Exercise 4.4 · Q6: Inverse of a $2\times2$ matrix

Basics you need

Solution

1. Compute $|A|$ and decide whether $A^{-1}$ exists

   $$\begin{aligned} |A| &= \begin{vmatrix} -1 & 5 \\ -3 & 2 \end{vmatrix} = (-1)\cdot 2 - 5\cdot (-3) \\ &= -2 + 15 = 13 \end{aligned}$$

   $|A| = 13 \neq 0$⁠, so $A$ is non-singular and $A^{-1}$ exists.

   Theorem 4 says a square matrix is invertible if and only if it is non-singular, that is $|A|\neq0$⁠. Checking this first tells us whether the rest of the work makes sense: if $|A|$ were 0, the answer would simply be "no inverse".
   Watch the signs: $a_{11}a_{22} = (-1)\cdot2 = -2$ and $a_{12}a_{21} = 5\cdot(-3) = -15$⁠, so $|A| = -2 - (-15) = -2 + 15 = 13$⁠. Basics: algebra the chapter leans on

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
   $a_{11} = -1$⁠; delete row 1 and column 1	$M_{11} = 2$	$A_{11} = +(2) = 2$
   $a_{12} = 5$⁠; delete row 1 and column 2	$M_{12} = -3$	$A_{12} = -(-3) = 3$
   $a_{21} = -3$⁠; delete row 2 and column 1	$M_{21} = 5$	$A_{21} = -(5) = -5$
   $a_{22} = 2$⁠; delete row 2 and column 2	$M_{22} = -1$	$A_{22} = +(-1) = -1$

   Deleting row $i$ and column $j$ of a $2\times2$ matrix leaves the single entry diagonally opposite, and that 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 $\operatorname{adj}A$

   $$[A_{ij}] = \begin{bmatrix} 2 & 3 \\ -5 & -1 \end{bmatrix}$$ $$\operatorname{adj}A = \begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}$$

   Shortcut check: swapping the diagonal entries $-1$ and $2$ and changing the signs of $5$ and $-3$ gives the same $\operatorname{adj}A$⁠.

   The adjoint is the transpose of the cofactor matrix (Definition 3, Section 4.5.1). The Remark after Example 12 gives the same result in one move for $2\times2$ matrices; doing both catches slips.
   Transpose: row $i$ becomes column $i$⁠. Basics: transpose and identity

4. Divide by $|A|$

   $$A^{-1} = \frac{1}{|A|}\operatorname{adj}A = \frac{1}{13}\begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}$$

   Multiplying each entry by $\frac{1}{13}$ (none of these fractions can be reduced, since 13 is prime and divides none of 2, 5, 3, 1):

   $$A^{-1} = \begin{bmatrix} \frac{2}{13} & -\frac{5}{13} \\ \frac{3}{13} & -\frac{1}{13} \end{bmatrix}$$
   Theorem 4 gives $A^{-1} = \frac{1}{|A|}\operatorname{adj}A$⁠. Keeping the factor $\frac{1}{13}$ outside, as the textbook does, avoids fractions inside the matrix; both forms are the same matrix.
   Scalar multiplication acts on every entry (Chapter 3, Section 3.4.2). Lesson: inverse of a matrix

5. Verify $AA^{-1} = I$

   Since $A^{-1} = \frac{1}{13}\operatorname{adj}A$⁠, we have $AA^{-1} = \frac{1}{13}\bigl(A(\operatorname{adj}A)\bigr)$⁠. Compute $A(\operatorname{adj}A)=[p_{ij}]$⁠, row of $A$ times column of $\operatorname{adj}A$⁠:

   $$\begin{aligned} p_{11} &= (-1)\cdot 2 + 5\cdot 3 \\ &= -2 + 15 = 13 \\ p_{12} &= (-1)\cdot (-5) + 5\cdot (-1) \\ &= 5 - 5 = 0 \\ p_{21} &= (-3)\cdot 2 + 2\cdot 3 \\ &= -6 + 6 = 0 \\ p_{22} &= (-3)\cdot (-5) + 2\cdot (-1) \\ &= 15 - 2 = 13 \end{aligned}$$ $$AA^{-1} = \frac{1}{13}\begin{bmatrix} 13 & 0 \\ 0 & 13 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$

   The other order works too. With $(\operatorname{adj}A)A = [q_{ij}]$⁠:

   $$\begin{aligned} q_{11} &= 2\cdot (-1) + (-5)\cdot (-3) \\ &= -2 + 15 = 13 \\ q_{12} &= 2\cdot 5 + (-5)\cdot 2 \\ &= 10 - 10 = 0 \\ q_{21} &= 3\cdot (-1) + (-1)\cdot (-3) \\ &= -3 + 3 = 0 \\ q_{22} &= 3\cdot 5 + (-1)\cdot 2 \\ &= 15 - 2 = 13 \end{aligned}$$

   so $A^{-1}A = \frac{1}{13}\,(13\,I) = I$ as well.

   By definition (Chapter 3, Definition 6), $B$ is the inverse of $A$ when $AB = BA = I$⁠. Both products give $I$⁠, so the matrix of Step 4 is $A^{-1}$⁠. It is the only one, because the inverse of a matrix is unique (Chapter 3, Theorem 3).
   Pulling the scalar out: $A(kB) = k(AB)$⁠. Basics: matrix facts the chapter uses

Every fact used, and where it comes from