<s>[INST] A square matrix A is invertible if and only if det A ≠ 0. Use the theorem above to find all values of k for which A is invertible $$\begin{pmatrix} k & k & 0 \\ k^2 & 25 & k^2 \\ 0 & k & k \end{pmatrix}?$$
I did a sample question before this one:
$$\begin{pmatrix} k & k & 0 \\ k^2 & 16 & k^2 \\ 0 & k & k \end{pmatrix}?$$
And was able to get k ≠ -2√2, 0, 2√2. Not sure exactly how, can anyone help me guide my way through these two questions?
 [/INST] Do the same thing you did before. First, take the determinant of 
$$\begin{pmatrix} k & k & 0 \\ k^2 & 25 & k^2 \\ 0 & k & k \end{pmatrix}$$
to obtain the equation $25k^2-2k^4$. So if this equation is zero, then the matrix is not invertible. Then
$$
\begin{align}
25k^2-2k^4&=0 \\
k^2(25-2k^2)&=0 \\
k^2(5+\sqrt{2}k)(5-\sqrt{2}k) &=0
\end{align}
$$
So $k=0,\frac{5}{\sqrt{2}}$ or $-\frac{5}{\sqrt{2}}$. Therefore, if $k$ is any real number except $0,\frac{5}{\sqrt{2}},-\frac{5}{\sqrt{2}}$ then the matrix is invertible. 
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