NIMCET 2014 — Mathematics PYQ

Concept

A matrix whose determinant is NOT equal to $0$ is invertible.

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Calculation

Let $[A]=\left[\begin{array}{ccc}-1& amp;3& amp;2\\ 1& amp;K& amp;-3\\ 1& amp;4& amp;5\end{array}\right]$.

The determinant of $A$ is:

$\Rightarrow |A|=\left|\begin{array}{ccc}-1& amp;3& amp;2\\ 1& amp;K& amp;-3\\ 1& amp;4& amp;5\end{array}\right|$

Using $R_2\to R_1+R_2$ and $R_3\to R_1+R_3$, we get:

$\Rightarrow |A|=\left|\begin{array}{ccc}-1& amp;3& amp;2\\ 0& amp;K+3& amp;-1\\ 0& amp;7& amp;7\end{array}\right|$

Using $C_2\to C_2-C_3$, we get:

$\Rightarrow |A|=\left|\begin{array}{ccc}-1& amp;1& amp;2\\ 0& amp;K+4& amp;-1\\ 0& amp;0& amp;7\end{array}\right|$

Using $R_3\leftrightarrow R_1$, we get:

$\Rightarrow -|A|=\left|\begin{array}{ccc}0& amp;0& amp;7\\ 0& amp;K+4& amp;-1\\ -1& amp;1& amp;2\end{array}\right|$

Expanding along $R_1$, we get:

$\Rightarrow -|A|=0+0+7[0-(-1)(K+4)]$

$\Rightarrow |A|=-(K+4)$

For the matrix to be invertible, the determinant of $A$ must be non-zero:

$\Rightarrow |A|=-(K+4)\ne 0$

$\Rightarrow K\ne -4$.

Correct Option: 2