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  • Solve system of linear equations, using matrix method. x - y + 2z = 7, 3x + 4y - 5z = -5, 2x - y + 3z = 12.

Q : 14         Solve system of linear equations, using matrix method.

                  \small x-y+2z=7

                \small 3x+4y-5z=-5

                 \small 2x-y+3z=12

Answers (1)

best_answer

The given system of equations

  \small x-y+2z=7

 \small 3x+4y-5z=-5

 \small 2x-y+3z=12

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

A = \begin{bmatrix} 1 &-1 &2 \\ 3 & 4 &-5 \\ 2& -1 &3 \end{bmatrix}, X = \begin{bmatrix} x\\y \\z \end{bmatrix}  and\ B =\begin{bmatrix} 7\\ -5 \\ 12 \end{bmatrix}. 

we have, 

|A| =1(12-5) +1(9+10)+2(-3-8) = 7+19-22 = 4 \neq0.

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)^{12-5} = 7      A_{12} =(-1)^{1+2}(9+10) = -19

A_{13} =(-1)^{1+3}(-3-8) = -11      A_{21} =(-1)^{2+1}(-3+2) = 1

A_{22} =(-1)^{2+2}(3-4) = -1      A_{23} =(-1)^{2+3}(-1+2) = -1

A_{31} =(-1)^{3+1}(5-8) = -3     A_{32} =(-1)^{3+2}(-5-6) = 11

A_{33} =(-1)^{3+3}(4+3) = 7

(adjA) =\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}

A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{4}\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}

So, the solutions can be found by X = A^{-1}B =\frac{1}{4}\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}\begin{bmatrix} 7\\-5 \\ 12 \end{bmatrix}

\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 49-5-36\\-133+5+132 \\ -77+5+84 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 8\\4 \\ 12 \end{bmatrix}= \begin{bmatrix} 2\\1 \\ 3 \end{bmatrix}

Hence the solutions of the given system of equations;

x =2,\ y =1,\ and\ \ z=3.

 

 

Posted by

Divya Prakash Singh

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