Determinants
Solution 5.6.1
(See Exercise 5.1.) By Theorem 5.15, the matrix \(A\) is invertible if and only if \(\det A \ne 0\). Using Sarrus’ rule, cf. Lemma 5.11, we compute
\[
\begin{align*}
\det A & = a \cdot 1 \cdot 4 + b \cdot (-1) \cdot 1 + 3 \cdot 2 \cdot (-1) \\
& \phantom{=} - 3 \cdot 1 \cdot 1 - b \cdot 2 \cdot 4 - a \cdot (-1) \cdot (-1) \\
& = 4a - b - 6 - 3 - 8b - a \\
& = 3a - 9b - 9 \\
& = 3(a-3b-3).
\end{align*}
\]
Therefore, \(A\) is invertible precisely for
In this case, the inverse can be computed using the adjugate formula from , where the adjugate is defined in Definition 5.14. The cofactors are
\[
\begin{align*}
c_{11} & = \det \left ( \begin{array}{cc} 1 & -1 \\ -1 & 4 \end{array} \right ) = 3,
&
c_{12} & = - \det \left ( \begin{array}{cc} 2 & -1 \\ 1 & 4 \end{array} \right ) = -9,
&
c_{13} & = \det \left ( \begin{array}{cc} 2 & 1 \\ 1 & -1 \end{array} \right ) = -3.
\end{align*}
\]
etc. One obtains the adjugate matrix:
\[
\mathrm{adj}(A) = \left ( \begin{array}{ccc} 3 & -4b-3 & -b-3 \\ -9 & 4a-3 & a+6 \\ -3 & a+b & a-2b \end{array} \right ),
\]
and therefore
\[
A^{-1} = \frac{1}{3a-9b-9} \left ( \begin{array}{ccc} 3 & -4b-3 & -b-3 \\ -9 & 4a-3 & a+6 \\ -3 & a+b & a-2b \end{array} \right ).
\]
Solution 5.6.2
(See Exercise 5.2.) From \(A^2 = {\mathrm {id}}\) and the product formula in Proposition 5.18, we get
\[
\det(A^2) = \det(A)\det(A) = (\det A)^2.
\]
On the other hand,
\[
\det(A^2) = \det({\mathrm {id}}).
\]
By Theorem 5.5, we have \(\det({\mathrm {id}})=1\). Therefore
Since \(\det A \in {\bf R}\), this implies
Solution 5.6.3
(See Exercise 5.3.) Using the \(2 \times 2\) determinant formula from Definition 5.1, we get
\[
\begin{align*}
\det A & = \det \left ( \begin{array}{cc} \cos r & -\sin r \\ \sin r & \cos r \end{array} \right ) \\
& = (\cos r)(\cos r) - (-\sin r)(\sin r) \\
& = \cos^2 r + \sin^2 r \\
& = 1.
\end{align*}
\]
So every rotation matrix has determinant \(1\). (In particular, by Theorem 5.15, every rotation matrix is invertible.)
Solution 5.6.4
(See Exercise 5.4.) We compute the determinant of
\[
A = \left ( \begin{array}{cccc} 2 & 0 & 1 & 4 \\ -1 & 3 & 0 & 2 \\ 1 & 0 & 2 & -3 \\ 0 & -2 & 5 & 1 \end{array} \right )
\]
in two different ways.
First method (using Theorem 5.5): Set \(A_0 := A\). Using row additions (which do not change the determinant, cf. Theorem 5.5) we will bring \(A\) to a diagonal form: we consider
\[
\begin{align*}
A_1 &:= \left ( \begin{array}{cccc} 2 & 0 & 1 & 4 \\ 0 & 3 & \frac 12 & 4 \\ 0 & 0 & \frac 32 & -5 \\ 0 & -2 & 5 & 1 \end{array} \right ) \\[-0.5mm]
&\text{(obtained from $A_0$ by }R_2\leadsto R_2+\frac12R_1,\; R_3\leadsto R_3-\frac12R_1\text{)}, \\
A_2 &:= \left ( \begin{array}{cccc} 2 & 0 & 1 & 4 \\ 0 & 3 & \frac 12 & 4 \\ 0 & 0 & \frac 32 & -5 \\ 0 & 0 & \frac {16}3 & \frac {11}3 \end{array} \right )
\quad\text{(from $A_1$ by }R_4\leadsto R_4+\frac23R_2\text{)}, \\
A_3 &:= \left ( \begin{array}{cccc} 2 & 0 & 1 & 4 \\ 0 & 3 & \frac 12 & 4 \\ 0 & 0 & \frac 32 & -5 \\ 0 & 0 & 0 & \frac {193}9 \end{array} \right )
\quad\text{(from $A_2$ by }R_4\leadsto R_4-\frac{32}9R_3\text{)}.
\end{align*}
\]
Now eliminate the entries above the pivots in columns 3 and 4:
\[
\begin{align*}
&A_4 := \left ( \begin{array}{cccc} 2 & 0 & 0 & \frac{22}3 \\ 0 & 3 & 0 & \frac{17}3 \\ 0 & 0 & \frac32 & -5 \\ 0 & 0 & 0 & \frac{193}9 \end{array} \right )
\quad\text{(from $A_3$ by }R_2\leadsto R_2-\frac13R_3,\; R_1\leadsto R_1-\frac23R_3\text{)}, \\
&A_5 := \left ( \begin{array}{cccc} 2 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & \frac32 & 0 \\ 0 & 0 & 0 & \frac{193}9 \end{array} \right )
\quad\text{(from $A_4$ by }R_1\leadsto R_1-\frac{66}{193}R_4,\; R_2\leadsto R_2-\frac{51}{193}R_4,\; R_3\leadsto R_3+\frac{45}{193}R_4\text{)}.
\end{align*}
\]
All these steps are row additions, so
\[
\det A_0 = \det A_1 = \dots = \det A_5.
\]
For the diagonal matrix \(A_5\), multilinearity in the rows and \(\det({\mathrm{id}})=1\) from Theorem 5.5 give
\[
\det A_5 = 2 \cdot 3 \cdot \frac32 \cdot \frac{193}9 = 193.
\]
Hence
Second method (using cofactor expansion, Proposition 5.23): Expand along the second column (which is the most efficient choice, since this column contains two zeros), and use Sarrus’ rule (Lemma 5.11) to compute the \(3 \times 3\)-determinants:
\[
\begin{align*}
\det A
&= (+3) \det \left ( \begin{array}{ccc} 2 & 1 & 4 \\ 1 & 2 & -3 \\ 0 & 5 & 1 \end{array} \right )
+ (+2) \det \left ( \begin{array}{ccc} 2 & 1 & 4 \\ -1 & 0 & 2 \\ 1 & 2 & -3 \end{array} \right ) \\
&= 3 \cdot 47 + 2 \cdot 26 \\
&= 141 + 52 \\
&= 193.
\end{align*}
\]
Solution 5.6.5
(See Exercise 5.5.) According to Proposition 5.20, the determinant is the product of the diagonal entries, i.e., it equals \(3 \cdot 4 \cdot 5 = 60\).
Solution 5.6.6
(See Exercise 5.6.) The matrix \(\left ( \begin{array}{ccc} 1 & 5 & 8 \\ 40 & -9 & 1 \\ 0 & 0 & 0 \end{array} \right )\) has a zero row, so the determinant is \(0\) (cf. Remark 5.7). The second matrix, \(\left ( \begin{array}{ccc} 1 & 5 & 8 \\ 40 & -9 & 1 \\ 1 & 5 & 8 \end{array} \right )\) has two equal rows, so the determinant is \(0\) (cf. Remark 5.9).
Equivalently, both matrices are non-invertible (their rows are linearly dependent), and by Theorem 5.15 their determinants must be zero.
Solution 5.6.7
(See Exercise 5.7.) The first matrix
\[
A = \left ( \begin{array}{cccc} 3 & 26 & -9 & 3 \\ 0 & 3 & 1 & 28 \\ 0 & 0 & 2 & 71 \\ 0 & 0 & 0 & 3 \end{array} \right ),
\]
is an upper triangular matrix, so we can use Proposition 5.20 to compute its determinant:
\[
\det A = 3 \cdot 3 \cdot 2 \cdot 3 = 54.
\]
For the second matrix,
\[
B = \left ( \begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array} \right ),
\]
we use Sarrus’ rule (Lemma 5.11):
\[
\begin{align*}
\det B
&= 1\cdot 5\cdot 9 + 2\cdot 6\cdot 7 + 3\cdot 4\cdot 8
- 3\cdot 5\cdot 7 - 2\cdot 4\cdot 9 - 1\cdot 6\cdot 8 \\
&= 45 + 84 + 96 - 105 - 72 - 48 \\
&= 0.
\end{align*}
\]
Hence the determinants are \(54\) and \(0\).
Solution 5.6.8
(See Exercise 5.8.) By Theorem 5.15 and , \(A^{-1} = \frac{1}{\det A}\,\mathrm{adj}(A).\) For
\[
A_1 = \left ( \begin{array}{cc} 10 & 9 \\ 11 & 10 \end{array} \right ),
\]
we have
\[
\det A_1 = 10\cdot 10 - 9\cdot 11 = 1,
\]
and by the \(2\times2\) inverse formula (cf. Example 5.17)
\[
A_1^{-1} = \left ( \begin{array}{cc} 10 & -9 \\ -11 & 10 \end{array} \right ).
\]
For
\[
A_2 = \left ( \begin{array}{ccc} 5 & 2 & -1 \\ 0 & 0 & 1 \\ 6 & 2 & 3 \end{array} \right ),
\]
we have \(\det A_2 = 2 \ne 0\), and a brief cofactor computation gives
\[
\mathrm{adj}(A_2)=\left ( \begin{array}{ccc} -2 & -8 & 2 \\ 6 & 21 & -5 \\ 0 & 2 & 0 \end{array} \right ).
\]
Hence
\[
A_2^{-1}=\frac12\left ( \begin{array}{ccc} -2 & -8 & 2 \\ 6 & 21 & -5 \\ 0 & 2 & 0 \end{array} \right )
= \left ( \begin{array}{ccc} -1 & -4 & 1 \\ 3 & \frac{21}{2} & -\frac52 \\ 0 & 1 & 0 \end{array} \right ).
\]
This exercise can also be solved without using the adjugate formula, as follows: by Theorem 4.80 (similarly to Example 4.81), we can compute \(A^{-1}\) by reducing \((A \ | \ {\mathrm{id}})\) to \(({\mathrm{id}} \ | \ A^{-1})\). For \(A_2\):
\[
\begin{align*}
\left ( \begin{array}{ccc|ccc} 5 & 2 & -1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 6 & 2 & 3 & 0 & 0 & 1 \end{array} \right )
&\leadsto
\left ( \begin{array}{ccc|ccc} 5 & 2 & -1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & -\frac25 & \frac{21}{5} & -\frac65 & 0 & 1 \end{array} \right ) \\
&\leadsto
\left ( \begin{array}{ccc|ccc} 5 & 2 & -1 & 1 & 0 & 0 \\ 0 & -\frac25 & \frac{21}{5} & -\frac65 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 & 0 \end{array} \right ) \\
&\leadsto
\left ( \begin{array}{ccc|ccc} 5 & 2 & -1 & 1 & 0 & 0 \\ 0 & 1 & -\frac{21}{2} & 3 & 0 & -\frac52 \\ 0 & 0 & 1 & 0 & 1 & 0 \end{array} \right ) \\
&\leadsto
\left ( \begin{array}{ccc|ccc} 1 & 0 & 0 & -1 & -4 & 1 \\ 0 & 1 & 0 & 3 & \frac{21}{2} & -\frac52 \\ 0 & 0 & 1 & 0 & 1 & 0 \end{array} \right ).
\end{align*}
\]
Therefore,
\[
A_2^{-1}=\left ( \begin{array}{ccc} -1 & -4 & 1 \\ 3 & \frac{21}{2} & -\frac52 \\ 0 & 1 & 0 \end{array} \right ),
\]
in agreement with Method 1.