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 $A^{-1}$.

 
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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 \\
\end{array} \right] \\[6pt] & \xrightarrow{\substack{R_2-R_1 \\ R_3-xR_1}}
\left[\begin{array}{rrr|rrr}
1 & 1 & x & 1 &0 & 0 \\
0 & x-1 & 0 & -1 & 1 & 0 \\
0 & 0 & x-x^2 & -x & 0 & 1 \\
\end{array} \right] \xrightarrow[\frac{1}{x-x^2} R_3]{\frac{1}{x-1}R_2}
\left[\begin{array}{rrr|rrr}
1 & 1 & x & 1 &0 & 0 \\[8pt] 0 & 1 & 0 & \frac{-1}{x-1} & \frac{1}{x-1} & 0 \\[8pt] 0 & 0 & 1 & \frac{-1}{1-x} & 0 & \frac{1}{x-x^2} \\
\end{array} \right]\\[6pt] & \xrightarrow{R_1-R_2}
\left[\begin{array}{rrr|rrr}
1 & 0 & x & \frac{x}{x-1} & \frac{-1}{x-1} & 0 \\[8pt] 0 & 1 & 0 & \frac{-1}{x-1} & \frac{1}{x-1} & 0 \\[8pt] 0 & 0 & 1 & \frac{-1}{1-x} & 0 & \frac{1}{x-x^2} \\
\end{array} \right] \xrightarrow{R_1-xR_3}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & 0 & \frac{-1}{x-1} & \frac{-x}{x-x^2}\\[8pt] 0 & 1 & 0 & \frac{-1}{x-1} & \frac{1}{x-1} & 0 \\[8pt] 0 & 0 & 1 & \frac{-1}{1-x} & 0 & \frac{1}{x-x^2} \\
\end{array} \right].
\end{align*}

Now that we reduced the left $3\times 3$ matrix into the identity matrix, the right $3\times 3$ matrix is the inverse matrix of $A$.
(Note that when we applied elementary row operations, we divided by $x-1$ and $x-x^2$, and this is where we needed to assume $x \neq 0, 1$.)

We have
\begin{align*}
A^{-1}=\begin{bmatrix}
0 & \frac{-1}{x-1} & \frac{-x}{x-x^2}\\[8pt] \frac{-1}{x-1} & \frac{1}{x-1} & 0 \\[8pt] \frac{-1}{1-x} & 0 & \frac{1}{x-x^2} \\
\end{bmatrix}
=\frac{1}{x(1-x)}\begin{bmatrix}
0 & x & -x \\
x &-x &0 \\
-x & 0 & 1
\end{bmatrix}.
\end{align*}

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