Question

If $\mathrm{A}=\left[\begin{array}{ccc}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]$, then $\mathrm{A}(\text{adj }\mathrm{A}) =$

• A. $\left[\begin{array}{ccc} -1/3 & 0 & 0 \\ 0 & -1/3 & 0 \\ 0 & 0 & -1/3 \end{array}\right]$
• B. $\left[\begin{array}{ccc} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right]$
• C. $\left[\begin{array}{ccc} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]$
• D. $\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & 2 \\ 3 & 2 & 4 \end{array}\right]$

Hint:

Never spend time calculating all 9 cofactors to construct the adjugate matrix when a question asks for $A(\text{adj } A)$. The theorem $A(\text{adj } A) = |A|I$ ensures the answer must be a scalar diagonal matrix. Calculating the single scalar determinant $3$ points directly to option (B) in under twenty seconds!

Answer 1

The Correct Option is B

Step 1: Recall the key property.
For any square matrix, $A(\text{adj }A) = |A|\,I$, so we only need the determinant of $A$.
Step 2: Set up the determinant.
$|A| = \begin{vmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{vmatrix}$, expanded along the first row.
Step 3: Evaluate the minors.
First minor $(1)(4)-(2)(2) = 0$; second minor $(-1)(4)-(2)(1) = -6$; third minor $(-1)(2)-(1)(1) = -3$.
Step 4: Combine with signs.
$|A| = 1(0) - 2(-6) + 3(-3) = 0 + 12 - 9 = 3$.
Step 5: Apply the property.
$A(\text{adj }A) = 3I = 3\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Step 6: Write the final matrix.
This equals $\begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$, which is option 2 and matches the key.
\[ \boxed{A(\text{adj }A) = 3I} \]