Hi Thank you all very much for your answers to my previous questions, really appreciate them. Unfortunately, I have one more..

Find the values of a for which the system of equations has a unique solution.

3x-ay+2z = 4

x+2y-3z = 1

-x-y+z = 12

In matrix form,

\\1 & 2 & -3 \\-1 & -1 & 1 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix}" alt="\begin{bmatrix} 3 & -a & 2 \\1 & 2 & -3 \\-1 & -1 & 1 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix}" />

= \\1\\12 \end{bmatrix}" alt="\begin{bmatrix} 4 \\1\\12 \end{bmatrix}" />

I know I am supposed to solve it such that the determinant is not zero to get a unique solution..I know how to get the determinant of 3 by 3 matrices, I got a = -0.5 (or is it 0.5? ( I was confused with the negative signs)) for determinant =0. However, I was not taught the inverse of a 3 by 3 matrix, I was told to key it into the calculator to which I can do, but not here, since there is an unknown a..I tried substituting with a=-0.5 but I goe fractional entries, and I do not think I am doing it the right way anyway..

Please help me!

I have done such questions before but they were only asking for the inverse of a two by two matrix, and I could easily find the value of the unknown etc..