Question

x+2y+z=1, 2x-y+z=5,3x+y-z=0 using matrix method​

Answer 1

Given:

We have three linear equations x+2y+z=1, 2x-y+z=5,3x+y-z=0

To Find:

solution by matrix method?

Step-by-step explanation:

• we will solve these equations by using X=AB

• The system of equation are

                   x+2y+z=1,

                   2x-y+z=5

                   3x+y-z=0

• write these equation in form $A^-^1$X=B

 $A=\left[\begin{array}{ccc}1&2&1\\2&-1&1\\3&1&-1\end{array}\right],B=\left[\begin{array}{c}1\\5\\0\end{array}\right],X=\left[\begin{array}{c}1\\x\\z\end{array}\right]$

• Calculating ModA

     $\begin{vmatrix}A\end{vmatrix}=\begin{vmatrix}1&2&1\\2&-1&1\\3&1&-1\end{vmatrix}\\=1(1-1)-2(-2-3)+1(2+3)=15$

Since it is 15

• Thus system of equation has unique solution.

• Calculating $A^-^1$

      $adj A=\left[\begin{array}{ccc}A_1_1&A_2_1&A_3_1\\A_1_2&A_2_2&A_3_2\\A_1_3&A_2_3&A_3_3\end{array}\right]=\left[\begin{array}{ccc}0&3&3\\5&-4&1\\5&5&-5\end{array}\right]$

• Thus

     X=AB

             $X=\frac{1}{15}\left[\begin{array}{ccc}0&3&3\\5&-4&1\\5&5&-5\end{array}\right] \left[\begin{array}{c}1\\5\\0\end{array}\right] \\X=\frac{1}{15}\left[\begin{array}{c}15\\-15\\30\end{array}\right]\\\\X=\left[\begin{array}{c}1\\-1\\2}\end{array}\right]$

Hence, value of x=1,y=-1,z=2