Target Exam

CUET

Subject

Maths. Section A

Chapter

Matrices

Question:

If the matrix $A=\begin{bmatrix} 1 & -1 & 2\\3 & 1 & -2 \\1 & 0 & 3\end{bmatrix}$, the value of |adj A| is :

Options:

134

144

12

121

Correct Answer:

144

Explanation:

The correct answer is Option (2) → 144

Given

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

We use the property:

$|\text{adj } A| = |A|^{n-1}$

where $n = 3$.

So,

$|\text{adj } A| = |A|^2$

Now find $|A|$.

$|A| = \begin{vmatrix} 1 & 3 & 1 \\ -1 & 1 & 0 \\ 2 & -2 & 3 \end{vmatrix}$

Expanding along the first row:

$= 1 \begin{vmatrix} 1 & 0 \\ -2 & 3 \end{vmatrix} - 3 \begin{vmatrix} -1 & 0 \\ 2 & 3 \end{vmatrix} + 1 \begin{vmatrix} -1 & 1 \\ 2 & -2 \end{vmatrix}$

$= 1(3) - 3(-3) + 1(2 - 2)$

$= 3 + 9 + 0 = 12$

Hence,

$|\text{adj } A| = 12^2 = 144$