Exercise 4.4 · Q10: Inverse of a $3\times3$ matrix by the adjoint

Basics you need

Solution

1. Compute $|A|$ to see whether $A^{-1}$ exists

   Look for zeros. Column 1 is $1, 0, 3$ and row 2 is $0, 2, -3$⁠; each has one zero, and no row or column has two. Expand along column 1. Its signs, read down the first column of the pattern, are $+, -, +$⁠.

   $$\begin{aligned} |A| &= 1\cdot M_{11} - 0\cdot M_{21} + 3\cdot M_{31} \end{aligned}$$

   The three minors, each with its row and column deleted:

   $$\begin{aligned} M_{11} &= \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix}\\ &= 2\cdot4-(-3)(-2) = 2\\[1mm] M_{21} &= \begin{vmatrix} -1 & 2 \\ -2 & 4 \end{vmatrix}\\ &= (-1)4-2(-2) = 0\\[1mm] M_{31} &= \begin{vmatrix} -1 & 2 \\ 2 & -3 \end{vmatrix}\\ &= (-1)(-3)-2\cdot2 = -1 \end{aligned}$$

   So

   $$\begin{aligned} |A| &= 1\cdot2 - 0\cdot0 + 3\cdot(-1)\\ &= 2 - 0 - 3 = -1 \end{aligned}$$

   Since $|A| = -1 \neq 0$⁠, $A$ is non-singular, so $A^{-1}$ exists.

   The question says "if it exists", so existence must be settled first. Theorem 4 says a square matrix is invertible exactly when its determinant is non-zero. The zero in column 1 means one of the three products is $0$ whatever its minor is; we computed $M_{21}$ anyway because it is needed again for the cofactor $A_{21}$ in Step 3.
   A $2\times2$ determinant is the main diagonal product minus the other diagonal product, $ad - bc$⁠. Lesson: determinant of a square matrix

2. Cofactors of row 1

   Delete row 1 and the column of each element. Signs for row 1: $+, -, +$⁠.

   $$\begin{aligned} M_{11} &= \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} = 2 \quad (\text{Step 1})\\ A_{11} &= (-1)^{1+1}M_{11} = +2\\[2mm] M_{12} &= \begin{vmatrix} 0 & -3 \\ 3 & 4 \end{vmatrix} = 0\cdot4-(-3)\cdot3 = 9\\ A_{12} &= (-1)^{1+2}M_{12} = -9\\[2mm] M_{13} &= \begin{vmatrix} 0 & 2 \\ 3 & -2 \end{vmatrix} = 0(-2)-2\cdot3 = -6\\ A_{13} &= (-1)^{1+3}M_{13} = -6 \end{aligned}$$
   For $M_{12}$ we delete row 1 and column 2 of $A$⁠. What is left is rows 2 and 3 of columns 1 and 3: the entries $0, -3$ on top and $3, 4$ below. The cofactor then attaches the chessboard sign: position $(1,2)$ has $1+2 = 3$⁠, odd, so the sign is $-$ and $A_{12} = -9$⁠. Position $(1,3)$ has $1+3=4$⁠, even, so $A_{13} = M_{13} = -6$⁠: a plus sign keeps the minor's own sign, even when the minor is negative.
   Minor first, sign second. Mixing the two up (for example writing $A_{13} = +6$ because "the sign is +") is the most common slip. Basics: the sign pattern

3. Cofactors of row 2

   Signs for row 2: $-, +, -$⁠.

   $$\begin{aligned} M_{21} &= \begin{vmatrix} -1 & 2 \\ -2 & 4 \end{vmatrix} = 0 \quad (\text{Step 1})\\ A_{21} &= (-1)^{2+1}M_{21} = 0\\[2mm] M_{22} &= \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = 1\cdot4-2\cdot3 = -2\\ A_{22} &= (-1)^{2+2}M_{22} = -2\\[2mm] M_{23} &= \begin{vmatrix} 1 & -1 \\ 3 & -2 \end{vmatrix} = 1(-2)-(-1)3 = 1\\ A_{23} &= (-1)^{2+3}M_{23} = -1 \end{aligned}$$
   $M_{22}$ keeps the four corners of $A$ (delete the middle row and middle column). $M_{23}$ deletes row 2 and column 3, leaving columns 1 and 2 of rows 1 and 3. A minus sign on a zero minor still gives zero, so $A_{21} = 0$⁠.

4. Cofactors of row 3

   Signs for row 3: $+, -, +$⁠.

   $$\begin{aligned} M_{31} &= \begin{vmatrix} -1 & 2 \\ 2 & -3 \end{vmatrix} = -1 \quad (\text{Step 1})\\ A_{31} &= (-1)^{3+1}M_{31} = -1\\[2mm] M_{32} &= \begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix} = 1(-3)-2\cdot0 = -3\\ A_{32} &= (-1)^{3+2}M_{32} = 3\\[2mm] M_{33} &= \begin{vmatrix} 1 & -1 \\ 0 & 2 \end{vmatrix} = 1\cdot2-(-1)0 = 2\\ A_{33} &= (-1)^{3+3}M_{33} = 2 \end{aligned}$$
   A quick test that the nine cofactors are right: multiplying any row of $A$ by its own cofactors must give $|A|$⁠. Row 1: $1\cdot2 + (-1)(-9) + 2(-6)$ is $2 + 9 - 12 = -1$⁠. Row 3: $3(-1) + (-2)(3) + 4\cdot2$ is $-3 - 6 + 8 = -1$⁠. Both equal $|A|$⁠, as the Remark in Section 4.4 says they must.
   Sum of (element × its own cofactor) along any row or column is $|A|$⁠. Lesson: minors and cofactors

5. Matrix of cofactors, then transpose it to get $\operatorname{adj}A$

   Put each cofactor $A_{ij}$ in position $(i,j)$⁠:

   $$[A_{ij}] = \begin{bmatrix} 2 & -9 & -6 \\ 0 & -2 & -1 \\ -1 & 3 & 2 \end{bmatrix}$$

   The adjoint is its transpose: row 1 becomes column 1, row 2 becomes column 2, row 3 becomes column 3.

   $$\operatorname{adj}A = \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix}$$
   Why the transpose? Entry $(i,j)$ of $A(\operatorname{adj}A)$ is row $i$ of $A$ times column $j$ of $\operatorname{adj}A$⁠. After transposing, column $j$ of $\operatorname{adj}A$ holds the cofactors of row $j$⁠. So the diagonal entries are "row times its own cofactors", which is $|A|$⁠, and the off-diagonal entries are "row times another row's cofactors", which is $0$ (Note in Section 4.4). That is exactly $A(\operatorname{adj}A) = |A|\,I$ (Theorem 1). Without the transpose the rows and cofactors would be mismatched.
   Transpose: the $(i,j)$ entry moves to position $(j,i)$⁠. Basics: transpose and identity · Lesson: the adjoint

6. Divide by $|A|$

   $$A^{-1} = \frac{1}{|A|}\operatorname{adj}A = \frac{1}{-1}\begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix}$$

   Multiplying a matrix by $\frac{1}{-1} = -1$ changes the sign of every entry:

   $$A^{-1} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$$
   This is the formula from Theorem 4: since $A(\operatorname{adj}A) = |A|\,I$ and $|A|\neq0$⁠, dividing by the number $|A|$ gives $A\cdot\frac{1}{|A|}\operatorname{adj}A = I$⁠, and the same on the other side. Because $|A| = -1$⁠, every entry of the inverse is a whole number, which is unusual and handy.
   A scalar times a matrix multiplies every entry. Basics: scaling and adding matrices · Lesson: inverse of a matrix

7. Check: $A\,A^{-1} = I$

   Both matrices are $3\times3$⁠, so the product is defined and is $3\times3$⁠. Write $c_{ij}$ for entry $(i,j)$⁠: row $i$ of $A$ times column $j$ of $A^{-1}$⁠.

   $$\begin{aligned} c_{11} &= 1(-2)+(-1)(9)+2(6) = 1\\ c_{12} &= 1(0)+(-1)(2)+2(1) = 0\\ c_{13} &= 1(1)+(-1)(-3)+2(-2) = 0\\[1mm] c_{21} &= 0(-2)+2(9)+(-3)(6) = 0\\ c_{22} &= 0(0)+2(2)+(-3)(1) = 1\\ c_{23} &= 0(1)+2(-3)+(-3)(-2) = 0\\[1mm] c_{31} &= 3(-2)+(-2)(9)+4(6) = 0\\ c_{32} &= 3(0)+(-2)(2)+4(1) = 0\\ c_{33} &= 3(1)+(-2)(-3)+4(-2) = 1 \end{aligned}$$ $$A\,A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$$
   The check catches any slip in a cofactor, a sign or the transpose: a single wrong entry in $A^{-1}$ would spoil at least one $c_{ij}$⁠. For square matrices of the same order, $AB = I$ already forces $BA = I$ too (taking determinants, $|A||B| = 1$⁠, so $A$ is invertible and $B = A^{-1}$⁠), so one product is enough.
   Pair up, multiply, add. Basics: matrix facts the chapter uses

Every fact used, and where it comes from