Q : 10       Solve system of linear equations, using matrix method.

                 \small 5x+2y=3

                \small 3x+2y=5

Answers (1)
D Divya Prakash Singh

The given system of equations

 \small 5x+2y=3

 \small 3x+2y=5

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

A = \begin{bmatrix} 5 &2 \\ 3& 2 \end{bmatrix}X = \begin{bmatrix} x\\y \end{bmatrix}  and B = \begin{bmatrix} 3\\5 \end{bmatrix}

we have, 

|A| = 10-6=4 \neq 0.

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

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

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

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

\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 6-10\\ -9+25 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} -4\\16 \end{bmatrix}= \begin{bmatrix} -1\\4 \end{bmatrix}

Hence the solutions of the given system of equations;

x =-1 \ and\ y =4.

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