For What Values of $a$, Is the Matrix Nonsingular?

Nonsingular matrix and singular matrix problems and solutions

Problem 670

Determine the values of a real number $a$ such that the matrix
\[A=\begin{bmatrix}
3 & 0 & a \\
2 &3 &0 \\
0 & 18a & a+1
\end{bmatrix}\] is nonsingular.
 
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Solution.

We apply elementary row operations and obtain:
\begin{align*}
A=\begin{bmatrix}
3 & 0 & a \\
2 &3 &0 \\
0 & 18a & a+1
\end{bmatrix}
\xrightarrow{R_1-R_2}
\begin{bmatrix}
1 & -3 & a \\
2 &3 &0 \\
0 & 18a & a+1
\end{bmatrix}\\[6pt] \xrightarrow{R_2-2R_1}
\begin{bmatrix}
1 & -3 & a \\
0 &9 &-2a \\
0 & 18a & a+1
\end{bmatrix}
\xrightarrow{R_3-(2a)R_2}
\begin{bmatrix}
1 & -3 & a \\
0 &9 &-2a \\
0 & 0 & 4a^2+a+1
\end{bmatrix}.
\end{align*}

From this, we see that the matrix $A$ is nonsingular if and only if the $(3, 3)$-entry $4a^2+a+1$ is not zero.
By the quadratic formula, we see that
\[a=\frac{-1\pm \sqrt{-15}}{8}\] are solutions of $4a^2+a+1=0$.

Note that these are not real numbers. Thus, for any real number $a$, we have $4a^2+a+1\neq 0$.

Hence, we can divide the third row by this number, and eventually we can reduce it to the identity matrix.
So the rank of $A$ is $3$, and $A$ is nonsingular for any real number $a$.


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