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  • Solve system of linear equations, using matrix method. x - y + z = 4, 2x + y - 3z = 0, x + y + z = 2.

Q : 12      Solve system of linear equations, using matrix method.

                \small x-y+z=4

                \small 2x+y-3z=0

               \small x+y+z=2

Answers (1)

best_answer

The given system of equations

 \small x-y+z=4

 \small 2x+y-3z=0

 \small x+y+z=2

can be written in the matrix form of AX =B, where

A = \begin{bmatrix} 1 &-1 &1 \\ 2 & 1 &-3 \\ 1& 1 &1 \end{bmatrix}, X = \begin{bmatrix} x\\y \\z \end{bmatrix}  and\ B =\begin{bmatrix} 4\\ 0 \\ 2 \end{bmatrix}. 

we have, 

|A| =1(1+3)+1(2+3)+1(2-1) = 4+5+1= 10 \neq 0.

So, A is non-singular, Therefore, its inverse A^{-1} exists.

as we know A^{-1} = \frac{1}{|A|} (adjA)

Now, we will find the cofactors;

A_{11} =(-1)^{1+1}(1+3) = 4      A_{12} =(-1)^{1+2}(2+3) = -5

A_{13} =(-1)^{1+3}(2-1) = 1      A_{21} =(-1)^{2+1}(-1-1) = 2

A_{22} =(-1)^{2+2}(1-1) = 0      A_{23} =(-1)^{2+3}(1+1) = -2

A_{31} =(-1)^{3+1}(3-1) = 2     A_{32} =(-1)^{3+2}(-3-2) = 5

A_{33} =(-1)^{3+3}(1+2) = 3

(adjA) =\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}

A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{10}\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}

So, the solutions can be found by X = A^{-1}B =\frac{1}{10}\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}\begin{bmatrix} 4\\0 \\ 2 \end{bmatrix}

\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 16+0+4\\-20+0+10 \\ 4+0+6 \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 20\\-10 \\ 10 \end{bmatrix}= \begin{bmatrix} 2\\-1 \\ 1 \end{bmatrix}

Hence the solutions of the given system of equations;

x =2,\ y =-1,\ and\ \ z=1.

Posted by

Divya Prakash Singh

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