Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Matrices

Question:

If \(A=\begin{bmatrix}2 & 0 & 0\\-1 & 2 & 3\\3 & 3 & 5\end{bmatrix}\), then A(adj A) is equal to:

Options:

\(\begin{bmatrix}3 & 3 & 3\\3 & 3 & 3\\3 & 3 & 3\end{bmatrix}\)

\(\begin{bmatrix}3 & 0 & 0\\0 & 3 & 0\\0 & 0 & 3\end{bmatrix}\)

\(\begin{bmatrix}2 & 0 & 0\\0 & 2 & 0\\0 & 0 & 2\end{bmatrix}\)

\(\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\)

Correct Answer:

\(\begin{bmatrix}2 & 0 & 0\\0 & 2 & 0\\0 & 0 & 2\end{bmatrix}\)

Explanation:

Adj A = $A_{11} = (10-9)= 1$   $A_{12} = (-5-9)= 14$   $A_{13} = (-3-6)= -9$

$A_{21} = (0-0)= 0$   $A_{22} = (6-0)= 6$   $A_{23} = (6-0)= 6$

$A_{31} = (0-0)= 0$   $A_{32} = (6-0)= 6$   $A_{33} = (4-0)= 4$

$|A|=2(10-9)-0(-5-9)+0(-3-6)=2(1)-0+0⇒2≠0$

$(adj)A=\begin{bmatrix}1&0&0\\14&6&-6\\-9&-6&4\end{bmatrix}$

$A(adj)A=\begin{bmatrix}2 & 0 & 0\\-1 & 2 & 3\\3 & 3 & 5\end{bmatrix}\begin{bmatrix}1&0&0\\14&6&-6\\-9&-6&4\end{bmatrix}⇒\begin{bmatrix}1×2+0×-1+0×3 & 1×0+0×2+0×3 & 1×0+0×3+0×5\\14×2+6×-1+(-6)×3 & 14×0+6×2+(-6×3) & 14×0+6×3+(-6)×5\\-9×2+(-6×-1)+4×3 & -9×0+(-6)×3+4×3 & -9×0+(-6)×3+4×5\end{bmatrix}$

\(\begin{bmatrix}2 & 0 & 0\\0 & 2 & 0\\0 & 0 & 2\end{bmatrix}\)