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(C) The inverse $A^{-1}$ exists if and only if the determinant $|A| 
e 0$. Calculating the determinant: $|A| = \begin{vmatrix} x + \lambda & x & x \

Vedclass · Accepted Answer

$3x + \lambda 
e 0, \lambda 
e 0$

Vedclass · Answer

(C) The inverse $A^{-1}$ exists if and only if the determinant $|A| 
e 0$. Calculating the determinant: $|A| = \begin{vmatrix} x + \lambda & x & x \ x & x + \lambda & x \ x & x & x + \lambda \end{vmatrix}$ Applying the operation $C_1 ightarrow C_1 + C_2 + C_3$: $|A| = \begin{vmatrix} 3x + \lambda & x & x \ 3x + \lambda & x + \lambda & x \ 3x + \lambda & x & x + \lambda \end{vmatrix} = (3x + \lambda) \begin{vmatrix} 1 & x & x \ 1 & x + \lambda & x \ 1 & x & x + \lambda \end{vmatrix}$ Applying $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$: $|A| = (3x + \lambda) \begin{vmatrix} 1 & x & x \ 0 & \lambda & 0 \ 0 & 0 & \lambda \end{vmatrix} = (3x + \lambda)(\lambda^2)$ For $A^{-1}$ to exist,$|A| 
e 0$,which implies $\lambda^2(3x + \lambda) 
e 0$. Therefore,$\lambda 
e 0$ and $3x + \lambda 
e 0$.

Let $A = \begin{bmatrix} x + \lambda & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda \end{bmatrix}$,then $A^{-1}$ exists if

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