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Exercises 10.6 Exercises
Cofactor and adjoint matrices.
For each matrix:
Compute the associated cofactor matrix.
Compute the associated adjoint matrix.
Compute the product of the matrix and its adjoint.
1.
\(\displaystyle
\begin{abmatrix}{rrr}
-3 \amp 0 \amp 5 \\
-3 \amp 0 \amp 2 \\
5 \amp -1 \amp 5
\end{abmatrix}\)
Answer .
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
2 \amp 25 \amp 3 \\
-5 \amp -40 \amp -3 \\
0 \amp -9 \amp 0
\end{abmatrix}\)
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
2 \amp -5 \amp 0 \\
25 \amp -40 \amp -9 \\
3 \amp -3 \amp 0
\end{abmatrix}\)
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
-3 \amp 0 \amp 5 \\
-3 \amp 0 \amp 2 \\
5 \amp -1 \amp 5
\end{abmatrix}
\begin{abmatrix}{rrr}
2 \amp -5 \amp 0 \\
25 \amp -40 \amp -9 \\
3 \amp -3 \amp 0
\end{abmatrix}
=
\begin{bmatrix}
9 \amp 0 \amp 0 \\
0 \amp 9 \amp 0 \\
0 \amp 0 \amp 9
\end{bmatrix}\)
The determinant of the original matrix is \(9\text{,}\) so the matrix is invertible with inverse
\begin{equation*}
\frac{1}{9} \,
\begin{abmatrix}{rrr}
2 \amp -5 \amp 0 \\
25 \amp -40 \amp -9 \\
3 \amp -3 \amp 0
\end{abmatrix}\text{.}
\end{equation*}
2.
\(\displaystyle
\begin{abmatrix}{rrr}
-3 \amp 5 \amp 0 \\
-4 \amp 2 \amp 4 \\
4 \amp 2 \amp 5
\end{abmatrix}\)
Answer .
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
2 \amp 36 \amp -16 \\
-25 \amp -15 \amp 26 \\
20 \amp 12 \amp 14
\end{abmatrix}\)
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
2 \amp -25 \amp 20 \\
36 \amp -15 \amp 12 \\
-16 \amp 26 \amp 14
\end{abmatrix}\)
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
-3 \amp 5 \amp 0 \\
-4 \amp 2 \amp 4 \\
4 \amp 2 \amp 5
\end{abmatrix}
\begin{abmatrix}{rrr}
2 \amp -25 \amp 20 \\
36 \amp -15 \amp 12 \\
-16 \amp 26 \amp 14
\end{abmatrix}
=
\begin{bmatrix}
174 \amp 0 \amp 0 \\
0 \amp 174 \amp 0 \\
0 \amp 0 \amp 174
\end{bmatrix}\)
The determinant of the original matrix is \(174\text{,}\) so the matrix is invertible with inverse
\begin{equation*}
\frac{1}{174} \,
\begin{abmatrix}{rrr}
2 \amp -25 \amp 20 \\
36 \amp -15 \amp 12 \\
-16 \amp 26 \amp 14
\end{abmatrix}\text{.}
\end{equation*}
3.
\(\displaystyle
\begin{abmatrix}{rrr}
5 \amp -3 \amp 8 \\
1 \amp 3 \amp -2 \\
-5 \amp 4 \amp -9
\end{abmatrix}\)
Answer .
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
-19 \amp 19 \amp 19 \\
5 \amp -5 \amp -5 \\
-18 \amp 18 \amp 18
\end{abmatrix}\)
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
-19 \amp 5 \amp -18 \\
19 \amp -5 \amp 18 \\
19 \amp -5 \amp 18
\end{abmatrix}\)
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
5 \amp -3 \amp 8 \\
1 \amp 3 \amp -2 \\
-5 \amp 4 \amp -9
\end{abmatrix}
\begin{abmatrix}{rrr}
-19 \amp 5 \amp -18 \\
19 \amp -5 \amp 18 \\
19 \amp -5 \amp 18
\end{abmatrix}
=
\zerovec\)
The determinant of the original matrix is
\(0\text{,}\) so it is not invertible. Without computing the determinant, we can also see that the adjoint is not invertible because its RREF would have two rows of zeros. As an invertible matrix would have an invertible adjoint, the original matrix cannot be invertible if its adjoint is not invertible.
4.
\(\displaystyle
\begin{abmatrix}{rrr}
-5 \amp 2 \amp -3 \\
0 \amp 5 \amp -2 \\
0 \amp 0 \amp 4
\end{abmatrix}\)
Answer .
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
20 \amp 0 \amp 0 \\
-8 \amp -20 \amp 0 \\
11 \amp -10 \amp -25
\end{abmatrix}\)
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
20 \amp - 8 \amp 11 \\
0 \amp -20 \amp -10 \\
0 \amp 0 \amp -25
\end{abmatrix}\)
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
-5 \amp 2 \amp -3 \\
0 \amp 5 \amp -2 \\
0 \amp 0 \amp 4
\end{abmatrix}
\begin{abmatrix}{rrr}
20 \amp - 8 \amp 11 \\
0 \amp -20 \amp -10 \\
0 \amp 0 \amp -25
\end{abmatrix}
=
\begin{bmatrix}
-100 \amp 0 \amp 0 \\
0 \amp -100 \amp 0 \\
0 \amp 0 \amp -100
\end{bmatrix}\)
The determinant of the original matrix is \(-100\text{,}\) so the matrix is invertible with inverse
\begin{equation*}
- \frac{1}{100} \,
\begin{abmatrix}{rrr}
20 \amp - 8 \amp 11 \\
0 \amp -20 \amp -10 \\
0 \amp 0 \amp -25
\end{abmatrix}\text{.}
\end{equation*}
5.
\(\displaystyle
\begin{abmatrix}{rrr}
1 \amp 0 \amp 0 \\
-5 \amp -4 \amp 0 \\
3 \amp 1 \amp -5
\end{abmatrix}\)
Answer .
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
20 \amp -25 \amp 7 \\
0 \amp - 5 \amp -1 \\
0 \amp 0 \amp -4
\end{abmatrix}\)
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
20 \amp 0 \amp 0 \\
-25 \amp -5 \amp 0 \\
7 \amp -1 \amp -4
\end{abmatrix}\)
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
1 \amp 0 \amp 0 \\
-5 \amp -4 \amp 0 \\
3 \amp 1 \amp -5
\end{abmatrix}
\begin{abmatrix}{rrr}
20 \amp 0 \amp 0 \\
-25 \amp -5 \amp 0 \\
7 \amp -1 \amp -4
\end{abmatrix}
=
\begin{bmatrix}
20 \amp 0 \amp 0 \\
0 \amp 20 \amp 0 \\
0 \amp 0 \amp 20
\end{bmatrix}\)
The determinant of the original matrix is \(20\text{,}\) so the matrix is invertible with inverse
\begin{equation*}
\frac{1}{20} \,
\begin{abmatrix}{rrr}
20 \amp 0 \amp 0 \\
-25 \amp -5 \amp 0 \\
7 \amp -1 \amp -4
\end{abmatrix}\text{.}
\end{equation*}
6.
\(\displaystyle
\begin{abmatrix}{rrr}
3 \amp 0 \amp -1 \\
0 \amp 5 \amp 4 \\
-1 \amp 4 \amp 2
\end{abmatrix}\)
Answer .
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
-6 \amp -4 \amp 5 \\
-4 \amp 5 \amp -12 \\
5 \amp -12 \amp 15
\end{abmatrix}\)
Notice that the matrix of cofactors is symmetric, hence the adjoint matrix is equal to the matrix of cofactors in this case.
\(\displaystyle \displaystyle
\begin{abmatrix}{rrr}
3 \amp 0 \amp -1 \\
0 \amp 5 \amp 4 \\
-1 \amp 4 \amp 2
\end{abmatrix}
\begin{abmatrix}{rrr}
-6 \amp -4 \amp 5 \\
-4 \amp 5 \amp -12 \\
5 \amp -12 \amp 15
\end{abmatrix}
=
\begin{bmatrix}
-23 \amp 0 \amp 0 \\
0 \amp -23 \amp 0 \\
0 \amp 0 \amp -23
\end{bmatrix}\)
The determinant of the original matrix is \(-23\text{,}\) so the matrix is invertible with inverse
\begin{equation*}
- \frac{1}{23} \,
\begin{abmatrix}{rrr}
-6 \amp -4 \amp 5 \\
-4 \amp 5 \amp -12 \\
5 \amp -12 \amp 15
\end{abmatrix}\text{.}
\end{equation*}
7.
\(\displaystyle
\begin{abmatrix}{rrrr}
0 \amp 0 \amp 0 \amp 1 \\
1 \amp 1 \amp 0 \amp 2 \\
-2 \amp -1 \amp 0 \amp 5 \\
0 \amp 0 \amp 1 \amp -4
\end{abmatrix}\)
Answer .
\(\displaystyle \displaystyle
\begin{abmatrix}{rrrr}
-7 \amp 9 \amp -4 \amp -1 \\
1 \amp -2 \amp 0 \amp 0 \\
1 \amp -1 \amp 0 \amp 0 \\
0 \amp 0 \amp -1 \amp 0
\end{abmatrix}\)
\(\displaystyle \displaystyle
\begin{abmatrix}{rrrr}
-7 \amp 1 \amp 1 \amp 0 \\
9 \amp -2 \amp -1 \amp 0 \\
-4 \amp 0 \amp 0 \amp -1 \\
-1 \amp 0 \amp 0 \amp 0
\end{abmatrix}\)
\(\displaystyle \displaystyle
\begin{abmatrix}{rrrr}
0 \amp 0 \amp 0 \amp 1 \\
1 \amp 1 \amp 0 \amp 2 \\
-2 \amp -1 \amp 0 \amp 5 \\
0 \amp 0 \amp 1 \amp -4
\end{abmatrix}
\begin{abmatrix}{rrrr}
-7 \amp 1 \amp 1 \amp 0 \\
9 \amp -2 \amp -1 \amp 0 \\
-4 \amp 0 \amp 0 \amp -1 \\
-1 \amp 0 \amp 0 \amp 0
\end{abmatrix}
=
\begin{abmatrix}{rrrr}
-1 \amp 0 \amp 0 \amp 0 \\
0 \amp -1 \amp 0 \amp 0 \\
0 \amp 0 \amp -1 \amp 0 \\
0 \amp 0 \amp 0 \amp -1
\end{abmatrix}\)
The determinant of the original matrix is \(-1\text{,}\) so the matrix is invertible with inverse
\begin{equation*}
\begin{abmatrix}{rrrr}
7 \amp -1 \amp -1 \amp 0 \\
-9 \amp 2 \amp 1 \amp 0 \\
4 \amp 0 \amp 0 \amp 1 \\
1 \amp 0 \amp 0 \amp 0
\end{abmatrix}\text{.}
\end{equation*}
8. Adjoints of special matrices.
For each statement, construct a convincing argument that it is true.
(a) The adjoint of a diagonal matrix is always diagonal.
(b) The adjoint of an upper triangular matrix is always upper triangular.
(c) The adjoint of an lower triangular matrix is always lower triangular.
(d) The adjoint of a symmetric matrix is always symmetric.
Inverses of matrices with variable entries.
For each matrix:
State a condition on the variable \(x\) which must be satisfied in order for the matrix to be invertible.
9.
\(\displaystyle
\begin{bmatrix}
x + 1 \amp x - 1 \\
x \amp 0
\end{bmatrix}\)
Answer .
Invertibility requires
\(x (1 - x) \neq 0\text{.}\) That is, it requires
\(x \neq 0, 1\text{.}\)
In the case \(x \neq 0,1\text{,}\) the inverse matrix is
\begin{equation*}
\frac{1}{x (1 - x)} \,
\begin{bmatrix}
0 \amp 1 - x \\
-x \amp x + 1
\end{bmatrix}\text{.}
\end{equation*}
10.
\(\displaystyle
\begin{bmatrix}
x \amp 1 \amp x \\
x + 1 \amp x \amp x + 1 \\
x + 2 \amp 1 \amp x
\end{bmatrix}\)
Answer .
Invertibility requires \(x^2 - x - 1 \neq 0\text{.}\) That is, it requires
\begin{equation*}
x \neq \frac{1 \pm \sqrt{5}}{2} \text{.}
\end{equation*}
In the case \(x^2 - x - 1 \neq 0\text{,}\) the inverse matrix is
\begin{equation*}
\frac{-1}{2 (x^2 - x - 1)}
\begin{bmatrix}
x^2 - x - 1 \amp 0 \amp -x^2 + x + 1 \\
2 x + 2 \amp -2 x \amp 0 \\
- x^2 - x + 1 \amp 2 \amp x^2 - x - 1
\end{bmatrix}\text{.}
\end{equation*}
11.
\(\displaystyle
\begin{bmatrix}
2 x + 1 \amp 1 \amp x - 2 \amp x \\
x + 4 \amp x + 2 \amp 0 \amp -2 x + 1 \\
x + 1 \amp 1 \amp 2 x - 3 \amp 0 \\
x \amp 0 \amp 0 \amp x
\end{bmatrix}\)
Answer .
Invertibility requires
\(x (x + 1) {(x - 1)}^2 \neq 0\text{.}\) That is, it requires
\(x \neq 0, \pm 1\text{.}\)
In the case \(x \neq 0, \pm 1\text{,}\) the inverse matrix is
\begin{equation*}
\begin{bmatrix}
\frac{(x + 2) (2 x - 3)}{(x + 1) {(x - 1)}^2} \amp
- \frac{1}{(x + 1) (x - 1)} \amp
- \frac{x^2 - 4}{(x + 1) {(x - 1)}^2} \amp
- \frac{2 x^3 + 3 x^2 - 9 x + 1}{x (x + 1) {(x - 1)}^2}
\\
- \frac{3 (2 x - 3)}{{(x - 1)}^2} \amp
\frac{1}{x - 1} \amp
\frac{3 (x - 2)}{{(x - 1)}^2} \amp
\frac{8 x^2 - 12 x + 1}{x {(x - 1)}^2}
\\
- \frac{1}{x - 1} \amp
0 \amp
\frac{1}{x - 1} \amp
\frac{1}{x - 1}
\\
- \frac{(x + 2) (2 x - 3)}{(x + 1) {(x - 1)}^2} \amp
\frac{1}{(x + 1) (x - 1)} \amp
\frac{x^2 - 4}{(x + 1) {(x - 1)}^2} \amp
\frac{3 x^3 + 2 x^2 - 10 x + 2}{x (x + 1) {(x - 1)}^2}
\end{bmatrix}\text{.}
\end{equation*}
Cramer’s rule for \(2\) -variable systems.
12.
\(\displaystyle
\begin{sysofeqns}{rcrcr}
5 x \amp - \amp 12 y \amp = \amp 1 \\
2 x \amp - \amp 5 y \amp = \amp 1
\end{sysofeqns}
\text{.}\)
Solution .
Set
\begin{align*}
A \amp =
\begin{abmatrix}{rr}
5 \amp -12 \\
2 \amp - 5
\end{abmatrix}
\text{,}
\amp
\uvec{b} \amp =
\begin{bmatrix} 1 \\ 1 \end{bmatrix}
\text{.}
\end{align*}
Compute \(\det A = -1\text{.}\) Since \(\det A \neq 0\text{,}\) Cramer’s rule can be applied. Replace one column of \(A\) by \(\uvec{b}\) and a time and compute determinants.
\begin{align*}
A_x \amp =
\begin{abmatrix}{rr}
1 \amp -12 \\
1 \amp - 5
\end{abmatrix}
\amp
\det A_x \amp = 7\\
A_y \amp =
\begin{bmatrix}
5 \amp 1 \\
2 \amp 1
\end{bmatrix}
\amp
\det A_y \amp = 3
\end{align*}
Finally, apply Cramer’s rule:
\begin{align*}
x \amp = \frac{\det A_x}{\det A} \amp
y \amp = \frac{\det A_y}{\det A}\\
\amp = \frac{7}{-1} \amp
\amp = \frac{3}{-1}\\
\amp = - 7 \text{,} \amp
\amp = - 3 \text{.}
\end{align*}
13.
\(\displaystyle
\begin{sysofeqns}{rcrcr}
8 x \amp + \amp 7 y \amp = \amp 3 \\
7 x \amp + \amp 7 y \amp = \amp 1
\end{sysofeqns}
\text{.}\)
Solution .
Set
\begin{align*}
A \amp =
\begin{bmatrix}
8 \amp 7 \\
7 \amp 7
\end{bmatrix}
\text{,}
\amp
\uvec{b} \amp =
\begin{bmatrix} 3 \\ 1 \end{bmatrix}
\text{.}
\end{align*}
Compute \(\det A = 7\text{.}\) Since \(\det A \neq 0\text{,}\) Cramer’s rule can be applied. Replace one column of \(A\) by \(\uvec{b}\) and a time and compute determinants.
\begin{align*}
A_x \amp =
\begin{bmatrix}
3 \amp 7 \\
1 \amp 7
\end{bmatrix}
\amp
\det A_x \amp = 14\\
A_y \amp =
\begin{bmatrix}
8 \amp 3 \\
7 \amp 1
\end{bmatrix}
\amp
\det A_y \amp = -13
\end{align*}
Finally, apply Cramer’s rule:
\begin{align*}
x \amp = \frac{\det A_x}{\det A} \amp
y \amp = \frac{\det A_y}{\det A}\\
\amp = \frac{14}{7} \amp
\amp = - \frac{13}{7} \text{.}\\
\amp = 2 \text{,}
\end{align*}
14.
\(\displaystyle
\begin{sysofeqns}{rcrcr}
6 s \amp + \amp 6 t \amp = \amp 5 \\
7 s \amp + \amp 8 t \amp = \amp -8
\end{sysofeqns}
\text{.}\)
Solution .
Set
\begin{align*}
A \amp =
\begin{bmatrix}
6 \amp 6 \\
7 \amp 8
\end{bmatrix}
\text{,}
\amp
\uvec{b} \amp =
\begin{abmatrix}{r} 5 \\ -8 \end{abmatrix}
\text{.}
\end{align*}
Compute \(\det A = 6\text{.}\) Since \(\det A \neq 0\text{,}\) Cramer’s rule can be applied. Replace one column of \(A\) by \(\uvec{b}\) and a time and compute determinants.
\begin{align*}
A_s \amp =
\begin{abmatrix}{rr}
5 \amp 6 \\
-8 \amp 8
\end{abmatrix}
\amp
\det A_s \amp = 88\\
A_t \amp =
\begin{abmatrix}{rr}
6 \amp 5 \\
7 \amp -8
\end{abmatrix}
\amp
\det A_t \amp = -83
\end{align*}
Finally, apply Cramer’s rule:
\begin{align*}
s \amp = \frac{\det A_s}{\det A} \amp
t \amp = \frac{\det A_t}{\det A}\\
\amp = \frac{88}{6} \amp
\amp = - \frac{83}{6} \text{.}\\
\amp = \frac{44}{3} \text{,}
\end{align*}
15.
\(\displaystyle
\begin{sysofeqns}{rcrcr}
6 \alpha \amp - \amp 9 \beta \amp = \amp -6 \\
8 \alpha \amp + \amp 7 \beta \amp = \amp -4
\end{sysofeqns}
\text{.}\)
Solution .
Set
\begin{align*}
A \amp =
\begin{abmatrix}{rrr}
6 \amp -9 \\
8 \amp 7
\end{abmatrix}
\text{,}
\amp
\uvec{b} \amp =
\begin{abmatrix}{r} -6 \\ -4 \end{abmatrix}
\text{.}
\end{align*}
Compute \(\det A = 114\text{.}\) Since \(\det A \neq 0\text{,}\) Cramer’s rule can be applied. Replace one column of \(A\) by \(\uvec{b}\) and a time and compute determinants.
\begin{align*}
A_{\alpha} \amp =
\begin{abmatrix}{rr}
-6 \amp -9 \\
-4 \amp 7
\end{abmatrix}
\amp
\det A_{\alpha} \amp = -78\\
A_{\beta} \amp =
\begin{abmatrix}{rrr}
6 \amp -6 \\
8 \amp -4
\end{abmatrix}
\amp
\det A_{\beta} \amp = 24
\end{align*}
Finally, apply Cramer’s rule:
\begin{align*}
\alpha \amp = \frac{\det A_{\alpha}}{\det A} \amp
\beta \amp = \frac{\det A_{\beta}}{\det A}\\
\amp = - \frac{78}{114} \amp
\amp = \frac{24}{114}\\
\amp = - \frac{13}{19} \text{,} \amp
\amp = \frac{4}{19} \text{.}
\end{align*}
Cramer’s rule for larger systems.
16. Variable
\(x\) in
\(\displaystyle
\begin{sysofeqns}{rcrcrcr}
3 x \amp \amp 2 y \amp - \amp 9 z \amp = \amp 6 \\
-2 x \amp \amp \amp + \amp 3 z \amp = \amp 0 \\
x \amp + \amp y \amp - \amp 4 z \amp = \amp 9
\end{sysofeqns}
\text{.}\)
Solution .
Set
\begin{align*}
A \amp =
\begin{abmatrix}{rrr}
3 \amp 2 \amp -9 \\
-2 \amp 0 \amp 3 \\
1 \amp 1 \amp -4
\end{abmatrix}
\text{,}
\amp
\uvec{b} \amp =
\begin{abmatrix}{r} 6 \\ 0 \\ 9 \end{abmatrix}
\text{.}
\end{align*}
Compute \(\det A = -1\text{.}\) Since \(\det A \neq 0\text{,}\) Cramer’s rule can be applied. Targeting variable \(x\text{,}\) replace the first column in \(A\) with column \(\uvec{b}\) to obtain
\begin{equation*}
A_x =
\begin{abmatrix}{rrr}
6 \amp 2 \amp -9 \\
0 \amp 0 \amp 3 \\
9 \amp 1 \amp -4
\end{abmatrix}\text{.}
\end{equation*}
Now compute \(\det A_x = 36\text{.}\) Therefore, the one unique solution to the system includes value
\begin{equation*}
x = \frac{\det A_x}{\det A} = \frac{36}{-1} = -36 \text{.}
\end{equation*}
17. Variable
\(z\) in
\(\displaystyle
\begin{sysofeqns}{rcrcrcr}
7 x \amp \amp \amp - \amp z \amp = \amp 6 \\
-3 x \amp + \amp 3 y \amp + \amp z \amp = \amp 9 \\
-4 x \amp - \amp 1 y \amp + \amp z \amp = \amp -11
\end{sysofeqns}
\text{.}\)
Solution .
Set
\begin{align*}
A \amp =
\begin{abmatrix}{rrr}
7 \amp 0 \amp -1 \\
-3 \amp 3 \amp 1 \\
-4 \amp -1 \amp 1
\end{abmatrix}
\text{,}
\amp
\uvec{b} \amp =
\begin{abmatrix}{r} 6 \\ 9 \\ -11 \end{abmatrix}
\text{.}
\end{align*}
Compute \(\det A = 13\text{.}\) Since \(\det A \neq 0\text{,}\) Cramer’s rule can be applied. Targeting variable \(z\text{,}\) replace the first third in \(A\) with column \(\uvec{b}\) to obtain
\begin{equation*}
A_z =
\begin{abmatrix}{rrr}
7 \amp 0 \amp 6 \\
-3 \amp 3 \amp 9 \\
-4 \amp -1 \amp -11
\end{abmatrix}\text{.}
\end{equation*}
Now compute \(\det A_z = -78\text{.}\) Therefore, the one unique solution to the system includes value
\begin{equation*}
z = \frac{\det A_z}{\det A} = \frac{-78}{13} = -6 \text{.}
\end{equation*}
18. Variable
\(x_2\) in
\(\displaystyle
\begin{sysofeqns}{rcrcrcrcr}
x_1 \amp + \amp x_2 \amp \amp \amp - \amp 2 x_4 \amp = \amp -4 \\
2 x_1 \amp + \amp 3 x_2 \amp \amp \amp - \amp 6 x_4 \amp = \amp -1 \\
2 x_1 \amp + \amp 5 x_2 \amp + \amp x_3 \amp - \amp 11 x_4 \amp = \amp -5 \\
2 x_1 \amp + \amp 3 x_2 \amp \amp \amp - \amp 5 x_4 \amp = \amp 4
\end{sysofeqns}
\text{.}\)
Solution .
Set
\begin{align*}
A \amp =
\begin{abmatrix}{rrrr}
1 \amp 1 \amp 0 \amp -2 \\
2 \amp 3 \amp 0 \amp -6 \\
2 \amp 5 \amp 1 \amp -11 \\
2 \amp 3 \amp 0 \amp -5
\end{abmatrix}
\text{,}
\amp
\uvec{b} \amp =
\begin{abmatrix}{r} -4 \\ -1 \\ -5 \\ 4 \end{abmatrix}
\text{.}
\end{align*}
Compute \(\det A = 1\text{.}\) Since \(\det A \neq 0\text{,}\) Cramer’s rule can be applied. Targeting variable \(x_2\text{,}\) replace the second column in \(A\) with column \(\uvec{b}\) to obtain
\begin{equation*}
A_2 =
\begin{abmatrix}{rrrr}
1 \amp -4 \amp 0 \amp -2 \\
2 \amp -1 \amp 0 \amp -6 \\
2 \amp -5 \amp 1 \amp -11 \\
2 \amp 4 \amp 0 \amp -5
\end{abmatrix}\text{.}
\end{equation*}
Now compute \(\det A_2 = 17\text{.}\) Therefore, the one unique solution to the system includes value
\begin{equation*}
x_2 = \frac{\det A_2}{\det A} = \frac{17}{1} = 17 \text{.}
\end{equation*}
19. Variable
\(y\) in
\(\displaystyle
\begin{sysofeqns}{rcrcrcrcr}
3 w \amp + \amp 3 x \amp \amp \amp \amp \amp = \amp 1 \\
\amp + \amp 2 x \amp - \amp 2 y \amp \amp \amp = \amp -4 \\
- w \amp + \amp 1 x \amp - \amp y \amp - \amp 3 z \amp = \amp 2 \\
3 w \amp - \amp 3 x \amp \amp \amp - \amp 2 z \amp = \amp 0
\end{sysofeqns}
\text{.}\)
Solution .
Set
\begin{align*}
A \amp =
\begin{abmatrix}{rrrr}
3 \amp 3 \amp 0 \amp 0 \\
0 \amp 2 \amp -2 \amp 0 \\
-1 \amp 1 \amp -1 \amp -3 \\
3 \amp -3 \amp 0 \amp -2
\end{abmatrix}
\text{,}
\amp
\uvec{b} \amp =
\begin{abmatrix}{r} 1 \\ -4 \\ 2 \\ 0 \end{abmatrix}
\text{.}
\end{align*}
Compute \(\det A = -120\text{.}\) Since \(\det A \neq 0\text{,}\) Cramer’s rule can be applied. Targeting variable \(y\text{,}\) replace the third column in \(A\) with column \(\uvec{b}\) to obtain
\begin{equation*}
A_y =
\begin{abmatrix}{rrrr}
3 \amp 3 \amp 1 \amp 0 \\
0 \amp 2 \amp -4 \amp 0 \\
-1 \amp 1 \amp 2 \amp -3 \\
3 \amp -3 \amp 0 \amp -2
\end{abmatrix}\text{.}
\end{equation*}
Now compute \(\det A_y = -310\text{.}\) Therefore, the one unique solution to the system includes value
\begin{equation*}
y = \frac{\det A_y}{\det A} = \frac{-310}{-120} = \frac{31}{12} \text{.}
\end{equation*}
Determinants of products.
For each collection of matrices:
Compute the determinant of each matrix individually.
Compute the product of the matrices and the determinant of the result.
Verify that the determinant of the product is the product of the determinants.
20.
\(\displaystyle
A =
\begin{abmatrix}{rr}
7 \amp 0 \\
7 \amp -2
\end{abmatrix}
\text{,}
\quad
B =
\begin{abmatrix}{rr}
2 \amp 6 \\
4 \amp 9
\end{abmatrix}\)
Answer .
\(\det A = -14\text{,}\) \(\det B = -6\text{.}\)
\(\displaystyle
A B =
\begin{bmatrix}
14 \amp 42 \\
6 \amp 24
\end{bmatrix}\)
\(\displaystyle (\det A) (\det B) = (-14) (-6) = 84 = \det (A B) \)
21.
\(\displaystyle
A =
\begin{abmatrix}{rr}
1 \amp 4 \\
-4 \amp 2
\end{abmatrix}
\text{,}
\quad
B =
\begin{abmatrix}{rr}
-1 \amp 8 \\
3 \amp -8
\end{abmatrix}
\text{,}
\quad
C =
\begin{abmatrix}{rr}
-2 \amp -2 \\
-5 \amp 7
\end{abmatrix}\)
Answer .
\(\det A = 18\text{,}\) \(\det B = -16\text{,}\) \(\det C = -24\text{.}\)
\(\displaystyle
A B C =
\begin{abmatrix}{rr}
98 \amp -190 \\
220 \amp -356
\end{abmatrix}\)
\(\displaystyle (\det A) (\det B) (\det C) = (18) (-16) (-24) = 6912 = \det (A B C) \)
22.
\(\displaystyle
A =
\begin{abmatrix}{rrr}
-1 \amp 0 \amp -6 \\
0 \amp -6 \amp -6 \\
4 \amp 6 \amp -8
\end{abmatrix}
\text{,}
\quad
B =
\begin{abmatrix}{rrr}
-1 \amp 0 \amp 9 \\
-4 \amp -7 \amp 0 \\
-1 \amp -1 \amp 0
\end{abmatrix}\)
Answer .
\(\det A = -228\text{,}\) \(\det B = -27\text{.}\)
\(\displaystyle
A B =
\begin{abmatrix}{rrr}
7 \amp 6 \amp -9 \\
30 \amp 48 \amp 0 \\
-20 \amp -34 \amp 36
\end{abmatrix}\)
\(\displaystyle (\det A) (\det B) = (-228) (-27) = 6156 = \det (A B) \)
Determinants of inverses.
For each matrix:
Compute its determinant.
Compute its inverse and the determinant of the result. (You may wish to use
Procedure 9.2.2 for the larger matrices.)
Verify that the determinant of the inverse is the reciprocal of the determinant.
23.
\(\displaystyle
A =
\begin{abmatrix}{rr}
1 \amp 2 \\
6 \amp -9
\end{abmatrix}\)
Answer .
\(\displaystyle \det A = -21\)
\(\displaystyle
\inv{A} =
\begin{abmatrix}{rr}
- \frac{1}{21} \amp - \frac{2}{21} \\
- \frac{6}{21} \amp \frac{9}{21}
\end{abmatrix}\)
\(\det (\inv{A}) = - 1 / 21 \)
\(\displaystyle \det (\inv{A}) = 1 / (-21) = 1 / (\det A) \)
24.
\(\displaystyle
A =
\begin{abmatrix}{rrr}
1 \amp 0 \amp 0 \\
-2 \amp 2 \amp 0 \\
-1 \amp -2 \amp -2
\end{abmatrix}\)
Answer .
\(\displaystyle \det A = -4\)
\(\displaystyle
\inv{A} =
\begin{abmatrix}{rrr}
1 \amp 0 \amp 0 \\
1 \amp \frac{1}{2} \amp 0 \\
- \frac{3}{2} \amp - \frac{1}{2} \amp - \frac{1}{2}
\end{abmatrix}\)
\(\det (\inv{A}) = - 1 / 4 \)
\(\displaystyle \det (\inv{A}) = 1 / (-4) = 1 / (\det A) \)
25.
\(\displaystyle
A =
\begin{abmatrix}{rrrr}
1 \amp 1 \amp -2 \amp 0 \\
0 \amp 1 \amp -1 \amp -2 \\
0 \amp -1 \amp 0 \amp -2 \\
1 \amp 2 \amp -2 \amp 0
\end{abmatrix}\)
Answer .
\(\displaystyle \det A = 2\)
\(\displaystyle
\inv{A} =
\begin{abmatrix}{rrrr}
-2 \amp -2 \amp 2 \amp 3 \\
-1 \amp 0 \amp 0 \amp 1 \\
-2 \amp -1 \amp 1 \amp 2 \\
\frac{1}{2} \amp 0 \amp - \frac{1}{2} \amp - \frac{1}{2}
\end{abmatrix}\)
\(\det (\inv{A}) = 1 / 2 \)
\(\displaystyle \det (\inv{A}) = 1 / 2 = 1 / (\det A) \)
26.
From
Proposition 5.5.4 we have a general formula for the inverse of a
\(2 \times 2\) matrix:
\begin{equation*}
A = \begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix}
\quad \Rightarrow \quad
\inv{A} = \frac{1}{a d - b c} \begin{abmatrix}{rr} d \amp -b \\ -c \amp a \end{abmatrix}\text{.}
\end{equation*}
Algebra of the determinant.
Let \(A, B, C\) represent square matrices of the same size where it is known that
\begin{align*}
\det A \amp = -3 \text{,} \amp
\det B \amp = -1 \text{,} \amp
\det C \amp = 4 \text{.}
\end{align*}
Use this information to compute the following.
27. 28. 29. 30. 31. 32. 33.
\(\det (B^2 A B^3 \inv{A})\)
34.
\(\det (\inv{A} C \utrans{C} \inv{C} B)\)
