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  • Solve system of linear equations, using matrix method. 2x + y + z = 1, x - 2y - z = 3/2, 3y - 5z = 9.

Q : 11       Solve system of linear equations, using matrix method.

                 \small 2x+y+z=1

                 \small x-2y-z= \frac{3}{2}

                \small 3y-5z=9

Answers (1)

best_answer

The given system of equations

  \small 2x+y+z=1

 \small x-2y-z= \frac{3}{2}

 \small 3y-5z=9

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

A = \begin{bmatrix} 2 &1 &1 \\ 1 & -2 &-1 \\ 0& 3 &-5 \end{bmatrix}, X = \begin{bmatrix} x\\y \\z \end{bmatrix}  and B =\begin{bmatrix} 1\\ \\ \frac{3}{2} \\ \\ 9 \end{bmatrix}

we have, 

|A| =2(10+3)-1(-5-0)+1(3-0) = 26+5+3 = 34 \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}(10+3) = 13      A_{12} =(-1)^{1+2}(-5-0) = 5

A_{13} =(-1)^{1+3}(3-0) = 3      A_{21} =(-1)^{2+1}(-5-3) = 8

A_{22} =(-1)^{2+2}(-10-0) = -10      A_{23} =(-1)^{2+3}(6-0) = -6

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

A_{33} =(-1)^{3+3}(-4-1) = -5

(adjA) =\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}

A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{34}\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}

So, the solutions can be found by X = A^{-1}B = \frac{1}{34}\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}\begin{bmatrix} 1\\\frac{3}{2} \\ 9 \end{bmatrix}

\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{34}\begin{bmatrix} 13+12+9\\5-15+27 \\ 3-9-45 \end{bmatrix} = \frac{1}{34}\begin{bmatrix} 34\\17 \\ -51 \end{bmatrix}= \begin{bmatrix} 1\\\frac{1}{2} \\ -\frac{3}{2} \end{bmatrix}

Hence the solutions of the given system of equations;

x =1,\ y =\frac{1}{2},\ and\ \ z=-\frac{3}{2}.

Posted by

Divya Prakash Singh

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