6-2 Matrix Multiplication, Inverses and Determinants
Find AB and BA, if possible.
1. A =
B=
SOLUTION: A=
;B=
A is a 1 × 2 matrix and B is a 2 × 2 matrix. Because the number of columns of A is equal to the number of rows of
B, AB exists.
To find the first entry of AB, find the sum of the products of the entries in row 1 of A and column 1 of B. Follow the
same procedure for row 2 column 1 of AB.
Because the number of columns of B is not equal to the number of rows of A, BA is undefined.
3. A =
B=
SOLUTION: A=
;B=
A is a 1 × 2 matrix and B is a 2 × 3 matrix. Because the number of columns of A is equal to the number of rows of
B, AB exists.
To find the first entry of AB, find the sum of the products of the entries in row 1 of A and column 1 of B. Follow the
same procedure for row 2 column 1 of AB and the remaining entry.
Because the number of columns of B is not equal to the number of rows of A, BA is undefined.
5. A =
B=
SOLUTION: A=
;B=
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A is a 3 × 1 matrix and B is a 2 × 3 matrix. Because the number of columns of A is not equal to the number of rows
6-2 Matrix Multiplication, Inverses and Determinants
Because the number of columns of B is not equal to the number of rows of A, BA is undefined.
5. A =
B=
SOLUTION: A=
;B=
A is a 3 × 1 matrix and B is a 2 × 3 matrix. Because the number of columns of A is not equal to the number of rows
of B, AB is undefined.
B is a 2 × 3 matrix and A is a 3 × 1 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.
7. A =
B=
SOLUTION: A=
;B=
A is a 2 × 2 matrix and B is a 2 × 3 matrix. Because the number of columns of A is equal to the number of rows of
B, AB exists.
To find the first entry of AB, find the sum of the products of the entries in row 1 of A and column 1 of B. Follow the
same procedure for row 2 column 1 of AB and the remaining entries.
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6-2 Matrix Multiplication, Inverses and Determinants
7. A =
B=
SOLUTION: A=
;B=
A is a 2 × 2 matrix and B is a 2 × 3 matrix. Because the number of columns of A is equal to the number of rows of
B, AB exists.
To find the first entry of AB, find the sum of the products of the entries in row 1 of A and column 1 of B. Follow the
same procedure for row 2 column 1 of AB and the remaining entries.
Because the number of columns of B is not equal to the number of rows of A, BA is undefined.
Determine whether A and B are inverse matrices.
19. A =
B=
SOLUTION: A=
;B=
If A and B are inverse matrices, then AB = BA = I.
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Because AB = BA = I, B = A
Page 3
−1
−1
and A = B .
6-2 Matrix Multiplication, Inverses and Determinants
Because the number of columns of B is not equal to the number of rows of A, BA is undefined.
Determine whether A and B are inverse matrices.
19. A =
B=
SOLUTION: A=
;B=
If A and B are inverse matrices, then AB = BA = I.
Because AB = BA = I, B = A
−1
−1
and A = B .
21. 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.
23. A =
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B=
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6-2 Matrix Multiplication, Inverses and Determinants
AB ≠ I, so A and B are not inverses.
23. A =
B=
SOLUTION: A=
;B =
If A and B are inverse matrices, then AB = BA = I.
Because AB = BA = I, B = A
−1
−1
and A = B .
25. A =
B=
SOLUTION: A=
;B =
If A and B are inverse matrices, then AB = BA = I.
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6-2 Matrix Multiplication,
−1 Inverses
−1 and Determinants
Because AB = BA = I, B = A
and A = B .
25. A =
B=
SOLUTION: A=
;B =
If A and B are inverse matrices, then AB = BA = I.
Because AB = BA = I, B = A
−1
−1
and A = B .
Find A -1, if it exists. If A -1 does not exist, write singular.
27. 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.
29. A =
SOLUTION: Create the doubly augmented matrix
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.
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6-2 A
Matrix
Multiplication,
and Determinants
row of 0s
has been formed, soInverses
the first 2 columns
cannot become the identity matrix. Therefore, A is singular.
29. A =
SOLUTION: Create the doubly augmented matrix
.
Apply elementary row operations to write the matrix in reduced row-echelon form.
The first two columns are the identity matrix. Therefore, A is invertible and A
Confirm that AA
−1
−1
=
.
−1
= A A = I.
31. A =
SOLUTION: eSolutions
Manual
Poweredaugmented
by Cognero
Create
the -doubly
matrix
.
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31. A =
6-2 Matrix Multiplication, Inverses and Determinants
SOLUTION: Create the doubly augmented matrix
.
Apply elementary row operations to write the matrix in reduced row-echelon form.
The first three columns are the identity matrix. Therefore, A is invertible and A
Confirm that AA
−1
−1
=
.
−1
= A A = I.
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The first three columns are the identity matrix. Therefore, A is invertible and A
−1
=
.
6-2 Matrix Multiplication, Inverses and Determinants
Confirm that AA
−1
−1
= A A = I.
33. A =
SOLUTION: Create the doubly augmented matrix
.
Apply elementary row operations to write the matrix in reduced row-echelon form.
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6-2 Matrix Multiplication, Inverses and Determinants
33. 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 3 columns cannot become the identity matrix. Therefore, A is singular.
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