Question

x+y+3z=10, x-y-z=-2, 2x+3y+4z=4 by matrix method

Answer 1

answer for your question......

Answer 2

The given equations are

x+y+3z = 10                          x-y-z = -2                          2x+3y+4z = 4

Converting the equations in matrix form we get,

A = $\left[\begin{array}{ccc}1&1&3\\1&-1&-1\\2&3&4\end{array}\right]$                 X= $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$         B= $\left[\begin{array}{ccc}10\\-2\\4\end{array}\right]$

First we will find the inverse of A

|A| = 1{-4-(-3)} - 1{4-(-2)} + 3{3-(-2)}

    = 1(-1) -1(6) + 3(5)

    = -1-6+15

    = 8 ≠ 0  Therefore, the inverse exists.

Now,

$A_{11}$ = -1                    $A_{12}$ = -6                            $A_{13}$ = 5

$A_{21}$ = 5                    $A_{22}$ = -2                            $A_{23}$ = -1

$A_{31}$ = 2                    $A_{32}$ = 4                             $A_{33}$  = -2

adj A = $\left[\begin{array}{ccc}-1&-6&5\\5&-2&-1\\2&4&-2\end{array}\right] ^{'}$ = $\left[\begin{array}{ccc}-1&5&2\\-6&-2&4\\5&-1&-2\end{array}\right]$

Therefore, $A^{-1}$ =   $\frac{adj A}{|A|} }$

                         =   $\frac{1}{8}$ $\left[\begin{array}{ccc}-1&5&2\\-6&-2&4\\5&-1&-2\end{array}\right]$

Now, $A^{-1}B = X$

        $\frac{1}{8}$$\left[\begin{array}{ccc}-1&5&2\\-6&-2&4\\5&-1&-2\end{array}\right]$ $\left[\begin{array}{ccc}10\\-2\\4\end{array}\right]$ = $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$

        $\frac{1}{8}$ $\left[\begin{array}{ccc}-16\\-40\\48\end{array}\right]$ = $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$

       $\left[\begin{array}{ccc}-2\\-5\\6\end{array}\right]$ = $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$

       

So, x= -2 ,  y= -5 ,   z= 6.