Q.If $A = \begin{pmatrix} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{pmatrix}$, then $A^{-1}$ exists if
(A) $\lambda = 2$
(B) $\lambda \neq 2$
(C) $\lambda \neq -2$
(D) None of these

Concept: Determinant Evaluation Using Identities — a matrix is invertible iff its determinant is non-zero.

Step 1: Compute $\det(A)$ by expanding along the first column:

Step 2: Evaluate the $2\times2$ determinants:

$\begin{vmatrix} 2 & 5 \\ 1 & 3 \end{vmatrix} = 6 - 5 = 1, \quad \begin{vmatrix} \lambda & -3 \\ 2 & 5 \end{vmatrix} = 5\lambda + 6$ …

$A^{-1}$ exists precisely when $\det A \neq 0$. Expand along the first column (which contains a zero):

Check by expanding along the second row:

$2\begin{vmatrix}2&-3\\1&3\end{vmatrix} - 5\begin{vmatrix}2&\lambda\\1&1\end{vmatrix} = 2(9) - 5(2-\lambda) = 5\lambda + 8,$ …

Method: Testing Invertibility via the Determinant

A square matrix has an inverse exactly when its determinant is nonzero, so any "does $A^{-1}$ exist" question reduces to computing $\det A$ (possibly containing a parameter) and finding the parameter value(s) that make it nonzero.

Steps

Step 1: Recall the invertibility criterion

Step 2: Expand $\det A$ along the row or column with the most zeros

Keep the parameter symbolic throughout the expansion; this yields $\det A$ as a linear (or higher-degree) expression in that parameter.

Step 3: Solve for the excluded (singular) value

Set the expression equal to zero and solve for the parameter — this is the exact value at which the matrix becomes singular and the inverse fails to exist. …

Common Mistakes

Mistake 1: Pattern-matching the answer to "$\lambda\ne2$" without actually computing the determinant

Why it's wrong: The number $2$ appears twice in the matrix (positions $(1,1)$ and $(2,2)$), tempting a guess that the singular condition is $\lambda=2$; the real condition, from $\det A=5\lambda+8$, is $\lambda=-\tfrac85$, which matches none of the given "$\lambda=2$"-style options. Correct approach: always compute $\det A$ explicitly as a function of the parameter — never infer the singular value from which numbers superficially look connected to the parameter.

Mistake 2: A sign error in the cofactor for the entry holding $\lambda$ …