Question

If \( A = \begin{bmatrix} 4 & \lambda & -3 0 & 2 & 5 1 & 1 & 3 \end{bmatrix} \), then \( A^{-1} \) exists if:

• A. \( \lambda = 2 \)
• B. \( \lambda = 0 \)
• C. \( \lambda \ne 2 \)
• D. \( \lambda \ne -2 \)

Hint:

A matrix is invertible iff its determinant is non-zeroAlways compute determinant carefully using expansion.

Answer 1

The Correct Option is D

Step 1: Condition for inverse.
Matrix \( A^{-1} \) exists if:
\[ \det(A) \ne 0 \]

Step 2: Expand determinant along first row.
\[ \det(A) = 4 \begin{vmatrix} 2 & 5 1 & 3 \end{vmatrix} - \lambda \begin{vmatrix} 0 & 5 1 & 3 \end{vmatrix} -3 \begin{vmatrix} 0 & 2 1 & 1 \end{vmatrix} \]

Step 3: Compute minors.
\[ \begin{vmatrix} 2 & 5 1 & 3 \end{vmatrix} = 6 - 5 = 1 \]
\[ \begin{vmatrix} 0 & 5 1 & 3 \end{vmatrix} = 0 - 5 = -5 \]
\[ \begin{vmatrix} 0 & 2 1 & 1 \end{vmatrix} = 0 - 2 = -2 \]

Step 4: Substitute values.
\[ \det(A) = 4(1) - \lambda(-5) - 3(-2) \]
\[ = 4 + 5\lambda + 6 \]
\[ = 10 + 5\lambda \]

Step 5: Factor determinant.
\[ \det(A) = 5(2+\lambda) \]

Step 6: Apply condition \( \det(A)\ne 0 \).
\[ 5(2+\lambda)\ne 0 \]
\[ \lambda \ne -2 \]

Step 7: Final conclusion.
\[ \boxed{\lambda \ne -2} \]