Matrix with variable x

Question

We have matrix

$$A=\begin{pmatrix}(x -2) & 3 & 1 \\ (x -4) & 3 & 2 \\ (x -6) & x & 3\end{pmatrix}$$

for which values ​​of the variable x

a) the matrix does not have an inverse matrix

b) and its vertical vectors are linearly independent

I have trouble understanding this problem so can someone explain this to me and maybe give a clue how start solving this? How to solve variable x that is in matrix? Thanks

Answer 1

Change the matrix to $3 \times 3$ matrix. $$A=\begin{pmatrix}(x -2) & 3 & 1 \\ (x -4) & 3 & 2 \\ (x -6) & x & 3\end{pmatrix}$$ Then there is two solution for part a and thus there is also solution for part b. I think the question is more make sense in this form.

You can try to find the determinant. $$\begin{vmatrix}(x -2) & 3 & 1 \\ (x -4) & 3 & 2 \\ (x -6) & x & 3\end{vmatrix}=\begin{vmatrix}(x -2) & 3 & 1 \\ -2 & 0 & 1 \\ -4 & x-3 & 2\end{vmatrix}=\begin{vmatrix}(x -2) & 3 & 1 \\ -2 & 0 & 1 \\ 0 & x-3 & 0\end{vmatrix}\\=-2(x-3)-(x-2)(x-3)=-x(x-3)$$

Answer 2

By the Rule of Sarrus we have $$ \det(A)=-(x-3)x. $$ So the matrix $A\in M_3(K)$ has an inverse if and only if $x\neq 0,3$ in the field $K$.