Example 18: Show $A^3-23A-40I=O$

Basics you need

Solution

1. Check that every term makes sense

   $A$ is square of order 3. So $AA$ is $(3\times3)(3\times3)$⁠, defined and $3\times3$⁠, and likewise $A^3=A\cdot A^2$ is $3\times3$⁠. The matrices $23A$ and $40I$ are $3\times3$ too, so the sum $A^3-23A-40I$ is defined and is $3\times3$⁠. Here $I=I_3$ and $O$ is the $3\times3$ zero matrix.

   Only square matrices have powers: for $AA$ we need columns of $A$ = rows of $A$⁠. The orders of $I$ and $O$ are not written, so we read them from the context: they must match the other terms.
   Identity matrix and zero matrix, Section 3.3 (vi) and (vii). Lesson: types of matrices

2. Compute $A^2=AA$

   Rows of $A$⁠: $[1\ \ 2\ \ 3]$⁠, $[3\ \ {-2}\ \ 1]$⁠, $[4\ \ 2\ \ 1]$⁠. Columns of $A$⁠: $(1,3,4)$⁠, $(2,-2,2)$⁠, $(3,1,1)$⁠. Write $A^2=[u_{ik}]$⁠:

   $$\begin{aligned} u_{11} &= 1\cdot 1 + 2\cdot 3 + 3\cdot 4\\ &= 1 + 6 + 12 = 19\\ u_{12} &= 1\cdot 2 + 2\cdot(-2) + 3\cdot 2\\ &= 2 - 4 + 6 = 4\\ u_{13} &= 1\cdot 3 + 2\cdot 1 + 3\cdot 1\\ &= 3 + 2 + 3 = 8\\ u_{21} &= 3\cdot 1 + (-2)\cdot 3 + 1\cdot 4\\ &= 3 - 6 + 4 = 1\\ u_{22} &= 3\cdot 2 + (-2)\cdot(-2) + 1\cdot 2\\ &= 6 + 4 + 2 = 12\\ u_{23} &= 3\cdot 3 + (-2)\cdot 1 + 1\cdot 1\\ &= 9 - 2 + 1 = 8\\ u_{31} &= 4\cdot 1 + 2\cdot 3 + 1\cdot 4\\ &= 4 + 6 + 4 = 14\\ u_{32} &= 4\cdot 2 + 2\cdot(-2) + 1\cdot 2\\ &= 8 - 4 + 2 = 6\\ u_{33} &= 4\cdot 3 + 2\cdot 1 + 1\cdot 1\\ &= 12 + 2 + 1 = 15 \end{aligned}$$ $$A^2 = \begin{bmatrix} 19 & 4 & 8 \\ 1 & 12 & 8 \\ 14 & 6 & 15 \end{bmatrix}$$
   $A^2$ is not the matrix of squared entries: its $(1,1)$ entry is 19, not $1^2=1$⁠. Each entry is a full row-times-column sum.
   Definition of the product, Section 3.4.5. Basics: pair up, multiply, add

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

   Rows of $A$ as before. Columns of $A^2$⁠: $(19,1,14)$⁠, $(4,12,6)$⁠, $(8,8,15)$⁠. Write $A^3=[v_{ik}]$⁠:

   $$\begin{aligned} v_{11} &= 1\cdot 19 + 2\cdot 1 + 3\cdot 14\\ &= 19 + 2 + 42 = 63\\ v_{12} &= 1\cdot 4 + 2\cdot 12 + 3\cdot 6\\ &= 4 + 24 + 18 = 46\\ v_{13} &= 1\cdot 8 + 2\cdot 8 + 3\cdot 15\\ &= 8 + 16 + 45 = 69\\ v_{21} &= 3\cdot 19 + (-2)\cdot 1 + 1\cdot 14\\ &= 57 - 2 + 14 = 69\\ v_{22} &= 3\cdot 4 + (-2)\cdot 12 + 1\cdot 6\\ &= 12 - 24 + 6 = -6\\ v_{23} &= 3\cdot 8 + (-2)\cdot 8 + 1\cdot 15\\ &= 24 - 16 + 15 = 23\\ v_{31} &= 4\cdot 19 + 2\cdot 1 + 1\cdot 14\\ &= 76 + 2 + 14 = 92\\ v_{32} &= 4\cdot 4 + 2\cdot 12 + 1\cdot 6\\ &= 16 + 24 + 6 = 46\\ v_{33} &= 4\cdot 8 + 2\cdot 8 + 1\cdot 15\\ &= 32 + 16 + 15 = 63 \end{aligned}$$ $$A^3 = \begin{bmatrix} 63 & 46 & 69 \\ 69 & -6 & 23 \\ 92 & 46 & 63 \end{bmatrix}$$
   We reuse $A^2$ rather than multiplying three matrices from scratch. Multiplying $A$ on the left of $A^2$ or on the right gives the same $A^3$⁠, by the associative law (Basic 4); the textbook uses $A\cdot A^2$⁠, and the check box below uses the other order.

4. Form $-23A$ and $-40I$

   Multiply every entry of $A$ by $-23$⁠, and every entry of $I$ by $-40$⁠:

   $$-23A = \begin{bmatrix} -23 & -46 & -69 \\ -69 & 46 & -23 \\ -92 & -46 & -23 \end{bmatrix}$$ $$-40I = \begin{bmatrix} -40 & 0 & 0 \\ 0 & -40 & 0 \\ 0 & 0 & -40 \end{bmatrix}$$
   Writing the subtraction as addition of negatives, $A^3+(-23A)+(-40I)$⁠, lets us add three matrices entry by entry without sign confusion. Note $-23\cdot(-2)=+46$ in the middle of $-23A$⁠. Since $I$ is zero off the diagonal, $-40I$ only affects the three diagonal entries.
   Scalar multiple $kA=[k\,a_{ij}]$ and the difference $X-Y=X+(-1)Y$⁠, Section 3.4.2. Lesson: addition and scalar multiplication

5. Add the three matrices entry by entry

   Write $A^3-23A-40I=[w_{ij}]$⁠, with $w_{ij}=v_{ij}-23a_{ij}-40\,\delta_{ij}$⁠, where $\delta_{ij}$ is 1 on the diagonal and 0 off it (the entries of $I$⁠):

   $$\begin{aligned} w_{11} &= 63 - 23 - 40 = 0\\ w_{12} &= 46 - 46 + 0 = 0\\ w_{13} &= 69 - 69 + 0 = 0\\ w_{21} &= 69 - 69 + 0 = 0\\ w_{22} &= -6 + 46 - 40 = 0\\ w_{23} &= 23 - 23 + 0 = 0\\ w_{31} &= 92 - 92 + 0 = 0\\ w_{32} &= 46 - 46 + 0 = 0\\ w_{33} &= 63 - 23 - 40 = 0 \end{aligned}$$ $$A^3 - 23A - 40I = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O$$
   Every one of the nine entries is 0, which is exactly what "$=O$" means. Off the diagonal, each entry of $A^3$ is exactly 23 times the matching entry of $A$ (for example $46=23\cdot2$ and $92=23\cdot4$⁠), so $-23A$ cancels it. On the diagonal, $v_{ii}-23a_{ii}$ is 40 each time ($63-23$⁠, $-6+46$⁠, $63-23$⁠), and $-40I$ cancels that.
   Equality with the zero matrix: every entry must be 0. Lesson: equality of matrices

6. What the identity tells us: an inverse for $A$ (beyond the textbook)

   This step is not part of the textbook's solution and is not needed for the answer. Rearranging, $A^3-23A=40I$⁠. The left side factors as $A(A^2-23I)$⁠, because $A\cdot A^2=A^3$ and $A\cdot(23I)=23A$⁠. So

   $$A\cdot\tfrac{1}{40}\big(A^2-23I\big) = I.$$

   Using the $A^2$ from Step 2, $A^2-23I$ subtracts 23 from each diagonal entry:

   $$A^2 - 23I = \begin{bmatrix} -4 & 4 & 8 \\ 1 & -11 & 8 \\ 14 & 6 & -8 \end{bmatrix}$$

   So the inverse of $A$ is this matrix divided by 40. (The same factoring on the other side, $(A^2-23I)A=A^3-23A$⁠, gives the product in the other order too.)

   This shows why such identities matter: a polynomial equation satisfied by $A$ with a non-zero constant term lets us write down $A^{-1}$ without any other method. For instance row 1 of $A$ times column 1 of $A^2-23I$ is $1\cdot(-4)+2\cdot1+3\cdot14=40$⁠, which divided by 40 gives the 1 of $I$⁠.
   Inverse of a square matrix, Section 3.7. Lesson: invertible matrices

Every fact used, and where it comes from