# Inverse Matrices Problem

• Oct 23rd 2010, 03:34 PM
Paymemoney
Inverse Matrices Problem
Hi
The following question, can someone tell me if this is the right method of doing it.
If, $\displaystyle X =\begin {bmatrix}x \\ y \\ z \\\end {bmatrix}$ and $\displaystyle D = \begin {bmatrix} 0\\ 0\\ c\\ \end{bmatrix}$, find conditions on a, b and c so the system of equations represented by CX = D has an infinite number of solutions.

previous question asked to find det C = $\displaystyle \begin{bmatrix}1 & 5 & 1 \\ 1 & 6 & -1 \\ 2 & a & b \\ \end{bmatrix}$

det C = b +2a-22

i made b=0 to find a which equals 11 therefore det C =0 and c=0 when i made the equation $\displaystyle X=C^{-1}D$

P.S
• Oct 24th 2010, 01:47 AM
HappyJoe
You need to remember the following fact:

A matrix $\displaystyle C$ has an inverse, if and only if $\displaystyle \text{det}(C)\neq 0$.

You are choosing $\displaystyle a$ and $\displaystyle b$, such that $\displaystyle \text{det}(C)=0$, in which case the matrix $\displaystyle C$ is not invertible, so the expression "$\displaystyle C^{-1}$" does not make sense in this case.
• Oct 24th 2010, 03:26 AM
Paymemoney
so how would you suggest i approach this question.
• Oct 24th 2010, 03:54 AM
HappyJoe
I would do it in the "old-fashioned" way. To solve the equation $\displaystyle CX=D$ for $\displaystyle X$, you row-reduce the augmented matrix

$\displaystyle \begin{bmatrix}1 & 5 & 1 & 0\\ 1 & 6 & -1 & 0 \\ 2 & a & b & c \\ \end{bmatrix}.$

For some values of $\displaystyle a$, $\displaystyle b$, and $\displaystyle c$, the last row may end up looking like

$\displaystyle \begin{bmatrix}0 & 0 & 0 & 1\\ \end{bmatrix},$

in which case there are no solutions. On the other hand, the last row could also end up looking like

$\displaystyle \begin{bmatrix}0 & 0 & 0 & 0\\ \end{bmatrix},$

in which case you need to consider, if the last variable (the $\displaystyle z$-variable) can be a free variable, thus giving rise to an infinite number of solutions.