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

Q : 8         Solve system of linear equations, using matrix method.

                 2x-y=-2

                3x+4y=3

Answers (1)

best_answer

The given system of equations

 2x-y=-2

 3x+4y=3

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

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

we have, 

|A| = 8+3=11 \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}{11}\begin{bmatrix} 4 &1 \\ -3& 2 \end{bmatrix}

So, the solutions can be found by X = A^{-1}B = \frac{1}{11}\begin{bmatrix} 4 &1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} -2\\3 \end{bmatrix}

\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -8+3\\ 6+6 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -5\\12 \end{bmatrix}= \begin{bmatrix} -\frac{5}{11}\\ \\-\frac{12}{11} \end{bmatrix}

Hence the solutions of the given system of equations;

x =\frac{-5}{11} \ and\ y =\frac{12}{11}.

Posted by

Divya Prakash Singh

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