# Find All Values of $x$ so that a Matrix is Singular ## Problem 168

Let
$A=\begin{bmatrix} 1 & -x & 0 & 0 \\ 0 &1 & -x & 0 \\ 0 & 0 & 1 & -x \\ 0 & 1 & 0 & -1 \end{bmatrix}$ be a $4\times 4$ matrix. Find all values of $x$ so that the matrix $A$ is singular. Add to solve later

## Hint.

Use the fact that a matrix is singular if and only if the determinant of the matrix is zero.

To compute the determinant, use a cofactor expansion.

## Solution.

We use the fact that a matrix is singular if and only if the determinant of the matrix is zero.

We compute the determinant of $A$ as follows.
\begin{align*}
&\det(A)= \begin{vmatrix}
1 & -x & 0 & 0 \\
0 &1 & -x & 0 \\
0 & 0 & 1 & -x \\
0 & 1 & 0 & -1
\end{vmatrix}\\[6pt] &=\begin{vmatrix}
1 & -x & 0 \\
0 &1 &-x \\
1 & 0 & -1
\end{vmatrix}
-0\begin{vmatrix}
-x & 0 & 0 \\
0 &1 &-x \\
1 & 0 & -1
\end{vmatrix}+0\begin{vmatrix}
-x & 0 & 0 \\
1 &-x &0 \\
1 & 0 & -1
\end{vmatrix}+0\begin{vmatrix}
-x & 0 & 0 \\
1 &-x &0 \\
0 & 1 & -x
\end{vmatrix}
\\[6pt] & \text{ by the first column cofactor expansion}\\[6pt] &=\begin{vmatrix}
1 & -x & 0 \\
0 &1 &-x \\
1 & 0 & -1
\end{vmatrix}\\[6pt] &=\begin{vmatrix}
1 & -x\\
0& -1
\end{vmatrix}-0\begin{vmatrix}
-x & 0\\
0& -1
\end{vmatrix}+\begin{vmatrix}
-x & 0\\
1& -x
\end{vmatrix}\\[6pt] & \text{ by the first column cofactor expansion}\\[6pt] &=-1+x^2.
\end{align*}

Therefore we have $\det(A)=x^2-1$. Thus $\det(A)=0$ if and only if $x=\pm 1$.
We conclude that the matrix $A$ is singular if and only if $x=\pm 1$.

## Comment.

You may use the rule of Sarrus to compute the $3\times 3$ determinant instead of the cofactor expansion if you like so. Add to solve later

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