Question

find the system of equation x-y+3z=5,4x+2y-z=0,-x+3y+z=5 by using matrix inversion method​

Answer 1

Given:

The system of equation x-y+3z=5, 4x+2y-z=0, -x+3y+z=5.

To Find:

find the system of equation using matrix inversion method.

Solution:

given system of equation is -

$x-y+3z=5,\\4x+2y-z=0,\\-x+3y+z=5$

therefore,

$\[\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&3 \\ 4&2&{ - 1} \\ { - 1}&3&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5 \\ 0 \\ 5 \end{array}} \right]\]$

it forms :- AX = B

⇒ X = (A^-1)B

therefore, |A| = 16

$\[Adj(A) = \left[ {\begin{array}{*{20}{c}} 5&{ - 3}&{14} \\ {10}&4&{ - 2} \\ { - 5}&{13}&6 \end{array}} \right]\]$

so,

$X = \dfrac{1}{|A|}AdjA(B)$

$X = \dfrac{1}{{16}}\left[ {\begin{array}{*{20}{c}} 5&{ - 3}&{14} \\ {10}&4&{ - 2} \\ { - 5}&{13}&6 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5 \\ 0 \\ 5 \end{array}} \right] \hfill \\\\ \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \dfrac{1}{{16}}\left[ {\begin{array}{*{20}{c}} {25 + 0 + 70} \\ {50 + 0 - 10} \\ { - 25 + 0 + 30} \end{array}} \right] \hfill$

$\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \dfrac{1}{{16}}\left[ {\begin{array}{*{20}{c}} {95} \\ {40} \\ 5 \end{array}} \right] \hfill$

On comparing , we get

$x=\dfrac{95}{16}\\y=\dfrac{40}{16}\\and\\z=\dfrac{5}{16}$

Answer 2

Step-by-step explanation: