Name: Solving System of Linear Equations using Matrices
Uploaded: 2019-07-19
Duration: 11 min 30 s
Description: A system of linear equations can be represented in matrix form using its coefficient, variable and constant matrices. Learn about the inverse matrix method!

Question

Test by matrix method, whether the following system of equations are solvable and hence solve if possible: $\frac{x}{3} - \frac{2}{y} = 1; \frac{x}{4} + \frac{3}{y} = 3$ .

Accepted Answer

Given the system of linear equations are $\frac{x}{3} - \frac{2}{y} = 1; \frac{x}{4} + \frac{3}{y} = 3$

Put $x = a$ and $\frac{1}{y} = b$ , then we have: $\frac{a}{3} - 2 b = 1; \frac{a}{4} + 3 b = 3$

$\Rightarrow a - 6 b = 3; a + 12 b = 12$

Let $A = [\begin{matrix} 1 & - 6 \\ 1 & 12 \end{matrix}], X = [\begin{array}{l} a \\ b \end{array}], B = [\begin{array}{l} 3 \\ 12 \end{array}]$

Writing given system of linear equations in the form of $A X = B$

$[\begin{array}{l} 1 & - 6 \\ 1 & 12 \end{array}] [\begin{array}{l} a \\ b \end{array}] = [\begin{array}{l} 3 \\ 12 \end{array}]$

Now, $|A| = (12) \times (1) - (1) \times (- 6) = 12 + 6 = 18 \neq 0$

Thus, matrix $A$ is non-singular, then inverse of the matrix $A$ is unique and its exists.

And $adj A = [\begin{matrix} 12 & 6 \\ - 1 & 1 \end{matrix}]$ [If $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ , then $adj A = [\begin{matrix} d & - b \\ - c & a \end{matrix}]$ ]

We know that, $A^{- 1} = \frac{1}{| A |} (adj A)$

$\Rightarrow A^{- 1} = \frac{1}{18} [\begin{matrix} 12 & 6 \\ - 1 & 1 \end{matrix}]$

Now, $A X = B$ can be written as $X = A^{- 1} B$ , so we get: $X = \frac{1}{18} [\begin{matrix} 12 & 6 \\ - 1 & 1 \end{matrix}] [\begin{array}{l} 3 \\ 12 \end{array}]$

$\Rightarrow X = \frac{1}{18} [\begin{array}{l} 36 + 72 \\ - 3 + 12 \end{array}]$

$\Rightarrow X = [\begin{array}{l} 2 + 4 \\ \frac{1}{2} \end{array}]$

$\Rightarrow [\begin{array}{l} a \\ b \end{array}] = [\begin{array}{l} 6 \\ \frac{1}{2} \end{array}]$

Thus, $a = 6; b = \frac{1}{2}$ .

Hence, $x = 6; y = 2$ .

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