Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Algebra

Question:

If $A=\begin{bmatrix}1 & 2 & -1\\2 & 3& 0\\1 & -1 & 2 \end{bmatrix}$, then adj (A) is :

Options:

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

$\begin{bmatrix}6 & -4 & -5\\-3 & 3& 3\\3 & -2 & -1 \end{bmatrix}$

$\begin{bmatrix}6 & -3 & 3\\-4 & 3 & -2\\-5 & 3 & -1 \end{bmatrix}$

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

Correct Answer:

$\begin{bmatrix}6 & -3 & 3\\-4 & 3 & -2\\-5 & 3 & -1 \end{bmatrix}$

Explanation:

The correct answer is Option (3) → $\begin{bmatrix}6 & -3 & 3\\-4 & 3 & -2\\-5 & 3 & -1 \end{bmatrix}$

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

The co-factor matrices are,

$A_{11}=(3×2-(-1)0)=6$

$A_{12}=-(2×2-1×0)=-4$

$A_{13}=(2×-1-3×1)=-5$

$A_{21}=-(4-1)=-3$

$A_{22}=(1×2-1(-1)1)=3$

$A_{23}=-(1×-1-2×1)=3$

$A_{31}=(2×0+3×1)=3$

$A_{32}=-(1×0+2)=-2$

$A_{33}=(1×3-2×2)=-1$

$Adj\,A=\begin{bmatrix}6&-4&-5\\-3&3&3\\3&-2&-1\end{bmatrix}^T$

$=\begin{bmatrix}6 & -3 & 3\\-4 & 3 & -2\\-5 & 3 & -1 \end{bmatrix}$