Given
$$A=\begin{pmatrix}
5 & 6 & 6-x & 8 \\
2 & 2-x & 2 & 8 \\
6 & 6 & 2 & 8 \\
2 & 3 & 6 & 7
\end{pmatrix}\in\mathbb{R}^{4\times 4}$$
One must find all values of $x\in\mathbb{R}$ such that the matrix is not invertible.
I tried finding the determinant of $A$ and in the process got that $x = 4$ and $x = 6/192$. Therefore, $A1 = 1360; A2 = 816; A3 = 1146; A4 = 1088$. However when I calculate $\det(A) = 1360 – 816 + 1146 – 1088= 602$ and I do not get 0 so that I could prove it is not invertible.
Best Answer
HINT
Let's calculate $\det A$ as a function of x and then set $\det A=0$.
Note that adding a scalar multiple of one column/row to another column/row does not change the value of the determinant. Thus you can simplify the matrix A before to perform the calculation.
You should obtain:
$$\det A=2(-13 x^2 + 19 x - 4)$$
and from here find the values for which $\det A=0$.
Notably
$$\det A=\begin{vmatrix} 5 & 6 & 6-x & 8 \\ 2 & 2-x & 2 & 8 \\ 6 & 6 & 2 & 8 \\ 2 & 3 & 6 & 7 \end{vmatrix}= \begin{vmatrix} -1 & x & 6-x & 8 \\ x & -x & 2 & 8 \\ 0 & 4 & 2 & 8 \\ -1 & -3 & 6 & 7 \end{vmatrix}=\\=-1\begin{vmatrix} -x & 2 & 8 \\ 4 & 2 & 8 \\ -3 & 6 & 7 \end{vmatrix}-x\begin{vmatrix} x & 6-x & 8 \\ 4 & 2 & 8 \\ -3 & 6 & 7 \end{vmatrix}+1\begin{vmatrix} x & 6-x & 8 \\ -x & 2 & 8 \\ 4 & 2 & 8 \\ \end{vmatrix}=\\= -1\cdot(34x+136)-x\cdot(18x-72)+(128-8x^2)=-26x^2+38x-8 $$