Question

solve the following system of equestion by matrix method 3x -2y +3z = 8 , 2x +y -z = 1 , 4x -3y +2z =4​

Answer 1

Given:

Three equations are given -

$\[3x - 2y + 3z = 8\],$$\[2x + y - z = 1\]$ and $\[4x - 3y + 2z = 4\]$.

To Solve:

Solve the above given equation by metric method.

Solution:

Here,  

$\[A = \left( {\begin{array}{*{20}{c}} 3&{ - 2}&3 \\ 2&1&{ - 1} \\ 4&{ - 3}&2 \end{array}} \right)\]$

$\[ B = \left( {\begin{array}{*{20}{c}} 8 \\ 1 \\ 4 \end{array}} \right) \]$  and $\[X = \left({\begin{array}{*{20}{c}} x \\ y \\ z \end{array}}\right)\]$

Since, we know that- $\[AX = B\]$

Now,

$\[ \left( {\begin{array}{*{20}{c}} 3&{ - 2}&3 \\ 2&1&{ - 1} \\ 4&{ - 3}&2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 8 \\ 1 \\ 4 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right)\]$

$\left( {\begin{array}{*{20}{c}} {24 - 2 + 12} \\ {16 + 1 - 4} \\ {32 - 3 + 8} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) \hfill$

$\left( {\begin{array}{*{20}{c}} {34} \\ {13} \\ {37} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) \hfill$

On comparing, we get '

X=34, y=13 and z=37.

Hence,

$\[X = \left( {\begin{array}{*{20}{c}} {34} \\ {13} \\ {37} \end{array}} \right)\]$