Exercise 4.4 · Q4: Verify $A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I$⁠, $3\times3$

Basics you need

Solution

1. Find $|A|$ by expanding along column 2

   Column 2 is $(-1,\ 0,\ 0)$⁠: two of its three entries are 0. The signs down column 2 are $-,\ +,\ -$ (from the pattern below), so

   $$\begin{matrix} + & - & + \\ - & + & - \\ + & - & + \end{matrix}$$ $$\begin{aligned} |A| &= -a_{12}M_{12} + a_{22}M_{22} - a_{32}M_{32} \\ &= -(-1)\,M_{12} + 0\cdot M_{22} - 0\cdot M_{32} \\ &= 1\cdot\begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix} \\ &= 3\cdot 3 - (-2)\cdot 1 = 9 + 2 = 11 \end{aligned}$$

   So $|A| = 11$⁠.

   Any row or column gives the same determinant (Section 4.2.3), so we choose the one that makes the work smallest. With two zeros in column 2, only one minor has to be computed: the terms $0\cdot M_{22}$ and $0\cdot M_{32}$ vanish whatever those minors are. $M_{12}$ is found by deleting row 1 and column 2, which leaves rows $(3,\ -2)$ and $(1,\ 3)$⁠.
   Remarks (i) and (ii) of Section 4.2.3: expand along the line with most zeros, and use $+1$ or $-1$ according as $i+j$ is even or odd. Lesson: expanding a 3 × 3 determinant

2. Cofactors of the row 1 entries $1,\ -1,\ 2$

   Delete row 1 (leaving rows $(3,\ 0,\ -2)$ and $(1,\ 0,\ 3)$⁠) and one column at a time:

   $$\begin{aligned} M_{11} &= \begin{vmatrix} 0 & -2 \\ 0 & 3 \end{vmatrix} = 0\cdot 3 - (-2)\cdot 0 \\ &= 0 - 0 = 0 \\ A_{11} &= (-1)^{1+1}\,(0) = 0 \\ M_{12} &= \begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix} = 3\cdot 3 - (-2)\cdot 1 \\ &= 9 + 2 = 11 \\ A_{12} &= (-1)^{1+2}\,(11) = -11 \\ M_{13} &= \begin{vmatrix} 3 & 0 \\ 1 & 0 \end{vmatrix} = 3\cdot 0 - 0\cdot 1 \\ &= 0 - 0 = 0 \\ A_{13} &= (-1)^{1+3}\,(0) = 0 \end{aligned}$$
   $M_{11}$ and $M_{13}$ are 0 because each keeps column 2 of $A$ below row 1, which is $(0,\ 0)$⁠: a $2\times2$ determinant with a zero column is 0. $M_{12}$ is the minor we already met in Step 1, and $A_{12}=-11$ is negative because position $(1,2)$ carries a minus sign.
   Definitions 1 and 2, Section 4.4. Lesson: minors and cofactors

3. Cofactors of the row 2 entries $3,\ 0,\ -2$

   Delete row 2 (leaving rows $(1,\ -1,\ 2)$ and $(1,\ 0,\ 3)$⁠):

   $$\begin{aligned} M_{21} &= \begin{vmatrix} -1 & 2 \\ 0 & 3 \end{vmatrix} = (-1)\cdot 3 - 2\cdot 0 \\ &= -3 - 0 = -3 \\ A_{21} &= (-1)^{2+1}\,(-3) = 3 \\ M_{22} &= \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} = 1\cdot 3 - 2\cdot 1 \\ &= 3 - 2 = 1 \\ A_{22} &= (-1)^{2+2}\,(1) = 1 \\ M_{23} &= \begin{vmatrix} 1 & -1 \\ 1 & 0 \end{vmatrix} = 1\cdot 0 - (-1)\cdot 1 \\ &= 0 + 1 = 1 \\ A_{23} &= (-1)^{2+3}\,(1) = -1 \end{aligned}$$
   Row 2 has signs $-,\ +,\ -$⁠. So the negative minor $M_{21}=-3$ gives the positive cofactor $A_{21}=3$⁠, while $M_{23}=1$ gives $A_{23}=-1$⁠. The zero entry $a_{22}$ still has a cofactor, $A_{22}=1$⁠, which we need for the adjoint.

4. Cofactors of the row 3 entries $1,\ 0,\ 3$

   Delete row 3 (leaving rows $(1,\ -1,\ 2)$ and $(3,\ 0,\ -2)$⁠):

   $$\begin{aligned} M_{31} &= \begin{vmatrix} -1 & 2 \\ 0 & -2 \end{vmatrix} = (-1)\cdot (-2) - 2\cdot 0 \\ &= 2 - 0 = 2 \\ A_{31} &= (-1)^{3+1}\,(2) = 2 \\ M_{32} &= \begin{vmatrix} 1 & 2 \\ 3 & -2 \end{vmatrix} = 1\cdot (-2) - 2\cdot 3 \\ &= -2 - 6 = -8 \\ A_{32} &= (-1)^{3+2}\,(-8) = 8 \\ M_{33} &= \begin{vmatrix} 1 & -1 \\ 3 & 0 \end{vmatrix} = 1\cdot 0 - (-1)\cdot 3 \\ &= 0 + 3 = 3 \\ A_{33} &= (-1)^{3+3}\,(3) = 3 \end{aligned}$$
   Position $(3,2)$ has $i+j=5$⁠, odd, so $M_{32}=-8$ becomes $A_{32}=8$⁠. Positions $(3,1)$ and $(3,3)$ have even $i+j$ and keep their minors.

5. The matrix of cofactors and $\operatorname{adj}A$

   $$[A_{ij}] = \begin{bmatrix} 0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3 \end{bmatrix}$$

   Transpose (row $i$ becomes column $i$⁠):

   $$\operatorname{adj}A = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}$$
   Definition 3. Column 1 of $\operatorname{adj}A$ is $(0,\ -11,\ 0)$⁠, the cofactors of row 1; column 2 is $(3,\ 1,\ -1)$⁠, the cofactors of row 2; column 3 is $(2,\ 8,\ 3)$⁠, the cofactors of row 3.
   Transpose: the $(i,j)$ entry moves to $(j,i)$⁠. Basics: transpose and identity

6. Compute $A(\operatorname{adj}A)$⁠, entry by entry

   Both factors are $3\times3$⁠, so the product is defined and is $3\times3$⁠. Write $A(\operatorname{adj}A)=[p_{ij}]$⁠. Row 1 of $A$ is $(1,\ -1,\ 2)$⁠; multiply it by each column of $\operatorname{adj}A$⁠:

   $$\begin{aligned} p_{11} &= 1\cdot 0 + (-1)\cdot (-11) + 2\cdot 0 \\ &= 0 + 11 + 0 = 11 \\ p_{12} &= 1\cdot 3 + (-1)\cdot 1 + 2\cdot (-1) \\ &= 3 - 1 - 2 = 0 \\ p_{13} &= 1\cdot 2 + (-1)\cdot 8 + 2\cdot 3 \\ &= 2 - 8 + 6 = 0 \end{aligned}$$

   Row 2 of $A$ is $(3,\ 0,\ -2)$⁠:

   $$\begin{aligned} p_{21} &= 3\cdot 0 + 0\cdot (-11) + (-2)\cdot 0 \\ &= 0 + 0 + 0 = 0 \\ p_{22} &= 3\cdot 3 + 0\cdot 1 + (-2)\cdot (-1) \\ &= 9 + 0 + 2 = 11 \\ p_{23} &= 3\cdot 2 + 0\cdot 8 + (-2)\cdot 3 \\ &= 6 + 0 - 6 = 0 \end{aligned}$$

   Row 3 of $A$ is $(1,\ 0,\ 3)$⁠:

   $$\begin{aligned} p_{31} &= 1\cdot 0 + 0\cdot (-11) + 3\cdot 0 \\ &= 0 + 0 + 0 = 0 \\ p_{32} &= 1\cdot 3 + 0\cdot 1 + 3\cdot (-1) \\ &= 3 + 0 - 3 = 0 \\ p_{33} &= 1\cdot 2 + 0\cdot 8 + 3\cdot 3 \\ &= 2 + 0 + 9 = 11 \end{aligned}$$ $$A(\operatorname{adj}A) = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$$
   Look at what each sum is. In symbols, $$\begin{aligned} p_{11} &= a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} \\ p_{12} &= a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} \end{aligned}$$ The first is row 1 times its own cofactors, the expansion of $|A|$ along row 1, so it is 11. The second is row 1 times the cofactors of row 2: that is the expansion of a determinant whose rows 1 and 2 are equal, which is 0. Every diagonal entry is of the first kind and every off-diagonal entry of the second.
   Remark after Example 10 and the Note after it, Section 4.4. Basics: matrix facts the chapter uses

7. Compute $(\operatorname{adj}A)A$⁠, entry by entry

   Now the rows come from $\operatorname{adj}A$ and the columns from $A$⁠, which are $(1,\ 3,\ 1)$⁠, $(-1,\ 0,\ 0)$ and $(2,\ -2,\ 3)$⁠. Write $(\operatorname{adj}A)A=[q_{ij}]$⁠. Row 1 of $\operatorname{adj}A$ is $(0,\ 3,\ 2)$⁠:

   $$\begin{aligned} q_{11} &= 0\cdot 1 + 3\cdot 3 + 2\cdot 1 \\ &= 0 + 9 + 2 = 11 \\ q_{12} &= 0\cdot (-1) + 3\cdot 0 + 2\cdot 0 \\ &= 0 + 0 + 0 = 0 \\ q_{13} &= 0\cdot 2 + 3\cdot (-2) + 2\cdot 3 \\ &= 0 - 6 + 6 = 0 \end{aligned}$$

   Row 2 of $\operatorname{adj}A$ is $(-11,\ 1,\ 8)$⁠:

   $$\begin{aligned} q_{21} &= (-11)\cdot 1 + 1\cdot 3 + 8\cdot 1 \\ &= -11 + 3 + 8 = 0 \\ q_{22} &= (-11)\cdot (-1) + 1\cdot 0 + 8\cdot 0 \\ &= 11 + 0 + 0 = 11 \\ q_{23} &= (-11)\cdot 2 + 1\cdot (-2) + 8\cdot 3 \\ &= -22 - 2 + 24 = 0 \end{aligned}$$

   Row 3 of $\operatorname{adj}A$ is $(0,\ -1,\ 3)$⁠:

   $$\begin{aligned} q_{31} &= 0\cdot 1 + (-1)\cdot 3 + 3\cdot 1 \\ &= 0 - 3 + 3 = 0 \\ q_{32} &= 0\cdot (-1) + (-1)\cdot 0 + 3\cdot 0 \\ &= 0 + 0 + 0 = 0 \\ q_{33} &= 0\cdot 2 + (-1)\cdot (-2) + 3\cdot 3 \\ &= 0 + 2 + 9 = 11 \end{aligned}$$ $$(\operatorname{adj}A)A = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$$
   Matrix multiplication is not commutative in general, so this second product is a genuinely separate calculation. Here the roles are column-wise: $q_{11} = A_{11}a_{11} + A_{21}a_{21} + A_{31}a_{31}$ is column 1 of $A$ times its own cofactors, $|A|$ expanded along column 1; the off-diagonal entries pair a column with the cofactors of a different column and give 0.

8. Compare with $|A|\,I$

   $$\begin{aligned} |A|\,I &= 11\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} \end{aligned}$$

   The three matrices agree in every one of the nine positions, so

   $$A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|\,I = 11I.$$
   Equal matrices must have the same order and equal corresponding entries. All three are $3\times3$ with 11 in each diagonal position and 0 elsewhere. This verifies Theorem 1 for this matrix.
   $kI$ has $k$ on the diagonal and 0 elsewhere. Basics: scaling and adding matrices

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