# 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.

## 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. Sponsored Links ### More from my site • Find All the Values of x so that a Given 3\times 3 Matrix is Singular Find all the values of x so that the following matrix A is a singular matrix. \[A=\begin{bmatrix} x & x^2 & 1 \\ 2 &3 &1 \\ 0 & -1 & 1 \end{bmatrix}.   Hint. Use the fact that a matrix is singular if and only if its determinant is […]
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