Question

Find all values for k for which the matrix A=(k -k 3
0 k+1 1
8 -8 k-1)
is invertible

Answer 1

\[\left[\begin{matrix}k & -k & 3 \\ 0 & k+1 & 1 \\ k & -8 & k-1\end{matrix}\right]\]

Answer 2

We want the determinant to NOT be zero.
This is because if the determinant is zero, the matrix is NOT invertible; otherwise, it is.
The determinant of this matrix is
$$
k\left ((k+1)(k-1)+8\right )+k\left(-k(k-1)+24\right )=0
$$
Whatever k makes this true makes the determinant NOT invertible.

Does this make sense?

Can you solve for k?

Answer 3

Thanks @ybarrap ! I got k=0,2 is this correct?

Answer 4

yes, k =0, 2 are correct.
note that for these values, matrix will NOT be invertible

so for matrix to be invertible, \(k \ne 0,2\)

Answer 5

I need A to be invertible though, so how do i find possible k values that will make it invertible? @hartnn

Answer 6

for A to be invertible, k should not equal 0 or 2
All other real values of k are its possible values.