Question

Matrix $A=\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }1 & 0 & -k \\
   2 & 1 & 3 \\
   k & 0 & 1\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]$is invertible for
A. $k = 1$
B. $k = - 1$
C. $k = 0$
D. All real $k$

Answer 1

Hint: A square matrix is said to be invertible if the identity matrix results from multiplying the matrix by its inverse. If and only if the determinant is not equal to zero, a square matrix is invertible. Thus, we need to find the value of $k$ such that the determinant of the given matrix achieves any real value other than zero.

Complete step-by-step solution:
The determinant is a scalar quantity that is a function of the rows and columns of a square matrix.
To find the value of $k$ we will first find the value of the determinant of the given matrix in terms of $k$ .
Thus, we get, $\left| A \right| = 1(1 \times 1 - 0 \times 3) - 0(2 \times 1 - k \times 3) - k(2 \times 0 - k \times 1)$
Solving this, we get, $\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\left| A \right|=1(1)-0-k(-k)$
So,
                                                                          \[\left| A \right|=1+{{k}^{2}}\]

Now, we know that for the given matrix to be invertible, $\left| A \right| \ne 0$
That is, \[1+{{k}^{2}}\ne 0\]
Hence, ${{k}^{2}}\ne 1$
This is true for all real values of k.

Hence, option D. is the answer.

Note: A non-singular matrix has a determinant whose value is not zero, whereas a singular matrix has a determinate with a value of zero. Only a nonsingular matrix has an inverse matrix; the singular matrix lacks one. The terms singular or degenerate refer to a square matrix that cannot be inverted.