Exercise 4.4 · Q3: Verify $A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I$⁠, $2\times2$

Basics you need

Solution

1. Compute $|A|$

   $$\begin{aligned} |A| &= \begin{vmatrix} 2 & 3 \\ -4 & -6 \end{vmatrix} = 2\cdot (-6) - 3\cdot (-4) \\ &= -12 + 12 = 0 \end{aligned}$$

   So $|A|=0$⁠: $A$ is a singular matrix.

   We need $|A|$ for the right-hand side $|A|\,I$⁠. It is 0 because the rows are proportional: row 2 is $(-4,\ -6) = -2\cdot(2,\ 3)$⁠, so the two diagonal products $2\cdot(-6)$ and $3\cdot(-4)$ are equal and cancel.
   Definition 4, Section 4.5: $A$ is singular when $|A|=0$⁠. Lesson: determinant of a square matrix

2. The four cofactors

   For each entry, delete its row and its column; the single entry left over is the minor $M_{ij}$⁠. Then multiply by $(-1)^{i+j}$⁠, which is $+$ for positions $(1,1)$⁠, $(2,2)$ and $-$ for $(1,2)$⁠, $(2,1)$⁠:

   Element	Minor	Cofactor
   $a_{11} = 2$⁠; delete row 1 and column 1	$M_{11} = -6$	$A_{11} = +(-6) = -6$
   $a_{12} = 3$⁠; delete row 1 and column 2	$M_{12} = -4$	$A_{12} = -(-4) = 4$
   $a_{21} = -4$⁠; delete row 2 and column 1	$M_{21} = 3$	$A_{21} = -(3) = -3$
   $a_{22} = -6$⁠; delete row 2 and column 2	$M_{22} = 2$	$A_{22} = +(2) = 2$

   In a $2\times2$ matrix, deleting row $i$ and column $j$ leaves one entry, the one diagonally opposite, and that number is the minor. The positions $(1,2)$ and $(2,1)$ have odd $i+j$⁠, so their minors change sign.
   Definitions 1 and 2, Section 4.4. Basics: the + − + sign pattern

3. Build $\operatorname{adj}A$

   The matrix of cofactors and its transpose:

   $$[A_{ij}] = \begin{bmatrix} -6 & 4 \\ -3 & 2 \end{bmatrix}$$ $$\operatorname{adj}A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix}$$

   Shortcut check: swap $2$ and $-6$ on the diagonal, change the signs of $3$ and $-4$⁠. This gives the same $\operatorname{adj}A$⁠.

   The adjoint is the transpose of the cofactor matrix (Definition 3). Doing it both ways, by the definition and by the Remark after Example 12, guards against a slip before we use $\operatorname{adj}A$ in two products.

4. Compute $A(\operatorname{adj}A)$

   Both factors are $2\times2$⁠, so the product is defined and is $2\times2$⁠. Write $A(\operatorname{adj}A)=[p_{ij}]$⁠; entry $p_{ij}$ is row $i$ of $A$ times column $j$ of $\operatorname{adj}A$⁠:

   $$\begin{aligned} p_{11} &= 2\cdot (-6) + 3\cdot 4 \\ &= -12 + 12 = 0 \\ p_{12} &= 2\cdot (-3) + 3\cdot 2 \\ &= -6 + 6 = 0 \\ p_{21} &= (-4)\cdot (-6) + (-6)\cdot 4 \\ &= 24 - 24 = 0 \\ p_{22} &= (-4)\cdot (-3) + (-6)\cdot 2 \\ &= 12 - 12 = 0 \end{aligned}$$ $$A(\operatorname{adj}A) = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$$
   Each product pairs a row of $A$ with a column of $\operatorname{adj}A$⁠. The diagonal entries $p_{11}, p_{22}$ are expansions of $|A|$ along a row, so they equal $|A| = 0$⁠. The off-diagonal entries pair a row with the cofactors of the other row, and such sums are always 0 (Note in Section 4.4).
   Row times column; a negative times a negative is positive: $(-4)\cdot(-6) = 24$⁠. Basics: matrix facts the chapter uses

5. Compute $(\operatorname{adj}A)A$

   Now the rows come from $\operatorname{adj}A$ and the columns from $A$⁠. Write $(\operatorname{adj}A)A=[q_{ij}]$⁠:

   $$\begin{aligned} q_{11} &= (-6)\cdot 2 + (-3)\cdot (-4) \\ &= -12 + 12 = 0 \\ q_{12} &= (-6)\cdot 3 + (-3)\cdot (-6) \\ &= -18 + 18 = 0 \\ q_{21} &= 4\cdot 2 + 2\cdot (-4) \\ &= 8 - 8 = 0 \\ q_{22} &= 4\cdot 3 + 2\cdot (-6) \\ &= 12 - 12 = 0 \end{aligned}$$ $$(\operatorname{adj}A)A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$$
   Matrix multiplication is not commutative in general, so $(\operatorname{adj}A)A$ has to be computed on its own; the question asks us to verify both orders. Here the entries are column-wise expansions: $q_{11}$ and $q_{22}$ are $|A|$ expanded along a column, and $q_{12}, q_{21}$ use the cofactors of a different column.

6. Compute $|A|\,I$ and compare

   $$|A|\,I = 0\cdot\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$$

   All three matrices are the $2\times2$ zero matrix, entry for entry, so

   $$A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|\,I = O.$$
   Two matrices are equal when they have the same order and the same entries in every position. All three are $2\times2$ with every entry 0, so the identity of Theorem 1 is verified for this $A$⁠.
   Scalar multiple: every entry of $I$ is multiplied by 0. Basics: scaling and adding matrices

7. What the zero result means

   Theorem 1 holds for every square matrix, singular or not. When $|A|\neq0$ we may divide it by $|A|$ to get $A^{-1} = \frac{1}{|A|}\operatorname{adj}A$ (Theorem 4). Here $|A|=0$⁠, so that division is impossible, and indeed this $A$ has no inverse.

   If some $B$ had $AB=I$⁠, then $|A|\,|B| = |I| = 1$ (Theorem 3), which is impossible when $|A|=0$⁠. The zero products we found are the visible sign of this: $\operatorname{adj}A$ is not the zero matrix, yet $A$ times it gives $O$⁠, something an invertible matrix could never do.
   Theorem 4, Section 4.5: $A$ is invertible exactly when $|A|\neq0$⁠. Basics: why the inverse divides by |A|

Every fact used, and where it comes from