Solving systems of linear equations with an unknown 'a' using matrices and elementary row operations

Question

Came across this one the other day... while I can narrow 'a' down I can't seem to find an exact/ optimised figure. For example 'a' cannot equal 1/3, 'a' must be less than 0.5...

Anyway, here's the problem. I've got a 3x3 matrix by a 3x1 which equates to a 3x1.

{0.6, 0.2, a},{0.4, 0.3, a},{0.0, 0.5, 1-2a}.{x, y, z} = {1360, 1260, 2000}

Would appreciate any help available,

Kind regards,

1AM35

Answer 1

Hint: do Gaussian Elimination or Row-Reduced-Echelon-Form (RREF). Note that I changed all the decimals to fractions to get exact results. You will end up with the identity matrix on the left of your augmented matrix, with:

• $z = -\dfrac{940}{3a - 1}$
• $y = \dfrac{40(206 a - 53)}{3a -1}$
• $x = \dfrac{20(281a-78)}{3a-1}$

From this you can see that $a \ne \dfrac{1}{3}$, otherwise $a$ is a "free variable" that you can make anything you like.

You effectively found the inverse of the matrix on the LHS times the column vector on the RHS.