Question

Using matrices, solve the following system of equation.
$2x-3y+5z=11$

$3x+2y-4z=-5$

$x+y-2z=-3$

Answer 1

Let solution is $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$

Then, $AX=B$, where $A=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]$ and $B=\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right]$

$\to X=A^{-1}B$

We first find $A^{-1}$,

Step 1: Calculate the Matrix of Minors, let it be M

$M=\left[\begin{array}{ccc}0& -2& 1\\ 1& -9& 5\\ 2& -23& 13\end{array}\right]$

Step 2:then turn it into matrix of co-factors,

This is easy! Just apply a "checkerboard" of minuses to the "Matrix of Minors". In other words, we need to change the sign of alternate cells, like this:

$M=\left[\begin{array}{ccc}0& -2& 1\\ 1& -9& 5\\ 2& -23& 13\end{array}\right]$ with $\left[\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right]\to M=\left[\begin{array}{ccc}0& 2& 1\\ -1& -9& -5\\ 2& 23& 13\end{array}\right]$

Step 3:then the adjugate:

Now "Transpose" all elements of the previous matrix... in other words swap their positions over the diagonal (the diagonal stays the same):

$M=\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]$

Step 4:Multiply by $\frac{1}{Determinant}$

$Det\left(A\right)=-1$
$M=\frac{1}{Determinant}\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]=\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]$

Hence $A^{-1}=M$

So $X=MB=\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right]$

$=\left[\begin{array}{c}0\times 11+1\times -5+(-2)\times (-3)\\ (-2)\times 11+9\times (-5)+(-23)\times (-3)\\ (-1)\times 11+5\times (-5)+(-13)\times (-3)\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$

$\therefore x=1,y=2,z=3$