Linear Algebra

Find Values of $a$ so that Augmented Matrix Represents a Consistent System

Problem 249

Suppose that the following matrix $A$ is the augmented matrix for a system of linear equations.
\[A= \left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
2 &-1 & -2 & a^2 \\
-1 & -7 & -11 & a
\end{array} \right],\] where $a$ is a real number. Determine all the values of $a$ so that the corresponding system is consistent.

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Solution.

We apply the elementary row operations to $A$ as follows.
\begin{align*}
A= \left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
2 &-1 & -2 & a^2 \\
-1 & -7 & -11 & a
\end{array} \right] \xrightarrow{\substack{R_2-2R_1\\ R_3+R_1}}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
0 &-5 & -8 & a^2-8 \\
0 & -5 & -8 & a+4
\end{array} \right]\\
\xrightarrow{R_3-R_2}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
0 &-5 & -8 & a^2-8 \\
0 & 0 & 0 & -a^2+a+12
\end{array} \right].
\end{align*}
The last matrix is in row echelon form. (It’s not in reduced row echelon form.)
Then the system is consistent if and only if $-a^2+a+12$ is zero. (See Remark below.)
Since we can factor
\[0=-a^2+a+12=-(a+3)(a-4),\] we see that the system is consistent if and only if $a=-3, 4$.

Remark

If $-a^2+a+12 \neq 0$, then we divide the third row by $-a^2+a+12$ and obtain
\[ \left[\begin{array}{rrr|r}
0 & 0 & 0 & 1
\end{array} \right]\] in the third row.
The corresponding linear equation is
\[0x_1+0x_2+0x_3=1,\] and clearly, there is no solution to this equation. Hence the system is inconsistent.

On the other hand, if $-a^2+a+12 = 0$, then the system has solutions. (You just need to reduce the above matrix further or use the back substitution.)

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