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Discover Linear Algebra: A first course in linear algebra

Jeremy Sylvestre

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)
\(M_{11}\)
Answer.
\(M_{11} = 4\)
(b)
\(C_{11}\)
Answer.
\(C_{11} = M_{11} = 4\)
(c)
\(M_{21}\)
Answer.
\(M_{21} = 1\)
(d)
\(C_{21}\)
Answer.
\(C_{21} = -M_{21} = -1\)

2.

\(\displaystyle \begin{abmatrix}{rr} 3 \amp -6 \\ 8 \amp 7 \end{abmatrix}\)
(a)
\(M_{12}\)
Answer.
\(M_{12} = 8\)
(b)
\(C_{12}\)
Answer.
\(C_{12} = -M_{12} = -8\)
(c)
\(M_{22}\)
Answer.
\(M_{22} = 3\)
(d)
\(C_{22}\)
Answer.
\(C_{22} = M_{22} = 3\)

3.

\(\displaystyle \begin{abmatrix}{rrr} 0 \amp 5 \amp -3 \\ -2 \amp 2 \amp -6 \\ 1 \amp -1 \amp 2 \end{abmatrix}\)
(a)
\(M_{11}\)
Answer.
\(\displaystyle M_{11} = \begin{avmatrix}{rr} 2 \amp -6 \\ -1 \amp 2 \end{avmatrix} = -2\)
(b)
\(C_{11}\)
Answer.
\(C_{11} = M_{11} = -2\)
(c)
\(M_{13}\)
Answer.
\(\displaystyle M_{13} = \begin{avmatrix}{rr} -2 \amp 2 \\ 1 \amp -1 \end{avmatrix} = 0\)
(d)
\(C_{13}\)
Answer.
\(C_{13} = M_{13} = 0\)
(e)
\(M_{31}\)
Answer.
\(\displaystyle M_{31} = \begin{avmatrix}{rr} 5 \amp -3 \\ 2 \amp -6 \end{avmatrix} = -24\)
(f)
\(C_{31}\)
Answer.
\(C_{31} = M_{31} = -24\)
(g)
\(M_{33}\)
Answer.
\(\displaystyle M_{33} = \begin{avmatrix}{rr} 0 \amp 5 \\ -2 \amp 2 \end{avmatrix} = 10\)
(h)
\(C_{33}\)
Answer.
\(C_{33} = M_{33} = 10\)

4.

\(\displaystyle \begin{abmatrix}{rrr} -2 \amp 2 \amp -1 \\ 3 \amp -3 \amp -4 \\ -2 \amp -3 \amp -2 \end{abmatrix}\)
(a)
\(M_{12}\)
Answer.
\(\displaystyle M_{12} = \begin{avmatrix}{rr} 3 \amp -4 \\ -2 \amp -2 \end{avmatrix} = -14\)
(b)
\(C_{12}\)
Answer.
\(C_{12} = -M_{12} = 14\)
(c)
\(M_{21}\)
Answer.
\(\displaystyle M_{21} = \begin{avmatrix}{rr} 2 \amp -1 \\ -3 \amp -2 \end{avmatrix} = -7\)
(d)
\(C_{21}\)
Answer.
\(C_{21} = -M_{21} = 7\)
(e)
\(M_{23}\)
Answer.
\(\displaystyle M_{23} = \begin{avmatrix}{rr} -2 \amp 2 \\ -2 \amp -3 \end{avmatrix} = 10\)
(f)
\(C_{23}\)
Answer.
\(C_{23} = -M_{23} = -10\)
(g)
\(M_{32}\)
Answer.
\(\displaystyle M_{32} = \begin{avmatrix}{rr} -2 \amp -1 \\ 3 \amp -4 \end{avmatrix} = 11\)
(h)
\(C_{32}\)
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)
\(M_{22}\)
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)
\(C_{22}\)
Answer.
\(C_{22} = M_{22}\)
(c)
\(M_{32}\)
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)
\(C_{32}\)
Answer.
\(C_{32} = -M_{32}\)

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)
\(M_{43}\)
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)
\(C_{43}\)
Answer.
\(C_{43} = -M_{43}\)
(c)
\(M_{13}\)
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)
\(C_{13}\)
Answer.
\(C_{13} = M_{13}\)

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)
Along the first row.
Answer.
\(-2 \cdot 4 - 1 \cdot 0 \)
(b)
Along the second row.
Answer.
\(-0 \cdot 1 + 4 \cdot (-2) \)

8.

\(\displaystyle \begin{abmatrix}{rr} 3 \amp -6 \\ 8 \amp 7 \end{abmatrix}\)
(a)
Along the first column.
Answer.
\(3 \cdot 7 - 8 \cdot (-6) \)
(b)
Along the second column.
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)
Along the third row.
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)
Along the second column.
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)
Along the third column.
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)
Along the first row.
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)
Along the first column.
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)
Along the second column.
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)
Along the second row.
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)
Along the first column.
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)
Along the second row.
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)
Along the third column.
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)
Along the fourth row.
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)
Along the first row.
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)
Along the second column.
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)
Along the third row.
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)
Along the fourth column.
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.
  1. A \(2 \times 2\) matrix \(M\text{,}\) the entries of which are all nonzero, but for which \(\det M = 0\text{.}\)
  2. A nonzero \(3 \times 3\) upper triangular matrix \(U\) for which \(\det U = 0\text{.}\)
  3. A nonzero \(3 \times 3\) diagonal matrix \(D\) for which \(\det D = 0\text{.}\)
  4. 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 \(-3\text{.}\)
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