How would I find k such that the following matrix is singular?

Question

\begin{pmatrix}-4&1&4\\ 4&-2&-3\\ -34+k&7&18\end{pmatrix}

I know that a singular matrix is not invertible and has a determinant of zero. But what I am confused about is getting the above matrix into the form required. Do I have to do it in reverse from how we would calculate the determinant of a matrix? If so, how?

Thank you!

Answer 1

As Gregory commented you must calculate the determinant and you will get an expression with $k$ as a variable.

Now, you know that the determinant of the matrix must be $0$ in order to be singular. So, solve $p(k)=0$ where $p(k)$ is the expression obtained by calculating the determinant of the matrix.

Answer 2

Alternative approach besides looking at determinant:

We can easily see that the last two columns are not multiple of each other.

If the first column is not in the span of the last two columns, then it is linearly dependent. Let $C_i$ denote the $i$-th column.

Let $a_2C_2+a_3C_3 = C_1$.

We can focus on the first two rows:

$$\begin{bmatrix} 1 & 4 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} a_2 \\ a_3 \end{bmatrix}= \begin{bmatrix} -4 \\ 4 \end{bmatrix} $$

$$\begin{bmatrix} a_2 \\ a_3 \end{bmatrix}= \begin{bmatrix} 1 & 4 \\ -2 & -3 \end{bmatrix} ^{-1}\begin{bmatrix} -4 \\ 4 \end{bmatrix} $$

We can solve for $a_2$ and $a_3$, and to make the system singular, we just have to set

$$-34+k=7a_2+18a_3 = \begin{bmatrix} 7 & 18 \end{bmatrix}\begin{bmatrix} a_2 \\ a_3 \end{bmatrix}$$

Hence $$k = 34+ \begin{bmatrix} 7 & 18 \end{bmatrix}\begin{bmatrix} 1 & 4 \\ -2 & -3 \end{bmatrix} ^{-1}\begin{bmatrix} -4 \\ 4 \end{bmatrix}$$