Section 10.4 Examples
Subsection 10.4.1 The \(2\times 2\) case
Let’s compute the adjoint of the general \(2\times 2\) matrix \(A = \left[\begin{smallmatrix} a \amp b \\ c \amp d \end{smallmatrix}\right] \text{.}\) First, the minors.
Minor determinants for \(2 \times 2\) matrices
In the matrix of cofactors for a \(2 \times 2\) matrix, the off-diagonal cofactors become negative, and then the adjoint is the transpose of that.
\begin{align*}
C_A \amp = \left[\begin{array}{rr} d \amp -c\\-b \amp a \end{array}\right] \amp
\adj A \amp = \left[\begin{array}{rr} d \amp -b\\-c \amp a \end{array}\right]
\end{align*}
The inverse of \(A\) is then the reciprocal of the determinant times the adjoint, so that
\begin{equation*}
\inv{A} = \frac{1}{a d - b c} \, \left[\begin{array}{rr} d \amp -b \\ -c \amp a \end{array}\right] \text{,}
\end{equation*}
as promised in Proposition 5.5.4.
Subsection 10.4.2 Computing an inverse using the adjoint
As mentioned, using the adjoint to compute the inverse of a matrix is not very efficient for matrices larger than \(2\times 2\text{.}\) In most cases, you are better off using the row reduction method. However, there are situations where you might want to use the adjoint instead, as in the example below.
Example 10.4.1. Using the adjoint to compute an inverse.
It can be tedious to row reduce a matrix with variable entries. Consider
\begin{equation*}
X = \begin{bmatrix} x \amp 1 \amp -1 \\ x-1 \amp 0 \amp x \\ 0 \amp x \amp 1 \end{bmatrix} \text{.}
\end{equation*}
To row reduce \(X\text{,}\) our first step would be to obtain a leading one in the first column. We might choose to perform \(R_1\to\frac{1}{x}R_1\text{,}\) except that this operation would be invalid in the case that \(x=0\text{.}\) Or we might choose to perform \(R_2\to\frac{1}{x-1}R_2\text{,}\) except that this operation would be invalid in the case that \(x=1\text{.}\) So to row reduce \(X\) we would need to consider three different cases, \(x=0\text{,}\) \(x=1\text{,}\) and \(x\neq 0,1\text{,}\) performing different row reduction sequences in each of these cases. And when we get to the point of trying to obtain a leading one in the second column, we might discover there are even more cases to consider.
So instead we will attempt to compute the inverse of \(X\) using the adjoint. First, the minors.
\begin{align*}
M_{11} \amp=
\left\lvert\begin{smallmatrix} 0 \amp x \\ x \amp 1 \end{smallmatrix}\right\rvert = -x^2
\amp
M_{12} \amp=
\left\lvert\begin{smallmatrix} x-1 \amp x\\ 0 \amp 1 \end{smallmatrix}\right\rvert = x-1
\amp
M_{13} \amp=
\left\lvert\begin{smallmatrix} x-1 \amp 0\\ 0 \amp x \end{smallmatrix}\right\rvert = x^2-x\\
M_{21} \amp=
\left\lvert\begin{smallmatrix} 1 \amp -1 \\ x \amp 1 \end{smallmatrix}\right\rvert = 1+x
\amp
M_{22} \amp=
\left\lvert\begin{smallmatrix} x \amp -1 \\ 0 \amp 1 \end{smallmatrix}\right\rvert = x
\amp
M_{23} \amp=
\left\lvert\begin{smallmatrix} x \amp 1 \\ 0 \amp x \end{smallmatrix}\right\rvert = x^2\\
M_{31} \amp=
\left\lvert\begin{smallmatrix} 1 \amp -1 \\ 0 \amp x \end{smallmatrix}\right\rvert = x
\amp
M_{32} \amp=
\left\lvert\begin{smallmatrix} x \amp -1 \\ x-1 \amp x \end{smallmatrix}\right\rvert = x^2+x-1
\amp
M_{33} \amp=
\left\lvert\begin{smallmatrix} x \amp 1 \\ x-1 \amp 0 \end{smallmatrix}\right\rvert = 1-x
\end{align*}
We obtain the matrix of cofactors by making certain minor determinants negative, according to the \(3\times 3\) pattern of cofactor signs, and then the adjoint is the transpose.
\begin{align*}
C_X \amp= \begin{bmatrix}
-x^2 \amp 1-x \amp x^2-x \\
-1-x \amp x \amp -x^2 \\
x \amp -x^2-x+1 \amp 1-x
\end{bmatrix}\\
\\
\adj X \amp= \begin{bmatrix}
-x^2 \amp -1-x \amp x \\
1-x \amp x \amp -x^2-x+1 \\
x^2-x \amp -x^2 \amp 1-x
\end{bmatrix}
\end{align*}
To compute the inverse of \(X\text{,}\) we still need its determinant. But we already have all the cofactors, so a cofactor expansion will be easy. Let’s do a cofactor expansion of \(\det X\) along the third row. (Remember that the cofactors already have the appropriate signs, so we are just summing “entry times cofactor” terms.)
\begin{equation*}
\det X = 0x + x(-x^2-x+1) + 1(1-x) = 1 - x^3 - x^2
\end{equation*}
Finally, we obtain a formula for the inverse of \(X\) that is valid for every value of \(x\) for which the determinant is nonzero,
\begin{equation*}
\inv{X} =
\frac{1}{1 - x^3 - x^2}
\begin{bmatrix}
-x^2 \amp -1-x \amp x \\
1-x \amp x \amp -x^2-x+1 \\
x^2-x \amp -x^2 \amp 1-x
\end{bmatrix}.
\end{equation*}
Subsection 10.4.3 Cramer’s rule
Example 10.4.2. Using Cramer’s rule to compute individual variable values in a system of equations.
Consider the system
\begin{equation*}
\left\{\begin{array}{rcrcrcrcr}
x_1 \amp - \amp x_2 \amp + \amp 2x_3 \amp + \amp x_4 \amp = \amp 1,\\
2x_1 \amp \amp \amp + \amp x_3 \amp + \amp x_4 \amp = \amp 1,\\
\amp \amp x_2 \amp \amp \amp - \amp 3x_4 \amp = \amp 0,\\
x_1 \amp - \amp 2x_2 \amp - \amp x_3 \amp \amp \amp = \amp 1,\\
\end{array}\right.
\end{equation*}
with coefficient matrix and vector of constants,
\begin{align*}
A \amp= \left[\begin{array}{rrrr}
1 \amp -1 \amp 2 \amp 1 \\
2 \amp 0 \amp 1 \amp 1 \\
0 \amp 1 \amp 0 \amp -3 \\
1 \amp -2 \amp -1 \amp 0
\end{array}\right],
\amp
\uvec{b} \amp = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 1 \end{bmatrix}.
\end{align*}
Conveniently, we have already computed \(\det A = -27\) in Example 8.4.8 (and again in Example 9.3.1). Since \(\det A \neq 0\text{,}\) we know that \(A\) is invertible and so the system has one unique solution. Suppose we want to know the value of \(x_2\) in the solution. We can form the matrix \(A_2\text{,}\) where the second column of \(A\) is replaced by \(\uvec{b}\text{,}\)
\begin{equation*}
A_2 = \left[\begin{array}{rrrr}
1 \amp 1 \amp 2 \amp 1 \\
2 \amp 1 \amp 1 \amp 1 \\
0 \amp 0 \amp 0 \amp -3 \\
1 \amp 1 \amp -1 \amp 0
\end{array}\right]\text{,}
\end{equation*}
and compute \(\det A_2\) by a cofactor expansion along the third row (expanding the corresponding \(3\times 3\) minor determinant along the first row),
\begin{align*}
\det A_2 \amp
= -(-3)\left\lvert\begin{array}{rrr}
1 \amp 1 \amp 2 \\
2 \amp 1 \amp 1 \\
1 \amp 1 \amp -1
\end{array}\right\rvert\\
\amp = 3\left(
1 \left\lvert\begin{array}{rr} 1 \amp 1 \\ 1 \amp -1 \end{array}\right\rvert
- 1 \left\lvert\begin{array}{rr} 2 \amp 1 \\ 1 \amp -1 \end{array}\right\rvert
+ 2 \left\lvert\begin{array}{rr} 2 \amp 1 \\ 1 \amp 1 \end{array}\right\rvert
\right)\\
\amp = 3\bbrac{(-1-1)-(-2-1)+2(2-1)}\\
\amp = 9\text{.}
\end{align*}
Thus, the value of \(x_2\) in the one unique solution to the system is
\begin{equation*}
x_2 = \frac{\det A_2}{\det A} = \frac{9}{-27} = -\frac{1}{3} \text{.}
\end{equation*}
If we also want to know the value of \(x_4\text{,}\) we form the matrix \(A_4\text{,}\) where the fourth column of \(A\) is replaced by \(\uvec{b}\text{,}\)
\begin{equation*}
A_4 = \left[\begin{array}{rrrr}
1 \amp -1 \amp 2 \amp 1 \\
2 \amp 0 \amp 1 \amp 1 \\
0 \amp 1 \amp 0 \amp 0 \\
1 \amp -2 \amp -1 \amp 1
\end{array}\right],
\end{equation*}
and compute \(\det A_4\) (again by a cofactor expansion along the third row, followed by an expansion along the first row of the corresponding \(3\times 3\) minor determinant),
\begin{align*}
\det A_4 \amp
= -1 \left[\begin{array}{rrr}
1 \amp 2 \amp 1 \\
2 \amp 1 \amp 1 \\
1 \amp -1 \amp 1
\end{array}\right]\\
\amp = -1\left(
1 \left[\begin{array}{rr} 1 \amp 1 \\ -1 \amp 1 \end{array}\right]
- 2 \left[\begin{array}{rr} 2 \amp 1 \\ 1 \amp 1 \end{array}\right]
+ 1 \left[\begin{array}{rr} 2 \amp 1 \\ 1 \amp -1 \end{array}\right]
\right)\\
\amp = -\bbrac{(1+1) - 2(2-1) + (-2-1)}\\
\amp = 3.
\end{align*}
Thus, the value of \(x_4\) in the one unique solution to the system is
\begin{equation*}
x_4 = \frac{\det A_4}{\det A} = \frac{3}{-27} = -\frac{1}{9} \text{.}
\end{equation*}
