# For Which Choices of $x$ is the Given Matrix Invertible?

## Problem 394

Determine the values of $x$ so that the matrix
$A=\begin{bmatrix} 1 & 1 & x \\ 1 &x &x \\ x & x & x \end{bmatrix}$ is invertible.
For those values of $x$, find the inverse matrix $A^{-1}$.

## Solution.

We use the fact that a matrix is invertible if and only if its determinant is nonzero.
So we compute the determinant of the matrix $A$.

We have
\begin{align*}
&\det(A)=\begin{vmatrix}
1 & 1 & x \\
1 &x &x \\
x & x & x
\end{vmatrix}\\
&=(1)\begin{vmatrix}
x & x\\
x& x
\end{vmatrix}-(1)\begin{vmatrix}
1 & x\\
x& x
\end{vmatrix}+x\begin{vmatrix}
1 & x\\
x& x
\end{vmatrix} && \text{by the first row cofactor expansion.}\\
&=(x^2-x^2)-(x-x^2)+x(x-x^2)\\
&=(x-1)(x-x^2)\\
&=x(x-1)^2.
\end{align*}

Thus, the determinant $\det(A)$ is zero if and only if $x=0, 1$.
Hence the matrix $A$ is invertible if and only if $x\neq 0, 1$.

Next, we suppose that $x \neq 0, 1$ and find the inverse matrix of $A$.
We reduce the augmented matrix $[A\mid I]$ as follows.
We have
\begin{align*}
&[A\mid I]= \left[\begin{array}{rrr|rrr}
1 & 1 & x & 1 &0 & 0 \\
1 & x & x & 0 & 1 & 0 \\
x & x & x & 0 & 0 & 1 \\

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