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Exercises 8.6 Exercises
Minors and cofactors.
In each case, calculate the requested minors and cofactors of the given matrix. For the larger matrices in
Exercise 5 and
Exercise 6 , express the requested minors as unevaluated determinant matrices and cofactors in terms of the minor.
1.
\(\displaystyle
\begin{abmatrix}{rr}
-2 \amp 1 \\
0 \amp 4
\end{abmatrix}\)
(a) (b) (c) (d)
Answer .
\(C_{21} = -M_{21} = -1\)
2.
\(\displaystyle
\begin{abmatrix}{rr}
3 \amp -6 \\
8 \amp 7
\end{abmatrix}\)
(a) (b)
Answer .
\(C_{12} = -M_{12} = -8\)
(c) (d) 3.
\(\displaystyle
\begin{abmatrix}{rrr}
0 \amp 5 \amp -3 \\
-2 \amp 2 \amp -6 \\
1 \amp -1 \amp 2
\end{abmatrix}\)
(a)
Answer .
\(\displaystyle
M_{11} =
\begin{avmatrix}{rr}
2 \amp -6 \\
-1 \amp 2
\end{avmatrix}
= -2\)
(b) (c)
Answer .
\(\displaystyle
M_{13} =
\begin{avmatrix}{rr}
-2 \amp 2 \\
1 \amp -1
\end{avmatrix}
= 0\)
(d) (e)
Answer .
\(\displaystyle
M_{31} =
\begin{avmatrix}{rr}
5 \amp -3 \\
2 \amp -6
\end{avmatrix}
= -24\)
(f)
Answer .
\(C_{31} = M_{31} = -24\)
(g)
Answer .
\(\displaystyle
M_{33} =
\begin{avmatrix}{rr}
0 \amp 5 \\
-2 \amp 2
\end{avmatrix}
= 10\)
(h) 4.
\(\displaystyle
\begin{abmatrix}{rrr}
-2 \amp 2 \amp -1 \\
3 \amp -3 \amp -4 \\
-2 \amp -3 \amp -2
\end{abmatrix}\)
(a)
Answer .
\(\displaystyle
M_{12} =
\begin{avmatrix}{rr}
3 \amp -4 \\
-2 \amp -2
\end{avmatrix}
= -14\)
(b)
Answer .
\(C_{12} = -M_{12} = 14\)
(c)
Answer .
\(\displaystyle
M_{21} =
\begin{avmatrix}{rr}
2 \amp -1 \\
-3 \amp -2
\end{avmatrix}
= -7\)
(d) (e)
Answer .
\(\displaystyle
M_{23} =
\begin{avmatrix}{rr}
-2 \amp 2 \\
-2 \amp -3
\end{avmatrix}
= 10\)
(f)
Answer .
\(C_{23} = -M_{23} = -10\)
(g)
Answer .
\(\displaystyle
M_{32} =
\begin{avmatrix}{rr}
-2 \amp -1 \\
3 \amp -4
\end{avmatrix}
= 11\)
(h)
Answer .
\(C_{32} = -M_{32} = -11\)
5.
\(\displaystyle
\begin{abmatrix}{rrrr}
4 \amp -3 \amp 5 \amp 6 \\
-1 \amp -6 \amp 2 \amp 2 \\
3 \amp 0 \amp 0 \amp 5 \\
4 \amp 6 \amp -3 \amp 0
\end{abmatrix}\)
(a)
Answer .
\(\displaystyle
M_{22} =
\begin{avmatrix}{rrr}
4 \amp 5 \amp 6 \\
3 \amp 0 \amp 5 \\
4 \amp -3 \amp 0
\end{avmatrix}\)
(b) (c)
Answer .
\(\displaystyle
M_{32} =
\begin{avmatrix}{rrr}
4 \amp 5 \amp 6 \\
-1 \amp 2 \amp 2 \\
4 \amp -3 \amp 0
\end{avmatrix}\)
(d) 6.
\(\displaystyle
\begin{abmatrix}{rrrr}
0 \amp 6 \amp 0 \amp 5 \\
-4 \amp 5 \amp 4 \amp 1 \\
2 \amp -5 \amp 3 \amp 4 \\
-2 \amp 6 \amp -4 \amp 3
\end{abmatrix}\)
(a)
Answer .
\(\displaystyle
M_{43} =
\begin{avmatrix}{rrr}
0 \amp 6 \amp 5 \\
-4 \amp 5 \amp 1 \\
2 \amp -5 \amp 4 \\
\end{avmatrix}\)
(b) (c)
Answer .
\(\displaystyle
M_{13} =
\begin{avmatrix}{rrr}
-4 \amp 5 \amp 1 \\
2 \amp -5 \amp 4 \\
-2 \amp 6 \amp 3
\end{avmatrix}\)
(d)
Cofactor expansions.
In each case, express the requested cofactor expansion as a linear combination of minor determinants, bringing the cofactor signs to the front of each term. For matrices larger than
\(2 \times 2\text{,}\) do not evaluate the minor determinants.
7.
\(\displaystyle
\begin{abmatrix}{rr}
-2 \amp 1 \\
0 \amp 4
\end{abmatrix}\)
(a)
Answer .
\(-2 \cdot 4 - 1 \cdot 0 \)
(b)
Answer .
\(-0 \cdot 1 + 4 \cdot (-2) \)
8.
\(\displaystyle
\begin{abmatrix}{rr}
3 \amp -6 \\
8 \amp 7
\end{abmatrix}\)
(a)
Answer .
\(3 \cdot 7 - 8 \cdot (-6) \)
(b)
Answer .
\(- (-6) \cdot 8 + 7 \cdot 3 \)
9.
\(\displaystyle
\begin{abmatrix}{rrr}
0 \amp 5 \amp -3 \\
-2 \amp 2 \amp -6 \\
1 \amp -1 \amp 2
\end{abmatrix}\)
(a)
Answer .
\(\displaystyle
1 \cdot
\begin{avmatrix}{rr}
5 \amp -3 \\
2 \amp -6
\end{avmatrix}
- (-1) \cdot
\begin{avmatrix}{rr}
0 \amp -3 \\
-2 \amp -6
\end{avmatrix}
+ 2 \cdot
\begin{avmatrix}{rr}
0 \amp 5 \\
-2 \amp 2
\end{avmatrix}\)
(b)
Answer .
\(\displaystyle
-5 \cdot
\begin{avmatrix}{rr}
-2 \amp -6 \\
1 \amp 2
\end{avmatrix}
+ 2 \cdot
\begin{avmatrix}{rr}
0 \amp -3 \\
1 \amp 2
\end{avmatrix}
- (-1) \cdot
\begin{avmatrix}{rr}
0 \amp -3 \\
-2 \amp -6
\end{avmatrix}\)
(c)
Answer .
\(\displaystyle
-3 \cdot
\begin{avmatrix}{rr}
-2 \amp 2 \\
1 \amp -1
\end{avmatrix}
- (-6) \cdot
\begin{avmatrix}{rr}
0 \amp 5 \\
1 \amp -1
\end{avmatrix}
+ 2 \cdot
\begin{avmatrix}{rr}
0 \amp 5 \\
-2 \amp 2
\end{avmatrix}\)
10.
\(\displaystyle
\begin{abmatrix}{rrr}
-2 \amp 2 \amp -1 \\
3 \amp -3 \amp -4 \\
-2 \amp -3 \amp -2
\end{abmatrix}\)
(a)
Answer .
\(\displaystyle
-2 \cdot
\begin{avmatrix}{rr}
-3 \amp -4 \\
-3 \amp -2
\end{avmatrix}
- 2 \cdot
\begin{avmatrix}{rr}
3 \amp -4 \\
-2 \amp -2
\end{avmatrix}
+ (-1) \cdot
\begin{avmatrix}{rr}
3 \amp -3 \\
-2 \amp -3
\end{avmatrix}\)
(b)
Answer .
\(\displaystyle
-2 \cdot
\begin{avmatrix}{rr}
-3 \amp -4 \\
-3 \amp -2
\end{avmatrix}
- 3 \cdot
\begin{avmatrix}{rr}
2 \amp -1 \\
-3 \amp -2
\end{avmatrix}
+ (-2) \cdot
\begin{avmatrix}{rr}
2 \amp -1 \\
-3 \amp -4 \\
\end{avmatrix}\)
(c)
Answer .
\(\displaystyle
-2 \cdot
\begin{avmatrix}{rr}
3 \amp -4 \\
-2 \amp -2
\end{avmatrix}
+ (-3) \cdot
\begin{avmatrix}{rr}
-2 \amp -1 \\
-2 \amp -2
\end{avmatrix}
- (-3) \cdot
\begin{avmatrix}{rr}
-2 \amp -1 \\
3 \amp -4 \\
\end{avmatrix}\)
(d)
Answer .
\(\displaystyle
-3 \cdot
\begin{avmatrix}{rr}
2 \amp -1 \\
-3 \amp -2
\end{avmatrix}
+ (-3) \cdot
\begin{avmatrix}{rr}
-2 \amp -1 \\
-2 \amp -2
\end{avmatrix}
- (-4) \cdot
\begin{avmatrix}{rr}
-2 \amp 2 \\
-2 \amp -3
\end{avmatrix}\)
11.
\(\displaystyle
\begin{abmatrix}{rrrr}
4 \amp -3 \amp 5 \amp 6 \\
-1 \amp -6 \amp 2 \amp 2 \\
3 \amp 0 \amp 0 \amp 5 \\
4 \amp 6 \amp -3 \amp 0
\end{abmatrix}\)
(a)
Answer .
\begin{equation*}
4 \cdot
\begin{avmatrix}{rrr}
-6 \amp 2 \amp 2 \\
0 \amp 0 \amp 5 \\
6 \amp -3 \amp 0
\end{avmatrix}
- (-1) \cdot
\begin{avmatrix}{rrr}
-3 \amp 5 \amp 6 \\
0 \amp 0 \amp 5 \\
6 \amp -3 \amp 0
\end{avmatrix}
+ 3 \cdot
\begin{avmatrix}{rrr}
-3 \amp 5 \amp 6 \\
-6 \amp 2 \amp 2 \\
6 \amp -3 \amp 0
\end{avmatrix}
- 4 \cdot
\begin{avmatrix}{rrr}
-3 \amp 5 \amp 6 \\
-6 \amp 2 \amp 2 \\
0 \amp 0 \amp 5 \\
\end{avmatrix}
\end{equation*}
(b)
Answer .
\begin{equation*}
- (-1) \cdot
\begin{avmatrix}{rrr}
-3 \amp 5 \amp 6 \\
0 \amp 0 \amp 5 \\
6 \amp -3 \amp 0
\end{avmatrix}
+ (-6) \cdot
\begin{avmatrix}{rrr}
4 \amp 5 \amp 6 \\
3 \amp 0 \amp 5 \\
4 \amp -3 \amp 0
\end{avmatrix}
- 2 \cdot
\begin{avmatrix}{rrr}
4 \amp -3 \amp 6 \\
3 \amp 0 \amp 5 \\
4 \amp 6 \amp 0
\end{avmatrix}
+ 2 \cdot
\begin{avmatrix}{rrr}
4 \amp -3 \amp 5 \\
3 \amp 0 \amp 0 \\
4 \amp 6 \amp -3
\end{avmatrix}
\end{equation*}
(c)
Answer .
\begin{equation*}
5 \cdot
\begin{avmatrix}{rrr}
-1 \amp -6 \amp 2 \\
3 \amp 0 \amp 5 \\
4 \amp 6 \amp 0
\end{avmatrix}
- 2 \cdot
\begin{avmatrix}{rrr}
4 \amp -3 \amp 6 \\
3 \amp 0 \amp 5 \\
4 \amp 6 \amp 0
\end{avmatrix}
+ 0 \cdot
\begin{avmatrix}{rrr}
4 \amp -3 \amp 6 \\
-1 \amp -6 \amp 2 \\
4 \amp 6 \amp 0
\end{avmatrix}
- (-3) \cdot
\begin{avmatrix}{rrr}
4 \amp -3 \amp 6 \\
-1 \amp -6 \amp 2 \\
3 \amp 0 \amp 5
\end{avmatrix}
\end{equation*}
(d)
Answer .
\begin{equation*}
-4 \cdot
\begin{avmatrix}{rrr}
-3 \amp 5 \amp 6 \\
-6 \amp 2 \amp 2 \\
0 \amp 0 \amp 5
\end{avmatrix}
+ 6 \cdot
\begin{avmatrix}{rrr}
4 \amp 5 \amp 6 \\
-1 \amp 2 \amp 2 \\
3 \amp 0 \amp 5
\end{avmatrix}
- (-3) \cdot
\begin{avmatrix}{rrr}
4 \amp -3 \amp 6 \\
-1 \amp -6 \amp 2 \\
3 \amp 0 \amp 5
\end{avmatrix}
+ 0 \cdot
\begin{avmatrix}{rrr}
4 \amp -3 \amp 5 \\
-1 \amp -6 \amp 2 \\
3 \amp 0 \amp 0
\end{avmatrix}
\end{equation*}
12.
\(\displaystyle
\begin{abmatrix}{rrrr}
0 \amp 6 \amp 0 \amp 5 \\
-4 \amp 5 \amp 4 \amp 1 \\
2 \amp -5 \amp 3 \amp 4 \\
-2 \amp 6 \amp -4 \amp 3
\end{abmatrix}\)
(a)
Answer .
\begin{equation*}
0 \cdot
\begin{avmatrix}{rrr}
5 \amp 4 \amp 1 \\
-5 \amp 3 \amp 4 \\
6 \amp -4 \amp 3
\end{avmatrix}
- 6 \cdot
\begin{avmatrix}{rrr}
-4 \amp 4 \amp 1 \\
2 \amp 3 \amp 4 \\
-2 \amp -4 \amp 3
\end{avmatrix}
+ 0 \cdot
\begin{avmatrix}{rrr}
-4 \amp 5 \amp 1 \\
2 \amp -5 \amp 4 \\
-2 \amp 6 \amp 3
\end{avmatrix}
- 5 \cdot
\begin{avmatrix}{rrr}
-4 \amp 5 \amp 4 \\
2 \amp -5 \amp 3 \\
-2 \amp 6 \amp -4
\end{avmatrix}
\end{equation*}
(b)
Answer .
\begin{equation*}
-6 \cdot
\begin{avmatrix}{rrr}
-4 \amp 4 \amp 1 \\
2 \amp 3 \amp 4 \\
-2 \amp -4 \amp 3
\end{avmatrix}
+ 5 \cdot
\begin{avmatrix}{rrr}
0 \amp 0 \amp 5 \\
2 \amp 3 \amp 4 \\
-2 \amp -4 \amp 3
\end{avmatrix}
- (-5) \cdot
\begin{avmatrix}{rrr}
0 \amp 0 \amp 5 \\
-4 \amp 4 \amp 1 \\
-2 \amp -4 \amp 3
\end{avmatrix}
+ 6 \cdot
\begin{avmatrix}{rrr}
0 \amp 0 \amp 5 \\
-4 \amp 4 \amp 1 \\
2 \amp 3 \amp 4
\end{avmatrix}
\end{equation*}
(c)
Answer .
\begin{equation*}
2 \cdot
\begin{avmatrix}{rrr}
6 \amp 0 \amp 5 \\
5 \amp 4 \amp 1 \\
6 \amp -4 \amp 3
\end{avmatrix}
- (-5) \cdot
\begin{avmatrix}{rrr}
0 \amp 0 \amp 5 \\
-4 \amp 4 \amp 1 \\
-2 \amp -4 \amp 3
\end{avmatrix}
+ 3 \cdot
\begin{avmatrix}{rrr}
0 \amp 6 \amp 5 \\
-4 \amp 5 \amp 1 \\
-2 \amp 6 \amp 3
\end{avmatrix}
- 4 \cdot
\begin{avmatrix}{rrr}
0 \amp 6 \amp 0 \\
-4 \amp 5 \amp 4 \\
-2 \amp 6 \amp -4
\end{avmatrix}
\end{equation*}
(d)
Answer .
\begin{equation*}
-5 \cdot
\begin{avmatrix}{rrr}
-4 \amp 5 \amp 4 \\
2 \amp -5 \amp 3 \\
-2 \amp 6 \amp -4
\end{avmatrix}
+ 1 \cdot
\begin{avmatrix}{rrr}
0 \amp 6 \amp 0 \\
2 \amp -5 \amp 3 \\
-2 \amp 6 \amp -4
\end{avmatrix}
- 4 \cdot
\begin{avmatrix}{rrr}
0 \amp 6 \amp 0 \\
-4 \amp 5 \amp 4 \\
-2 \amp 6 \amp -4
\end{avmatrix}
+ 3 \cdot
\begin{avmatrix}{rrr}
0 \amp 6 \amp 0 \\
-4 \amp 5 \amp 4 \\
2 \amp -5 \amp 3
\end{avmatrix}
\end{equation*}
Choosing a cofactor expansion.
For each matrix, choose a row or column for which calculating the cofactor expansion would require the fewest minor determinant calculations.
13.
\(\displaystyle
\begin{abmatrix}{rrr}
-4 \amp -3 \amp -6 \\
0 \amp 5 \amp -9 \\
-4 \amp 9 \amp -9
\end{abmatrix}\)
Answer .
A cofactor expansion along either the first column or the second row would involve only two minor determinant calculations, rather than three.
14.
\(\displaystyle
\begin{abmatrix}{rrr}
-7 \amp 1 \amp -4 \\
-3 \amp -8 \amp 3 \\
5 \amp 0 \amp 0
\end{abmatrix}\)
Answer .
A cofactor expansion along the third row would involve only one minor determinant calculation, rather than three.
15.
\(\displaystyle
\begin{abmatrix}{rrrr}
-1 \amp -1 \amp 9 \amp 0 \\
0 \amp 3 \amp -2 \amp 8 \\
-2 \amp -7 \amp 4 \amp -2 \\
-3 \amp -3 \amp 7 \amp 5
\end{abmatrix}\)
Answer .
A cofactor expansion along any of the first or second rows or the first or fourth columns would involve only three minor determinant calculations, rather than four.
16.
\(\displaystyle
\begin{abmatrix}{rrrr}
4 \amp 0 \amp 2 \amp 1 \\
-2 \amp 0 \amp -1 \amp -9 \\
4 \amp -9 \amp 8 \amp 5 \\
3 \amp 9 \amp 5 \amp 6
\end{abmatrix}\)
Answer .
A cofactor expansion along the second column would involve only two minor determinant calculations, rather than four.
Computing determinants.
Compute the determinant of each matrix.
17.
\(\displaystyle \begin{bmatrix} -9 \end{bmatrix} \)
Answer .
\(\displaystyle \det \begin{bmatrix} -9 \end{bmatrix} = -9 \)
18.
\(\displaystyle \begin{bmatrix} 0 \end{bmatrix} \)
Answer .
\(\displaystyle \det \begin{bmatrix} 0 \end{bmatrix} = 0 \)
19.
\(\displaystyle
\begin{abmatrix}{rr}
-5 \amp -4 \\
1 \amp 2
\end{abmatrix}\)
Solution .
Using the general
\(a d - b c\) formula for the determinant of a
\(2 \times 2\) matrix (see
Subsection 8.3.3 ), we have
\begin{equation*}
\begin{avmatrix}{rr}
-5 \amp -4 \\
1 \amp 2
\end{avmatrix}
= (-5) \cdot 2 - (-4) \cdot 1
= -6\text{.}
\end{equation*}
20.
\(\displaystyle
\begin{bmatrix}
2 \amp 5 \\
1 \amp 2
\end{bmatrix}\)
Solution .
Using the general
\(a d - b c\) formula for the determinant of a
\(2 \times 2\) matrix (see
Subsection 8.3.3 ), we have
\begin{equation*}
\begin{vmatrix}
2 \amp 5 \\
1 \amp 2
\end{vmatrix}
= 2 \cdot 2 - 5 \cdot 1
= -1\text{.}
\end{equation*}
21.
\(\displaystyle
\begin{abmatrix}{rr}
-5 \amp 6 \\
4 \amp 5
\end{abmatrix}\)
Solution .
Using the general
\(a d - b c\) formula for the determinant of a
\(2 \times 2\) matrix (see
Subsection 8.3.3 ), we have
\begin{equation*}
\begin{avmatrix}{rr}
-5 \amp 6 \\
4 \amp 5
\end{avmatrix}
= (-5) \cdot 5 - 6 \cdot 4
= -49\text{.}
\end{equation*}
22.
\(\displaystyle
\begin{bmatrix}
1 \amp 6 \\
6 \amp 0
\end{bmatrix}\)
Solution .
Using the general
\(a d - b c\) formula for the determinant of a
\(2 \times 2\) matrix (see
Subsection 8.3.3 ), we have
\begin{equation*}
\begin{vmatrix}
1 \amp 6 \\
6 \amp 0
\end{vmatrix}
= 1 \cdot 0 - 6 \cdot 6
= -36\text{.}
\end{equation*}
23.
\(\displaystyle
\begin{abmatrix}{rrr}
1 \amp -1 \amp 4 \\
-2 \amp 5 \amp -1 \\
4 \amp -1 \amp -3
\end{abmatrix}\)
Solution .
Let \(A\) represent the matrix. Expanding along the first row, we have
\begin{align*}
\det A \amp =
\begin{avmatrix}{lll}
\phantom{-}{\color{red}\mathbf{1}}^+ \amp {\color{red}\mathbf{-1}}^- \amp \phantom{-}{\color{red}\mathbf{4}}^+ \\
-2 \amp \phantom{-}5 \amp -1 \\
\phantom{-}4 \amp -1 \amp -3
\end{avmatrix}\\
\amp =
1 \cdot
\begin{avmatrix}{rr}
5 \amp -1 \\
-1 \amp -3
\end{avmatrix}
- (-1) \cdot
\begin{avmatrix}{rr}
-2 \amp -1 \\
4 \amp -3
\end{avmatrix}
+ 4 \cdot
\begin{avmatrix}{rr}
-2 \amp 5 \\
4 \amp -1
\end{avmatrix}\\
\amp =
\bbrac{5 \cdot (-3) - (-1) \cdot (-1)}
+ \bbrac{(-2) \cdot (-3) - (-1) \cdot 4}\\
\amp \phantom{=} \quad
+ 4 \bbrac{(-2) \cdot (-1) - 5 \cdot 4}\\
\amp = -78 \text{.}
\end{align*}
To compute the minor determinants in this cofactor expansion we have used the general
\(a d - b c\) formula for the determinant of a
\(2 \times 2\) matrix. (See
Subsection 8.3.3 .)
24.
\(\displaystyle
\begin{abmatrix}{rrr}
1 \amp 3 \amp 7 \\
1 \amp 1 \amp 0 \\
-1 \amp -4 \amp -11
\end{abmatrix}\)
Solution .
Let \(A\) represent the matrix. Expand along the second row so that there are only two cofactors to compute:
\begin{align*}
\det A \amp =
\begin{avmatrix}{lll}
\phantom{-}1 \amp \phantom{-}3 \amp \phantom{-1}7 \\
\phantom{-}{\color{red}\mathbf{1}}^- \amp \phantom{-}{\color{red}\mathbf{1}}^+ \amp \phantom{-}{\color{red}\mathbf{0}}^- \\
-1 \amp -4 \amp -11
\end{avmatrix}\\
\amp =
-1 \cdot
\begin{avmatrix}{rr}
3 \amp 7 \\
-4 \amp -11
\end{avmatrix}
+ 1 \cdot
\begin{avmatrix}{rr}
1 \amp 7 \\
-1 \amp -11
\end{avmatrix}\\
\amp =
- \bbrac{3 \cdot (-11) - 7 \cdot (-4)}
+ \bbrac{1 \cdot (-11) - 7 \cdot (-1)}\\
\amp = 1 \text{.}
\end{align*}
To compute the minor determinants in this cofactor expansion we have used the general
\(a d - b c\) formula for the determinant of a
\(2 \times 2\) matrix. (See
Subsection 8.3.3 .)
25.
\(\displaystyle
\begin{abmatrix}{rrrr}
-2 \amp 0 \amp 3 \amp 7 \\
-3 \amp -1 \amp 2 \amp -2 \\
3 \amp 1 \amp -2 \amp 3 \\
1 \amp 0 \amp -1 \amp -3
\end{abmatrix}\)
Solution .
Let \(A\) represent the matrix. Choose the second column for a cofactor expansion since it will involve the fewest minor determinant calculations:
\begin{align*}
\det A \amp =
\begin{avmatrix}{rrrr}
-2 \amp {\color{red}\mathbf{ 0}}^- \amp 3 \amp 7 \\
-3 \amp {\color{red}\mathbf{-1}}^+ \amp 2 \amp -2 \\
3 \amp {\color{red}\mathbf{ 1}}^- \amp -2 \amp 3 \\
1 \amp {\color{red}\mathbf{ 0}}^+ \amp -1 \amp -3
\end{avmatrix}\\
\amp =
(-1) \cdot
\begin{avmatrix}{rrr}
-2 \amp 3 \amp 7 \\
3 \amp -2 \amp 3 \\
1 \amp -1 \amp -3
\end{avmatrix}
- 1 \cdot
\begin{avmatrix}{rrr}
-2 \amp 3 \amp 7 \\
-3 \amp 2 \amp -2 \\
1 \amp -1 \amp -3
\end{avmatrix}
\end{align*}
Let’s compute these two
\(3 \times 3\) minor determinants separately; call the first matrix
\(A_1\) and the second
\(A_2\text{.}\) Neither has any zero entries, but the third row in each contains a couple of ones, so we’ll choose to expand along that row in each. In both expansions, we will use the general
\(a d - b c\) formula for the resulting
\(2 \times 2\) minor determinants (see
Subsection 8.3.3 ).
\begin{align*}
\det A_1 \amp =
\begin{avmatrix}{lll}
-2 \amp \phantom{-}3 \amp \phantom{-}7 \\
\phantom{-}3 \amp -2 \amp \phantom{-}3 \\
{\color{red}\mathbf{1}}^+ \amp {\color{red}\mathbf{-1}}^- \amp {\color{red}\mathbf{-3}}^+
\end{avmatrix}\\
\amp =
1 \cdot
\begin{avmatrix}{rr}
3 \amp 7 \\
-2 \amp 3 \\
\end{avmatrix}
- (-1) \cdot
\begin{avmatrix}{rr}
-2 \amp 7 \\
3 \amp 3 \\
\end{avmatrix}
+ (-3) \cdot
\begin{avmatrix}{rr}
-2 \amp 3 \\
3 \amp -2
\end{avmatrix}\\
\amp =
\bbrac{3 \cdot 3 - 7 \cdot (-2)}
+ \bbrac{(-2) \cdot 3 - 7 \cdot 3}
- 3 \bbrac{(-2) \cdot (-2) - 3 \cdot 3}\\
\amp = 11
\end{align*}
\begin{align*}
\det A_2 \amp =
\begin{avmatrix}{lll}
-2 \amp \phantom{-}3 \amp \phantom{-}7 \\
-3 \amp \phantom{-}2 \amp -2 \\
\phantom{-}{\color{red}\mathbf{1}}^+ \amp {\color{red}\mathbf{-1}}^- \amp {\color{red}\mathbf{-3}}^+
\end{avmatrix}\\
\amp =
1 \cdot
\begin{avmatrix}{rr}
3 \amp 7 \\
2 \amp -2 \\
\end{avmatrix}
- (-1) \cdot
\begin{avmatrix}{rr}
-2 \amp 7 \\
-3 \amp -2 \\
\end{avmatrix}
+ (-3) \cdot
\begin{vmatrix}
-2 \amp 3 \\
-3 \amp 2
\end{vmatrix}\\
\amp =
\bbrac{3 \cdot (-2) - 7 \cdot 2}
+ \bbrac{(-2) \cdot (-2) - 7 \cdot (-3)}
- 3 \bbrac{(-2) \cdot 2 - 3 \cdot (-3)}\\
\amp = -10
\end{align*}
Substituting these results into our original cofactor expansion for \(\det A\text{,}\) we have
\begin{align*}
\det A \amp = (-1) \cdot \det A_1 - 1 \cdot \det A_2 \\
\amp = (-1) \cdot 11 - 1 \cdot (-10) \\
\amp = -1 \text{.}
\end{align*}
26.
\(\displaystyle
\begin{bmatrix}
1 \amp 0 \amp 1 \amp 0 \\
0 \amp 1 \amp 0 \amp 1 \\
1 \amp 0 \amp 1 \amp 0 \\
0 \amp 1 \amp 0 \amp 1
\end{bmatrix}\)
Solution .
Let \(A\) represent the matrix. Any row or column we choose will contain two zeros, so just expand along the first row:
\begin{align*}
\det A \amp =
\begin{vmatrix}
{\color{red}\mathbf{1}}^+ \amp {\color{red}\mathbf{0}}^- \amp {\color{red}\mathbf{1}}^+ \amp {\color{red}\mathbf{0}}^- \\
0 \amp 1 \amp 0 \amp 1 \\
1 \amp 0 \amp 1 \amp 0 \\
0 \amp 1 \amp 0 \amp 1
\end{vmatrix}\\
\amp =
1 \cdot
\begin{vmatrix}
1 \amp 0 \amp 1 \\
0 \amp 1 \amp 0 \\
1 \amp 0 \amp 1
\end{vmatrix}
+ 1 \cdot
\begin{vmatrix}
0 \amp 1 \amp 1 \\
1 \amp 0 \amp 0 \\
0 \amp 1 \amp 1
\end{vmatrix}
\end{align*}
Let’s compute these two
\(3 \times 3\) minor determinants separately; call the first matrix
\(A_1\) and the second
\(A_2\text{.}\) The second row in each contains a couple of zeros, so we’ll choose to expand along that row in each. In both expansions, we will use the general
\(a d - b c\) formula for the resulting
\(2 \times 2\) minor determinants (see
Subsection 8.3.3 ).
\begin{align*}
\det A_1 \amp =
\begin{avmatrix}{lll}
1 \amp 0 \amp 1 \\
{\color{red}\mathbf{0}}^- \amp {\color{red}\mathbf{1}}^+ \amp {\color{red}\mathbf{0}}^- \\
1 \amp 0 \amp 1
\end{avmatrix}
\amp
\det A_2 \amp =
\begin{avmatrix}{lll}
0 \amp 1 \amp 1 \\
{\color{red}\mathbf{1}}^- \amp {\color{red}\mathbf{0}}^+ \amp {\color{red}\mathbf{0}}^- \\
0 \amp 1 \amp 1
\end{avmatrix}\\
\amp =
1 \cdot
\begin{vmatrix}
1 \amp 1 \\
1 \amp 1 \\
\end{vmatrix}
\amp
\amp =
-1 \cdot
\begin{vmatrix}
1 \amp 1 \\
1 \amp 1
\end{vmatrix}\\
\amp = (1 \cdot 1 - 1 \cdot 1)
\amp
\amp = - (1 \cdot 1 - 1 \cdot 1)\\
\amp = 0 \amp \amp = 0
\end{align*}
Substituting these results into our original cofactor expansion for \(\det A\text{,}\) we have
\begin{align*}
\det A \amp = 1 \cdot \det A_1 + 1 \cdot \det A_2 \\
\amp = 1 \cdot 0 + 1 \cdot 0 \\
\amp = 0 \text{.}
\end{align*}
27.
\(\displaystyle
\begin{abmatrix}{rrrrr}
-4 \amp 0 \amp 0 \amp 0 \amp 0 \\
-1 \amp -1 \amp 0 \amp 0 \amp 0 \\
-6 \amp 4 \amp 2 \amp 0 \amp 0 \\
1 \amp -2 \amp 0 \amp 1 \amp 0 \\
-5 \amp 6 \amp -2 \amp -4 \amp 2
\end{abmatrix}\)
Solution .
Let
\(A\) represent the matrix. Notice that
\(A\) is lower triangular, so by
Statement 1 of
Proposition 8.5.2 its determinant is simply the product of its diagonal entries:
\begin{equation*}
\det A
= (-4) \cdot (-1) \cdot 2 \cdot 1 \cdot 2
= 16\text{.}
\end{equation*}
28.
\(\displaystyle
\begin{abmatrix}{rrrrrr}
4 \amp 5 \amp 0 \amp 2 \amp -5 \amp -1 \\
0 \amp 5 \amp -5 \amp 1 \amp -4 \amp -4 \\
0 \amp 0 \amp -1 \amp -2 \amp 0 \amp -3 \\
0 \amp 0 \amp 0 \amp 4 \amp 1 \amp 3 \\
0 \amp 0 \amp 0 \amp 0 \amp -3 \amp -4 \\
0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 3
\end{abmatrix}\)
Solution .
Let
\(A\) represent the matrix. Notice that
\(A\) is upper triangular, so by
Statement 1 of
Proposition 8.5.2 its determinant is simply the product of its diagonal entries:
\begin{equation*}
\det A
= 4 \cdot 5 \cdot (-1) \cdot 4 \cdot (-3) \cdot 3
= 720\text{.}
\end{equation*}
29.
\(\displaystyle
\begin{abmatrix}{ccccccr}
2 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \\
0 \amp 2 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \\
0 \amp 0 \amp 4 \amp 0 \amp 0 \amp 0 \amp 0 \\
0 \amp 0 \amp 0 \amp 5 \amp 0 \amp 0 \amp 0 \\
0 \amp 0 \amp 0 \amp 0 \amp 4 \amp 0 \amp 0 \\
0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 3 \amp 0 \\
0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp -5
\end{abmatrix}\)
Solution .
Let
\(A\) represent the matrix. Notice that
\(A\) is diagonal, so by
Statement 1 of
Proposition 8.5.2 its determinant is simply the product of its diagonal entries:
\begin{equation*}
\det A
= 2 \cdot 2 \cdot 4 \cdot 5 \cdot 4 \cdot 3 \cdot (-5)
= -4800\text{.}
\end{equation*}
30.
\(\displaystyle
\begin{bmatrix}
5 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \\
0 \amp 5 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \\
0 \amp 0 \amp 5 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \\
0 \amp 0 \amp 0 \amp 5 \amp 0 \amp 0 \amp 0 \amp 0 \\
0 \amp 0 \amp 0 \amp 0 \amp 5 \amp 0 \amp 0 \amp 0 \\
0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 5 \amp 0 \amp 0 \\
0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 5 \amp 0 \\
0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 5
\end{bmatrix}\)
Solution .
Let
\(A\) represent the matrix. Notice that
\(A = 5 I\) is scalar, so by
Statement 2 of
Proposition 8.5.2 its determinant is simply the power of the common diagonal entry value, with exponent equal to the size of the square matrix:
\begin{equation*}
\det A = 5^8 = 390625 \text{.}
\end{equation*}
31. Counterintuitive determinant examples.
Provide example matrices with the requested properties.
A \(2 \times 2\) matrix \(M\text{,}\) the entries of which are all non zero, but for which \(\det M = 0\text{.}\)
A nonzero \(3 \times 3\) upper triangular matrix \(U\) for which \(\det U = 0\text{.}\)
A nonzero \(3 \times 3\) diagonal matrix \(D\) for which \(\det D = 0\text{.}\)
A pair of \(2 \times 2\) matrices \(A\) and \(B\) for which both \(\det A = 0\) and \(\det B = 0\) but \(\det (A + B) \neq 0\text{.}\) (This demonstrates that \(\det (A + B)\) is not equal to \(\det A + \det B\) in general.)
Variable determinants.
Each matrix below contains one or more entries involving a parameter. Compute the determinant as a formula in that parameter. Then determine all values of the parameter for which the determinant is equal to
\(0\text{.}\)
32.
\(\displaystyle
\begin{abmatrix}{cr}
k \amp -4 \\
k + 3 \amp 2
\end{abmatrix}\)
Solution .
Using the general
\(a d - b c\) formula for the determinant of a
\(2 \times 2\) matrix (see
Subsection 8.3.3 ), we have
\begin{equation*}
\begin{avmatrix}{cr}
k \amp -4 \\
k + 3 \amp 2
\end{avmatrix}
= k \cdot 2 - (-4) \cdot (k + 3)
= 6 k + 12\text{.}
\end{equation*}
This determinant will equal \(0\) when
\begin{align*}
6 k + 12 \amp = 0 \\
6 k \amp = -12 \\
k \amp = -2 \text{.}
\end{align*}
33.
\(\displaystyle
\begin{bmatrix}
k + 1 \amp 2 \\
4 \amp k - 1
\end{bmatrix}\)
Solution .
Using the general
\(a d - b c\) formula for the determinant of a
\(2 \times 2\) matrix (see
Subsection 8.3.3 ), we have
\begin{equation*}
\begin{vmatrix}
k + 1 \amp 2 \\
4 \amp k - 1
\end{vmatrix}
= (k + 1) (k - 1) - 2 \cdot 4
= k^2 - 9\text{.}
\end{equation*}
This determinant will equal \(0\) when \(k = \pm 3\text{.}\)
34.
\(\displaystyle
\begin{abmatrix}{rrr}
x - 3 \amp 0 \amp -1 \\
5 \amp x - 5 \amp -3 \\
4 \amp 0 \amp x - 7
\end{abmatrix}\)
Solution .
Let \(A\) represent the matrix. Expanding along the second row so that there are is only cofactor to compute:
\begin{align*}
\det A \amp =
\begin{avmatrix}{clc}
x - 3 \amp {\color{red}\mathbf{ 0 }}^- \amp -1 \\
5 \amp {\color{red}\mathbf{x - 5}}^+ \amp -3 \\
4 \amp {\color{red}\mathbf{ 0 }}^- \amp x - 7
\end{avmatrix}\\
\amp =
(x - 5) \cdot
\begin{avmatrix}{rr}
x - 3 \amp -1 \\
4 \amp x - 7
\end{avmatrix}\\
\amp = (x - 5) \bbrac{ (x - 3) (x - 7) - (-1) \cdot 4 } \\
\amp = (x - 5)^3 \text{.}
\end{align*}
This determinant will equal \(0\) when \(x = 5 \text{.}\)
35.
\(\displaystyle
\begin{bmatrix}
z - 2 \amp 1 \amp 3 \amp 0 \\
0 \amp z + 1 \amp 2 \amp -2 \\
0 \amp 0 \amp z - 7 \amp 3 \\
0 \amp 0 \amp 0 \amp z + 3
\end{bmatrix}\)
Solution .
Let
\(A\) represent the matrix. This is an upper triangular matrix, so by
Statement 1 of
Proposition 8.5.2 its determinant is simply the product of its diagonal entries:
\begin{equation*}
\det A = (z - 2) (z + 1) (x - 7) (z + 3) \text{.}
\end{equation*}
This determinant will equal zero when \(z\) is equal to one of \(2\text{,}\) \(-1\text{,}\) \(7\text{,}\) or \(-3\text{.}\)
