The matrix $\begin{pmatrix}1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1\end{pmatrix}$is not invertible if a has the value
A. 2
B. 1
C. 0
D. -1

Hint:
We are given a matrix that is not invertible. Recall what an invertible matrix is.

Complete Step-by-Step Answer:
Let A =$\begin{pmatrix}1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1\end{pmatrix}$ , which is not invertible.
We know that an invertible matrix is a matrix that has an inverse and its determinant is non-zero.
We have that A is a non-invertible matrix, which means A does not have an inverse and its determinant is zero.
Therefore, we have $|A| = 0$.
$\implies\begin{vmatrix}1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1 \end{vmatrix} = 0$
$\implies1(2\times1-5\times1)-a(1\times1-5\times2)+2(1\times1-2\times2) = 0$
$\implies1(2-5)-a(1-10)+2(1-4)=0$
$\implies1\times(-3)-a\times(-9)+2\times(-3)=0$
$\implies-3+9a-6 = 0$
$\implies9a-9=0$
$\implies9a = 9$
$\implies a = \dfrac{9}{9}$
$\implies a = 1$
Therefore, the value of a is 1. So, the answer is Option B.

Note:
Another name for non-invertible matrix is singular matrix. A singular matrix is a rare case, almost all matrices are non-singular, that is they have an inverse. Non-square matrices do not have an inverse, but they might have a right inverse or left inverse. We know that the system of equations can be written as, AX=B. If A is a singular matrix, then we cannot find a solution for this system of equations.