Exercise 4.4 · Q17 (MCQ): $|\operatorname{adj}A|$ for a non-singular $3\times3$ matrix

Basics you need

Solution

1. Why $A(\operatorname{adj}A) = |A|\,I$ for every $3\times3$ matrix

   Let $A = [a_{ij}]$ be any $3\times3$ matrix with cofactors $A_{ij}$⁠. Column $j$ of $\operatorname{adj}A$ is $A_{j1}, A_{j2}, A_{j3}$ (the cofactors of row $j$⁠), so entry $(i,j)$ of $A(\operatorname{adj}A)$ is

   $$a_{i1}A_{j1} + a_{i2}A_{j2} + a_{i3}A_{j3}.$$
   • If $i = j$⁠: this is row $i$ times its own cofactors, the expansion of $|A|$ along row $i$⁠. So every diagonal entry is $|A|$⁠.
   • If $i \neq j$⁠: replace row $j$ of $A$ by a copy of row $i$⁠, and call the new determinant $D$⁠. The cofactors of row $j$ do not involve row $j$ itself, so they are the same in $D$ as in $|A|$⁠, and expanding $D$ along row $j$ gives exactly the sum above. But $D$ has two equal rows. Expand $D$ along its third row $k$ (the one that is neither $i$ nor $j$⁠): each minor keeps rows $i$ and $j$⁠, which are equal, so it has the form $$\begin{vmatrix} x & y \\ x & y \end{vmatrix} = xy - yx = 0.$$ Hence $D = 0$⁠, and every off-diagonal entry is $0$⁠.
   $$A(\operatorname{adj}A) = \begin{bmatrix} |A| & 0 & 0 \\ 0 & |A| & 0 \\ 0 & 0 & |A| \end{bmatrix} = |A|\,I$$
   The book states this as Theorem 1 and verifies it with the facts in Basic 2. The argument above also shows why the Note in Section 4.4 is true for $3\times3$ determinants: "a row times another row's cofactors" is a determinant with a repeated row, and such a determinant is $0$⁠. This holds whether or not $|A| = 0$⁠.
   Lesson: minors and cofactors · Lesson: the adjoint

2. Take the determinant of both sides

   Equal matrices have equal determinants, so

   $$\bigl|A(\operatorname{adj}A)\bigr| = \bigl|\,|A|\,I\,\bigr|.$$

   $A$ and $\operatorname{adj}A$ are both $3\times3$⁠, so by Theorem 3 the left side is $|A|\,|\operatorname{adj}A|$⁠.

   Theorem 3 turns the determinant of a product into a product of two numbers. That is the step that lets us isolate the unknown number $|\operatorname{adj}A|$⁠.
   $|XY| = |X||Y|$⁠. Basics: matrix facts the chapter uses

3. Evaluate the right side: $\bigl|\,|A|\,I\,\bigr| = |A|^3$

   Write $k = |A|$⁠, an ordinary number. Then $kI$ is diagonal with $k, k, k$⁠. Expand along row 1, which has two zeros:

   $$\begin{aligned} \begin{vmatrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{vmatrix} &= k\begin{vmatrix} k & 0 \\ 0 & k \end{vmatrix} - 0 + 0\\ &= k\,(k\cdot k - 0\cdot0) = k^3 \end{aligned}$$

   So the right side is $|A|^3$⁠, and the equation of Step 2 becomes

   $$|A|\;|\operatorname{adj}A| = |A|^3.$$
   A common slip is to say $|kI| = k$⁠. The scalar $k$ multiplies all three rows, so it comes out three times. This agrees with Basic 4 applied to $I$⁠: $|kI| = k^3|I| = k^3 \cdot 1$⁠.
   Lesson: expanding a 3 × 3 determinant

4. Divide by $|A|$⁠, which is not zero

   $A$ is non-singular, so $|A| \neq 0$⁠, and we may divide both sides by $|A|$⁠:

   $$|\operatorname{adj}A| = \frac{|A|^3}{|A|} = |A|^2.$$

   This holds for every non-singular $3\times3$ matrix, so the answer is (B). For a square matrix of any order $n$⁠, the same steps give $|A|\,|\operatorname{adj}A| = |A|^n$ and hence $|\operatorname{adj}A| = |A|^{n-1}$⁠, as the book notes after Theorem 3.

   Here is where "nonsingular" in the question is used. Without $|A| \neq 0$⁠, the equation $|A|\,|\operatorname{adj}A| = |A|^3$ would read $0 = 0$ and tell us nothing about $|\operatorname{adj}A|$⁠.
   Cancel a factor only when it is non-zero. Basics: algebra the chapter leans on

5. Examine every option

   Test matrix 1, a diagonal matrix:

   $$A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \quad |A| = 2\cdot1\cdot1 = 2$$

   Its cofactors are $A_{11} = 1\cdot1 - 0 = 1$⁠, $A_{22} = 2\cdot1 - 0 = 2$⁠, $A_{33} = 2\cdot1 - 0 = 2$⁠, and the six others are $0$⁠, because each of those minors has a row or a column of zeros. So

   $$\operatorname{adj}A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}, \quad |\operatorname{adj}A| = 4$$

   Test matrix 2: the matrix of Exercise 4.4 Q10, with $|A| = -1$ and

   $$\operatorname{adj}A = \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix}$$

   Expanding along row 1, signs $+, -, +$⁠: the middle term is $0$ times its cofactor, so it is left out, and the third term has sign $+$ but its entry, in position $(1,3)$ of $\operatorname{adj}A$⁠, is $-1$⁠, which is why it carries a minus sign:

   $$\begin{aligned} |\operatorname{adj}A| &= 2\bigl((-2)2 - 3(-1)\bigr)\\ &\quad - \bigl((-9)(-1) - (-2)(-6)\bigr)\\ &= 2(-1) - (-3) = 1 \end{aligned}$$

   So Test 1 has $|A| = 2$⁠, $|\operatorname{adj}A| = 4$⁠, and Test 2 has $|A| = -1$⁠, $|\operatorname{adj}A| = 1$⁠.

   Option	Value in Test 1; Test 2	Verdict
   (A) $|A|$	$2$⁠; $-1$	Wrong: $2 \neq 4$⁠. (This is the rule for order 2.)
   (B) $|A|^2$	$4$⁠; $1$	Correct in both, and proved in Steps 1–4.
   (C) $|A|^3$	$8$⁠; $-1$	Wrong: $8 \neq 4$⁠. ($|A|^3$ is the determinant of $A(\operatorname{adj}A)$⁠.)
   (D) $3|A|$	$6$⁠; $-3$	Wrong: $6 \neq 4$⁠. (It confuses $|kI|$ with $3k$⁠.)

   Steps 1–4 prove (B) for every non-singular $3\times3$ matrix. To reject the others, one explicit matrix for which an option gives the wrong value is enough, because each option claims to hold for every such matrix. Test 1 already rejects (A), (C) and (D); Test 2 shows the sign is right too, since $|A|^2$ is never negative.
   One counterexample defeats a "for all" claim. Basics: algebra the chapter leans on

6. A note: singular matrices obey the same rule

   The question assumes $|A| \neq 0$⁠, but $|\operatorname{adj}A| = |A|^2$ is true even when $|A| = 0$⁠, where it says $|\operatorname{adj}A| = 0$⁠. Suppose $|A| = 0$ but $|\operatorname{adj}A| \neq 0$⁠. By Step 1, $A(\operatorname{adj}A) = 0\cdot I = O$⁠. Since $|\operatorname{adj}A| \neq 0$⁠, $\operatorname{adj}A$ would then be invertible (Theorem 4), and multiplying on the right by its inverse gives $A = O$⁠. But every cofactor of the zero matrix is $0$⁠, so $\operatorname{adj}A = O$ and $|\operatorname{adj}A| = 0$⁠, a contradiction. So $|\operatorname{adj}A| = 0$ whenever $|A| = 0$⁠.

   This is not needed for the answer, but it shows the division in Step 4 is the only place the hypothesis matters: the formula itself survives, it just needs a different proof when $|A| = 0$⁠.

Every fact used, and where it comes from