Q.If $A = \begin{bmatrix} \lambda & 1 \\ -1 & -\lambda \end{bmatrix}$ then $A^{-1}$ does not exist if $\lambda =$ (A) $0$ (B) $\pm 1$ (C) $2$ (D) $3$

A square matrix $A$ is called singular if its determinant is zero, $|A| = 0$, and non-singular if $|A| \neq 0$. This distinction is the single most important gatekeeping condition in this chapter: a matrix $X$ satisfying $AX = I$ can exist only when $A$ is non-singular. If $|A| = 0$, no matrix $X$ can undo $A$'s action, because a singular matrix collapses vectors onto a lower-dimensional subspace, and information lost in that collapse can never be recovered by multiplying with any other matrix. Geometrically, when a $2\times2$ matrix is singular its two rows (or columns), viewed as vectors, are parallel/proportional — precisely the situation where trying to "solve" using that matrix collapses to zero useful information. Checking singularity is always the first step before attempting to find an inverse: compute $|A|$; if it is $0$, stop and report that $A^{-1}$ does not exist; only if $|A| \neq 0$ does it make sense to proceed to the elementary-transformation method or the adjoint meth …