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

Question

$$\begin{pmatrix} k & k & 0 \\ k^2 & 25 & k^2 \\ 0 & k & k \end{pmatrix}?$$

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?

Answer 1

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.

Answer 2

While the question is intended as an exercise in calculating the determinant of a matrix, note that one can also determine the answer by simple inspection (as one can do so in a recent question):

• When is there a column of zeros?
• When is the second column the sum of the first and the third columns?