Find All Values of $x$ such that the Matrix is Invertible

Inverse Matrices Problems and Solutions

Problem 721

Given any constants $a,b,c$ where $a\neq 0$, find all values of $x$ such that the matrix $A$ is invertible if
\[
A=
\begin{bmatrix}
1 & 0 & c \\
0 & a & -b \\
-1/a & x & x^{2}
\end{bmatrix}
.
\]

 
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Solution.

We know that $A$ is invertible precisely when $\det(A)\neq 0$. We therefore compute, by expanding along the first row,
\begin{align*}
\det(A)
&=
1
\begin{vmatrix}
a & -b \\ x & x^{2}
\end{vmatrix}
+c
\begin{vmatrix}
0 & a \\ -1/a & x
\end{vmatrix}
=
1(ax^{2}+bx)
+c(0+1)
\\
&=
ax^{2}+bx+c
.
\end{align*}
Thus $\det(A)\neq 0$ when $ax^{2}+bx+c\neq 0$. We know by the quadratic formula that $ax^{2}+bx+c=0$ precisely when
\[
x=
\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2}
.
\] Therefore, $A$ is invertible so long as $x$ satisfies both of the following inequalities:
\[
x\neq
\dfrac{-b+\sqrt{b^{2}-4ac}}{2}
,\quad
x\neq
\dfrac{-b-\sqrt{b^{2}-4ac}}{2}
.
\]


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