# Question #47482

Mar 11, 2017

See below

#### Explanation:

Your $n \times n$, $\textcolor{red}{A} \textcolor{b l u e}{m a t h b f x} = \textcolor{g r e e n}{m a t h b f b}$, system in matrix form is:

$\textcolor{red}{\left(\begin{matrix}1 & 2 & - 3 \\ 3 & - 1 & 5 \\ 4 & 1 & {a}^{2} - 14\end{matrix}\right)} \textcolor{b l u e}{\left(\begin{matrix}x \\ y \\ z\end{matrix}\right)} = \textcolor{g r e e n}{\left(\begin{matrix}4 \\ 2 \\ a + 2\end{matrix}\right)}$

When there is no obvious linear pattern - in terms of the row and column vectors dependence - the easiest thing to do is to start row reducing. I cannot get Gaussian elimination working on Socratic so will just stack the numbers up in brackets:

$\left(\begin{matrix}1 & 2 & - 3 \\ 3 & - 1 & 5 \\ 4 & 1 & {a}^{2} - 14\end{matrix}\right) \left(\begin{matrix}4 \\ 2 \\ a + 2\end{matrix}\right)$

$R 2 \to R 2 - 3 R 1 , R 3 \to R 3 - 4 R 1$

$\left(\begin{matrix}1 & 2 & - 3 \\ 0 & - 7 & 14 \\ 0 & - 7 & {a}^{2} - 2\end{matrix}\right) \left(\begin{matrix}4 \\ - 10 \\ a - 14\end{matrix}\right)$

$R 3 \to R 3 - R 2$

$\left(\begin{matrix}1 & 2 & - 3 \\ 0 & - 7 & 14 \\ 0 & 0 & {a}^{2} - 16\end{matrix}\right) \left(\begin{matrix}4 \\ - 10 \\ a - 4\end{matrix}\right)$

So we have this:

$\left(\begin{matrix}1 & 2 & - 3 \\ 0 & - 7 & 14 \\ 0 & 0 & {a}^{2} - 16\end{matrix}\right) \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}4 \\ - 10 \\ a - 4\end{matrix}\right)$

If we start at the bottom:

$\left({a}^{2} - 16\right) z = a - 4$

$\implies \left(a - 4\right) \left(a + 4\right) z = a - 4$

So we go through the scenarios:

• If $a = 4$, then the solution is everything in $\setminus m a t h c a l {R}^{2}$. There are an infinite number of solutions for z.

• If $a \ne 4$, that last equation becomes: $\left(a + 4\right) z = 1 \implies z = \frac{1}{a + 4}$. In that event, $x$ and $y$ are just linear combinations of $z$.

• However, if $a = - 4$, there is no solution to the matrix equation.

We can explore this a bit more.

If $a = 4$, we have this:

$\left(\begin{matrix}1 & 2 & - 3 \\ 0 & - 7 & 14 \\ 0 & 0 & 0\end{matrix}\right) \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}4 \\ - 10 \\ 0\end{matrix}\right)$. Look at the bottom row !

If $a = - 4$, we have this:

$\left(\begin{matrix}1 & 2 & - 3 \\ 0 & - 7 & 14 \\ 0 & 0 & 0\end{matrix}\right) \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}4 \\ - 10 \\ - 8\end{matrix}\right)$. Again, look at the bottom row. This version of $a$ that insists that $z = - 8$.