0
$\begingroup$

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

$x + y +3z = a$

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

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Gerry Myerson Apr 3 '14 at 8:40
  • $\begingroup$ I am guessing that you want to write x,y, z in terms of 'a'. Most likely by using Gauss-Jordan Elimination. $\endgroup$ – Dreamer78692 Apr 3 '14 at 8:42
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

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...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.