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.
$\endgroup$ 22 Answers
$\begingroup$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
$$\det A=\begin{vmatrix} 5 & 6 & 6-x & 8 \\ 2 & 2-x & 2 & 8 \\ 6 & 6 & 2 & 8 \\ 2 & 3 & 6 & 7 \end{vmatrix}= \begin{vmatrix} -1 & x & 6-x & 8 \\ x & -x & 2 & 8 \\ 0 & 4 & 2 & 8 \\ -1 & -3 & 6 & 7 \end{vmatrix}=\\=-1\begin{vmatrix} -x & 2 & 8 \\ 4 & 2 & 8 \\ -3 & 6 & 7 \end{vmatrix}-x\begin{vmatrix} x & 6-x & 8 \\ 4 & 2 & 8 \\ -3 & 6 & 7 \end{vmatrix}+1\begin{vmatrix} x & 6-x & 8 \\ -x & 2 & 8 \\ 4 & 2 & 8 \\ \end{vmatrix}=\\= -1\cdot(34x+136)-x\cdot(18x-72)+(128-8x^2)=-26x^2+38x-8 $$
$\endgroup$ 14 $\begingroup$Some properties of square matrices that may be useful in your case:
If two rows/columns of a matrix are proportional the matrix is $0$. For example:
$$\begin{pmatrix}
a & b & c & d \\
e & f & g & h \\
x & y & z & w \\
ka & kb & kc & kd
\end{pmatrix}=0$$
You can multiply any row/column of a matrix with a constant and add it to another row/column. For example:
$$\begin{pmatrix}
a & b & c & d \\
e & f & g & h \\
x & y & z & w \\
i & j & m & n
\end{pmatrix}=\begin{pmatrix}
a & b & c & d+kb \\
e & f & g & h+kf \\
x & y & z & w+ky \\
i & j & m & n+kj
\end{pmatrix} $$
If any row/column of a matrix is all $0$, then the matrix is $0$. For example:
$$\begin{pmatrix}
a & b & c & d \\
e & f & g & h \\
0 & 0 & 0 & 0 \\
i & j & m & n
\end{pmatrix}=0$$
By using the second property multiple times on multiple rows and columns you can get one of the rows/columns to be $0$ or be proportional and then you get that $detA=0$.
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