Exercise 4.4 · Q16: Verify $A^3-6A^2+9A-4I=O$⁠; hence find $A^{-1}$

Basics you need

Solution

1. Compute $A^2 = A\cdot A$

   $A$ is $3\times3$⁠, so $A\cdot A$ is $3\times3$⁠. Because $A$ is symmetric, its columns are the same as its rows: $(2,-1,1)$⁠, $(-1,2,-1)$⁠, $(1,-1,2)$⁠.

   $$\begin{aligned} c_{11} &= 2\cdot2 + (-1)(-1) + 1\cdot1 = 6\\ c_{12} &= 2(-1) + (-1)2 + 1(-1) = -5\\ c_{13} &= 2\cdot1 + (-1)(-1) + 1\cdot2 = 5\\[1mm] c_{21} &= (-1)2 + 2(-1) + (-1)1 = -5\\ c_{22} &= (-1)(-1) + 2\cdot2 + (-1)(-1) = 6\\ c_{23} &= (-1)1 + 2(-1) + (-1)2 = -5\\[1mm] c_{31} &= 1\cdot2 + (-1)(-1) + 2\cdot1 = 5\\ c_{32} &= 1(-1) + (-1)2 + 2(-1) = -5\\ c_{33} &= 1\cdot1 + (-1)(-1) + 2\cdot2 = 6 \end{aligned}$$ $$A^2 = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix}$$
   The result is symmetric ($c_{12} = c_{21}$⁠, $c_{13} = c_{31}$⁠, $c_{23} = c_{32}$⁠), as Basic 3 says it must be. That is a free check on three of the nine entries.
   Pair up, multiply, add. Basics: matrix facts the chapter uses

2. Compute $A^3 = A^2\cdot A$

   Rows of $A^2$ times the columns of $A$⁠:

   $$\begin{aligned} d_{11} &= 6\cdot2 + (-5)(-1) + 5\cdot1 = 22\\ d_{12} &= 6(-1) + (-5)2 + 5(-1) = -21\\ d_{13} &= 6\cdot1 + (-5)(-1) + 5\cdot2 = 21\\[1mm] d_{21} &= (-5)2 + 6(-1) + (-5)1 = -21\\ d_{22} &= (-5)(-1) + 6\cdot2 + (-5)(-1) = 22\\ d_{23} &= (-5)1 + 6(-1) + (-5)2 = -21\\[1mm] d_{31} &= 5\cdot2 + (-5)(-1) + 6\cdot1 = 21\\ d_{32} &= 5(-1) + (-5)2 + 6(-1) = -21\\ d_{33} &= 5\cdot1 + (-5)(-1) + 6\cdot2 = 22 \end{aligned}$$ $$A^3 = \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix}$$
   Again symmetric, as expected. Associativity means $A^2A = AA^2$⁠, so it does not matter which side the extra $A$ goes on.

3. Combine $A^3 - 6A^2 + 9A - 4I$

   $$6A^2 = \begin{bmatrix} 36 & -30 & 30 \\ -30 & 36 & -30 \\ 30 & -30 & 36 \end{bmatrix}$$ $$9A = \begin{bmatrix} 18 & -9 & 9 \\ -9 & 18 & -9 \\ 9 & -9 & 18 \end{bmatrix}$$

   $4I$ subtracts $4$ from the diagonal only. Entry by entry ($A^3$⁠, minus $6A^2$⁠, plus $9A$⁠, minus $4I$⁠):

   $$\begin{aligned} (1,1)&:\ 22 - 36 + 18 - 4 = 0\\ (1,2)&:\ -21 + 30 - 9 - 0 = 0\\ (1,3)&:\ 21 - 30 + 9 - 0 = 0\\[1mm] (2,1)&:\ -21 + 30 - 9 - 0 = 0\\ (2,2)&:\ 22 - 36 + 18 - 4 = 0\\ (2,3)&:\ -21 + 30 - 9 - 0 = 0\\[1mm] (3,1)&:\ 21 - 30 + 9 - 0 = 0\\ (3,2)&:\ -21 + 30 - 9 - 0 = 0\\ (3,3)&:\ 22 - 36 + 18 - 4 = 0 \end{aligned}$$

   So $A^3 - 6A^2 + 9A - 4I = O$, verified.

   All nine entries must be zero for the sum to be the zero matrix $O$⁠; one non-zero entry would make the statement false. The pattern repeats (all diagonal entries alike, and all off-diagonal entries alike apart from sign), so the nine lines are really three calculations, but every position is still checked.
   Basics: scaling and adding matrices

4. $|A| \neq 0$⁠, so $A^{-1}$ exists

   No entry is zero; expand along row 1. Signs $+, -, +$⁠.

   $$\begin{aligned} M_{11} &= \begin{vmatrix} 2 & -1 \\ -1 & 2 \end{vmatrix}\\ &= 2\cdot2-(-1)(-1) = 3\\[1mm] M_{12} &= \begin{vmatrix} -1 & -1 \\ 1 & 2 \end{vmatrix}\\ &= (-1)2-(-1)1 = -1\\[1mm] M_{13} &= \begin{vmatrix} -1 & 2 \\ 1 & -1 \end{vmatrix}\\ &= (-1)(-1)-2\cdot1 = -1 \end{aligned}$$ $$\begin{aligned} |A| &= 2\cdot3 - (-1)(-1) + 1\cdot(-1)\\ &= 6 - 1 - 1 = 4 \end{aligned}$$

   $|A| = 4 \neq 0$⁠, so $A^{-1}$ exists.

   Multiplying the equation by $A^{-1}$ in the next step needs $A^{-1}$ to exist (Theorem 4). The middle term is $a_{12}$ times its cofactor, $(-1)\cdot(-M_{12})$⁠, which is $(-1)\cdot1 = -1$⁠; writing it as $-a_{12}M_{12} = -(-1)(-1)$ gives the same $-1$⁠.
   Lesson: expanding a 3 × 3 determinant

5. Multiply the equation by $A^{-1}$

   Post-multiply every term of $AAA - 6AA + 9A - 4I = O$ by $A^{-1}$⁠, using $AA^{-1} = I$⁠:

   $$\begin{aligned} AA(AA^{-1}) - 6A(AA^{-1}) &\\ + 9AA^{-1} - 4IA^{-1} &= OA^{-1}\\ A^2 - 6A + 9I - 4A^{-1} &= O \end{aligned}$$ $$A^{-1} = \frac{1}{4}\left(A^2 - 6A + 9I\right)$$
   Distribute over each term, all on the right, and regroup by associativity, for example $(AAA)A^{-1}$ as $AA(AA^{-1})$⁠; then $AA^{-1} = I$⁠, $XI = X$⁠, $IA^{-1} = A^{-1}$⁠, $OA^{-1} = O$⁠. Moving $-4A^{-1}$ to the other side gives $4A^{-1} = A^2 - 6A + 9I$⁠, and we divide by the number 4.
   Basics: why the inverse divides by |A|

6. Compute $A^2 - 6A + 9I$ and $A^{-1}$

   $6A$ has rows $12, -6, 6$ / $-6, 12, -6$ / $6, -6, 12$⁠. Entry by entry ($A^2$⁠, minus $6A$⁠, plus $9I$ on the diagonal):

   $$\begin{aligned} (1,1)&:\ 6 - 12 + 9 = 3\\ (1,2)&:\ -5 + 6 + 0 = 1\\ (1,3)&:\ 5 - 6 + 0 = -1\\[1mm] (2,1)&:\ -5 + 6 + 0 = 1\\ (2,2)&:\ 6 - 12 + 9 = 3\\ (2,3)&:\ -5 + 6 + 0 = 1\\[1mm] (3,1)&:\ 5 - 6 + 0 = -1\\ (3,2)&:\ -5 + 6 + 0 = 1\\ (3,3)&:\ 6 - 12 + 9 = 3 \end{aligned}$$ $$A^{-1} = \frac{1}{4}\begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}$$
   Cross-check with the adjoint: $A^{-1} = \frac{1}{|A|}\operatorname{adj}A = \frac14\operatorname{adj}A$⁠, so the integer matrix must be $\operatorname{adj}A$⁠. Its first row should be the cofactors of column 1: $A_{11} = M_{11} = 3$⁠, $A_{21} = -M_{21}$ with $M_{21} = (-1)2 - 1(-1) = -1$⁠, so $A_{21} = 1$⁠, and $A_{31} = M_{31}$ with $M_{31} = (-1)(-1) - 1\cdot2 = -1$⁠. That is $3, 1, -1$ ✓.

7. Check: $AA^{-1} = I$

   Multiply $A$ by the integer matrix, then divide by 4:

   $$\begin{aligned} (1,1)&:\ 2\cdot3 + (-1)1 + 1(-1) = 4\\ (1,2)&:\ 2\cdot1 + (-1)3 + 1\cdot1 = 0\\ (1,3)&:\ 2(-1) + (-1)1 + 1\cdot3 = 0\\[1mm] (2,1)&:\ (-1)3 + 2\cdot1 + (-1)(-1) = 0\\ (2,2)&:\ (-1)1 + 2\cdot3 + (-1)1 = 4\\ (2,3)&:\ (-1)(-1) + 2\cdot1 + (-1)3 = 0\\[1mm] (3,1)&:\ 1\cdot3 + (-1)1 + 2(-1) = 0\\ (3,2)&:\ 1\cdot1 + (-1)3 + 2\cdot1 = 0\\ (3,3)&:\ 1(-1) + (-1)1 + 2\cdot3 = 4 \end{aligned}$$ $$AA^{-1} = \frac{1}{4}\begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} = I$$
   Before dividing, the product is $4I = |A|\,I$⁠, exactly $A(\operatorname{adj}A) = |A|\,I$ (Theorem 1), confirming the integer matrix is $\operatorname{adj}A$⁠.

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