Q.Check whether the following matrix is invertible or not. $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

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 method. The same test decides whether a system of linear equations $AX = B$ (with as many equations as unknowns) can be solved by the method of inversion: a singular coefficient matrix means the method of inversion cannot be applied, and generally signals that the equations are either dependent (infinitely many solutions) or inconsistent (no solution) rather than having the one unique solution the method of inversion is built to find.