
Matrices
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.
3xay+2z = 4
x+2y3z = 1
xy+z = 12
In matrix form,
$\displaystyle \begin{bmatrix} 3 & a & 2 \\1 & 2 & 3 \\1 & 1 & 1 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix}$
=$\displaystyle \begin{bmatrix} 4 \\1\\12 \end{bmatrix}$
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..

Re: Matrices
the determinant we want is:
$\displaystyle \begin{vmatrix}3&a&2\\1&2&3\\1&1&1 \end{vmatrix}$
for a 3x3 matrix, the easiest method to employ is the "rule of sarrus" (form 3 triples product by "wraparound diagonals" from upperleft to lower right, and sum these, form three more triple products from "wrap around diagonals" from upperright to lowerleft, subtract these from the first sum).
in the matrix at hand that gives us:
$\displaystyle (3)(2)(1) + (a)(3)(1) + (2)(1)(1)  (2)(2)(1)  (a)(1)(1)  (3)(3)(1)$
$\displaystyle = 6  3a  2 + 4 + a  9 = 2a  1$
if the determinant is nonzero, it means the matrix IS invertible, if we call our matrix A, then the unique solution is A^{1}(4,1,12).
the determinant will be nonzero whenever:
$\displaystyle 2a + 1 \neq 0$ that is, when $\displaystyle a \neq \frac{1}{2}$
note that what happens when a = 1/2, is the first column is: twice the second column plus the third. this means that A is of rank 2 (since the 2nd and 3rd columns are clearly LI), and that the image (column space) of A is:
span({(1/2,2,1),(2,3,1)}). in this particular case, when a = 1/2, the system has NO solution, as (4,1,12) is not IN the column space.
since the problem does not ask you to specifically give a solution when a ≠ 1/2, this is all that needs be done.

Re: Matrices
Thank you so much! I get why the answer is 'no solutions'!
Could I trouble you to elaborate more about the part where you identified A is rank 2, I don't remember learning that in school, and we were never taught to look at the relationship of entries within a matrix in such an intricate way!