Question

If matrix A is

$$\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix},$$

then value of $\lvert \operatorname{adj}A\rvert$ is

• A. 11
• B. 121
• C. 1331
• D. None

Answer 1

Answer: (B)

121

Answer 2

Step 1: Compute $\det A$.

$$\det A = 1\bigl(0\cdot3 -(-2)\cdot0\bigr) \;-\;(-1)\bigl(3\cdot3 -(-2)\cdot1\bigr) \;+\;2\bigl(3\cdot0 -0\cdot1\bigr) = 0 +1\cdot(9+2)+0 = 11.$$

Step 2: Use the property
For an $n\times n$ matrix $A$,

$$\det(\operatorname{adj}A) = (\det A)^{\,n-1}.$$

Here $n=3$, so

$$\lvert\operatorname{adj}A\rvert = (\det A)^{2} = 11^2 = 121.$$