# I need to solve for “a” in the following 3 equations

$x+2y +z = a^2$

$x + y +3z = a$

$3x +4y + 8z = 8$

I have studied matrices, but not when "a" is involved in finding the solution. Please assist

• The title says you need to solve for $a$, but I suspect you mean something else, since you can solve for $a$ by just writing $a=x+y+z$. What do you really want? – Gerry Myerson Apr 3 '14 at 8:40
• I am guessing that you want to write x,y, z in terms of 'a'. Most likely by using Gauss-Jordan Elimination. – Dreamer78692 Apr 3 '14 at 8:42

## 2 Answers

The determinant of the matrix is not zero, so $a$ can be anything.

$\begin{bmatrix}1 & 2 & 1 & |a^2 \\1 & 1 & 3 & |a\\ 3& 4 & 8 & |8\end{bmatrix}$

Now use matrix reduction method to solve, something like this:

$\begin{bmatrix}1 & 2 & 1 & |a^2 \\1 & 1 & 3 & |a\\ 3& 4 & 8 & |8\end{bmatrix}$ ~$\begin{bmatrix}1 & 2 & 1 & |a^2 \\0 & -1 & 2 & |a-a^2\\ 0 & -2 & 5 & |8 -3a^2\end{bmatrix}$ ~ $\begin{bmatrix}1 & 2 & 1 & |a^2 \\0 & -1 & 2 & |a-a^2\\ 0 & 0 & 1 & |8 -3a^2-2(a-a^2)\end{bmatrix}$

Now you solve step by step for x,y and z.

$z=8 -3a^2-2(a-a^2)=8-2a-a^2$

$-y+2z=a-a^2$

$y=2z-a+a^2=2(8-2a-a^2)-a+a^2=16-5a-a^2$

$x+2y+z=a^2$

$x=a^2-z-2y=a^2-8+2a+a^2-2(16-5a-a^2)=-40+12a+3a^2$

And now you can write down the solution as:

$x=-40+12a+3a^2$

$y=16-5a-a^2$

$z=8-2a-a^2$

Or in the vector form using base $1,a,a^2$... But this can be calculated of $a\in R$ as it's mentioned in the comment bellow system determinant is nonzero so $a$ can be anything...