan elimination to solve the system. First, use elementary row operations to transform A into I.

The solution of the equation is given by X.

Therefore, the solution is (1, −5, −6). ANSWER:

· = ; (1, −5, −6)

Determine whether A and B are inverse matrices.

20. A =

B =

SOLUTION:

A = ; B =

If A and B are inverse matrices, then AB = BA = I.

Because AB = BA = I, B = A−1 and A = B−1. ANSWER: yes

24. A =

B =

SOLUTION:

A = ;B =

If A and B are inverse matrices, then AB = BA = I.

AB ≠ I, so A and B are not inverses. ANSWER: no

Find A -1, if it exists. If A -1 does not exist, write singular.

28. A =

SOLUTION: Create the doubly augmented matrix .

Apply elementary row operations to write the matrix in reduced row-echelon form.

A row of 0s has been formed, so the first 2 columns cannot become the identity matrix. Therefore, A is singular. ANSWER: singular

Find the determinant of each matrix. Then find the inverse of the matrix, if it exists.

36.

6-2 SOLUTION: Matrix Multiplication, Inverses and Determinants Find the determinant.

det(A) = |A| = = ad − bc

Because the determinant is not 0, the matrix is invertible. Find the inverse matrix.

Confirm that AA−1 = A−1A = I.

eSolutions Manual - Powered by Cognero Page 6

ANSWER:

−23;

39.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

−11;

42.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

12; Find AB and BA, if possible.

8. A =

B =

SOLUTION:

A = ; B =

A is a 2 × 3 matrix and B is a 4 × 2 matrix. Because the number of columns of A is not equal to the number of rows of B, AB is undefined.

B is a 4 × 2 matrix and A is a 2 × 3 matrix. Because the number of columns of B is equal to the number of rows of A, BA exists.

To find the first entry of BA, find the sum of the products of the entries in row 1 of B and column 1 of A. Follow the same procedure for row 2 column 1 of BA and the remaining entries.

ANSWER:

AB is undefined; BA = .

Write each system of equations as a matrix equation, AX = B. Then use Gauss-Jordan elimination on the augmented matrix to solve they system.

11.

· =

Write the augmented matrix . Attach the matrix of constant terms to the end of the 3 × 3 matrix.

Use Gauss-Jordan elimination to solve the system. First, use elementary row operations to transform A into I.

The solution of the equation is given by X.

Therefore, the solution is (1, −5, −6). ANSWER:

· = ; (1, −5, −6)

Determine whether A and B are inverse matrices.

20. A =

B =

SOLUTION:

A = ; B =

If A and B are inverse matrices, then AB = BA = I.

Because AB = BA = I, B = A−1 and A = B−1. ANSWER: yes

24. A =

B =

SOLUTION:

A = ;B =

If A and B are inverse matrices, then AB = BA = I.

AB ≠ I, so A and B are not inverses. ANSWER: no

Find A -1, if it exists. If A -1 does not exist, write singular.

28. A =

SOLUTION: Create the doubly augmented matrix .

Apply elementary row operations to write the matrix in reduced row-echelon form.

A row of 0s has been formed, so the first 2 columns cannot become the identity matrix. Therefore, A is singular. ANSWER: singular

Find the determinant of each matrix. Then find the inverse of the matrix, if it exists.

36.

SOLUTION: Find the determinant.

det(A) = |A| = = ad − bc

Because the determinant is not 0, the matrix is invertible. Find the inverse matrix.

Confirm that AA−1 = A−1A = I.

6-2 Matrix Multiplication, Inverses and Determinants

ANSWER:

−23;

39.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

eSolutions Manual - Powered by Cognero Page 7

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

−11;

42.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

12; Find AB and BA, if possible.

8. A =

B =

SOLUTION:

A = ; B =

A is a 2 × 3 matrix and B is a 4 × 2 matrix. Because the number of columns of A is not equal to the number of rows of B, AB is undefined.

B is a 4 × 2 matrix and A is a 2 × 3 matrix. Because the number of columns of B is equal to the number of rows of A, BA exists.

To find the first entry of BA, find the sum of the products of the entries in row 1 of B and column 1 of A. Follow the same procedure for row 2 column 1 of BA and the remaining entries.

ANSWER:

AB is undefined; BA = .

Write each system of equations as a matrix equation, AX = B. Then use Gauss-Jordan elimination on the augmented matrix to solve they system.

11.

· =

Write the augmented matrix . Attach the matrix of constant terms to the end of the 3 × 3 matrix.

Use Gauss-Jordan elimination to solve the system. First, use elementary row operations to transform A into I.

The solution of the equation is given by X.

Therefore, the solution is (1, −5, −6). ANSWER:

· = ; (1, −5, −6)

Determine whether A and B are inverse matrices.

20. A =

B =

SOLUTION:

A = ; B =

If A and B are inverse matrices, then AB = BA = I.

Because AB = BA = I, B = A−1 and A = B−1. ANSWER: yes

24. A =

B =

SOLUTION:

A = ;B =

If A and B are inverse matrices, then AB = BA = I.

AB ≠ I, so A and B are not inverses. ANSWER: no

Find A -1, if it exists. If A -1 does not exist, write singular.

28. A =

SOLUTION: Create the doubly augmented matrix .

Apply elementary row operations to write the matrix in reduced row-echelon form.

A row of 0s has been formed, so the first 2 columns cannot become the identity matrix. Therefore, A is singular. ANSWER: singular

Find the determinant of each matrix. Then find the inverse of the matrix, if it exists.

36.

SOLUTION: Find the determinant.

det(A) = |A| = = ad − bc

Because the determinant is not 0, the matrix is invertible. Find the inverse matrix.

Confirm that AA−1 = A−1A = I.

ANSWER:

−23; 6-2 Matrix Multiplication, Inverses and Determinants

39.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

−11;

42. eSolutions Manual - Powered by Cognero Page 8

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

12; Find AB and BA, if possible.

8. A =

B =

SOLUTION:

A = ; B =

A is a 2 × 3 matrix and B is a 4 × 2 matrix. Because the number of columns of A is not equal to the number of rows of B, AB is undefined.

B is a 4 × 2 matrix and A is a 2 × 3 matrix. Because the number of columns of B is equal to the number of rows of A, BA exists.

To find the first entry of BA, find the sum of the products of the entries in row 1 of B and column 1 of A. Follow the same procedure for row 2 column 1 of BA and the remaining entries.

ANSWER:

AB is undefined; BA = .

Write each system of equations as a matrix equation, AX = B. Then use Gauss-Jordan elimination on the augmented matrix to solve they system.

11.

· =

Write the augmented matrix . Attach the matrix of constant terms to the end of the 3 × 3 matrix.

Use Gauss-Jordan elimination to solve the system. First, use elementary row operations to transform A into I.

The solution of the equation is given by X.

Therefore, the solution is (1, −5, −6). ANSWER:

· = ; (1, −5, −6)

Determine whether A and B are inverse matrices.

20. A =

B =

SOLUTION:

A = ; B =

If A and B are inverse matrices, then AB = BA = I.

Because AB = BA = I, B = A−1 and A = B−1. ANSWER: yes

24. A =

B =

SOLUTION:

A = ;B =

If A and B are inverse matrices, then AB = BA = I.

AB ≠ I, so A and B are not inverses. ANSWER: no

Find A -1, if it exists. If A -1 does not exist, write singular.

28. A =

SOLUTION: Create the doubly augmented matrix .

Apply elementary row operations to write the matrix in reduced row-echelon form.

A row of 0s has been formed, so the first 2 columns cannot become the identity matrix. Therefore, A is singular. ANSWER: singular

Find the determinant of each matrix. Then find the inverse of the matrix, if it exists.

36.

SOLUTION: Find the determinant.

det(A) = |A| = = ad − bc

Because the determinant is not 0, the matrix is invertible. Find the inverse matrix.

Confirm that AA−1 = A−1A = I.

ANSWER:

−23;

39.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

−11; 6-2 Matrix Multiplication, Inverses and Determinants

42.

SOLUTION: Find the determinant.

det(A) = |A| = =

Because the determinant is not 0, the matrix is invertible. Use a graphing calculator to find the inverse matrix.

Use the Frac feature under the MATH menu to write the inverse using fractions.

Therefore, A – 1 = .

ANSWER:

12;