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(D) matrix $A$ is non-singular if and only if its determinant $|A| 
eq 0$.Given the matrix $A = \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 3 & \lambda

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) matrix $A$ is non-singular if and only if its determinant $|A| 
eq 0$.Given the matrix $A = \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 3 & \lambda & 5 \end{bmatrix}$,we calculate its determinant:$|A| = 1(5 	imes 5 - 6 	imes \lambda) - 2(4 	imes 5 - 6 	imes 3) + 3(4 	imes \lambda - 5 	imes 3)$$|A| = 1(25 - 6\lambda) - 2(20 - 18) + 3(4\lambda - 15)$$|A| = 25 - 6\lambda - 2(2) + 12\lambda - 45$$|A| = 25 - 6\lambda - 4 + 12\lambda - 45$$|A| = 6\lambda - 24$For the matrix to be non-singular,we require $|A| 
eq 0$:$6\lambda - 24 
eq 0$$6\lambda 
eq 24$$\lambda 
eq 4$Therefore,$\lambda$ should not be equal to $4$.

In order that the matrix $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & \lambda & 5 \end{bmatrix}$ be non-singular,$\lambda$ should not be equal to:

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