# Find k such that the following matrix M is singular. $M=\begin{bmatrix}-4&0&-3\\-4&-4&1\\-14+k&-8&-1\end{bmatrix}$

Find k such that the following matrix M is singular.
$M=\begin{bmatrix}-4&0&-3\\-4&-4&1\\-14+k&-8&-1\end{bmatrix}$

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Charles Clute
A square matrix A is said to be singular if its determinant value is 0.
i.e $$\displaystyle{\left|{A}\right|}={0}$$
Here, the given matrix is $M=\begin{bmatrix}-4&0&-3\\-4&-4&1\\-14+k&-8&-1\end{bmatrix}$
So we must have
$$\displaystyle{\left|{M}\right|}={0}$$
$\begin{bmatrix}-4&0&-3\\-4&-4&1\\-14+k&-8&-1\end{bmatrix}=0$
Now, expanding the determinant by its 1st row
$$\displaystyle-{4}{\left({4}+{8}\right)}-{0}-{3}{\left({32}+{4}{\left(-{14}+{k}\right)}\right)}={0}$$
$$\displaystyle-{48}-{3}{\left({32}-{96}+{4}{k}\right)}={0}$$
$$\displaystyle-{48}-{3}{\left(-{64}+{4}{k}\right)}={0}$$
$$\displaystyle-{48}+{192}-{12}{k}={0}$$
$$\displaystyle{144}-{12}{k}={0}$$
$$\displaystyle{12}{k}={144}$$
$$\displaystyle{k}={12}$$
So ,the required value of k=12 for which the matric is singular.