Name: The Applications of Matrices
Uploaded: 2019-07-19
Duration: 7 min 47 s
Description: Let's learn the applications of matrices, such as solving homogenous and non-homogenous linear equations, the Gauss-Jordan method, and the reduction method.

Question

Solve the following equations by the methods of inversion

2 x - y = - 2, x + 5 y = 5

Accepted Answer

Given equations are $2 x - y = - 2$ and $x + 5 y = 5$

Use the given equations to get the corresponding matrix equation as follows:

$[\begin{matrix} 2 & - 1 \\ 1 & 5 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} - 2 \\ 5 \end{matrix}]$

Here, $[\begin{matrix} 2 & - 1 \\ 1 & 5 \end{matrix}] = A$ , $[\begin{matrix} x \\ y \end{matrix}] = X$ and $[\begin{matrix} - 2 \\ 5 \end{matrix}] = B$

Hence, the matrix equation is $A X = B$

Multiply the above matrix equation by $A^{- 1}$ , we get

$(A^{- 1} A) X = A^{- 1} B$

$I X = A^{- 1} B$ $(∵ A^{- 1} A = I)$

$X = A^{- 1} B$ -----(i)

Now find the $|A|$

$A = [\begin{matrix} 2 & - 1 \\ 1 & 5 \end{matrix}]$

$|A| = 10 + 1 = 11$

For the given matrix $A$ find minors and co-factors.

$A_{11} = {(- 1)}^{1 + 1} |5| = 5$

$A_{12} = {(- 1)}^{1 + 2} |1| = - 1$

$A_{21} = {(- 1)}^{2 + 1} |1| = - 1$

$A_{22} = {(- 1)}^{2 + 2} |2| = 2$

$∴$ $adj A = [\begin{matrix} 5 & - 1 \\ - 1 & 2 \end{matrix}]$

$∴$ $A^{- 1} = \frac{1}{|A|} (adj A)$

$A^{- 1} = \frac{1}{11} [\begin{matrix} 5 & - 1 \\ - 1 & 2 \end{matrix}]$

Use equation (i) to get

$X = \frac{1}{11} [\begin{matrix} 5 & - 1 \\ - 1 & 2 \end{matrix}] [\begin{matrix} - 2 \\ 5 \end{matrix}]$

$[\begin{matrix} x \\ y \end{matrix}] = \frac{1}{11} [\begin{matrix} 5 & - 1 \\ - 1 & 2 \end{matrix}] [\begin{matrix} - 2 \\ 5 \end{matrix}]$

$[\begin{matrix} x \\ y \end{matrix}] = \frac{1}{11} [\begin{matrix} 5 \times (- 2) + (- 1) \times 5 \\ (- 1) \times (- 2) + 2 \times 5 \end{matrix}]$

$[\begin{matrix} x \\ y \end{matrix}] = \frac{1}{11} [\begin{matrix} - 10 - 5 \\ 2 + 10 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} - \frac{15}{11} \\ \frac{12}{11} \end{matrix}]$

Hence, by equality of matrices we get $x = - \frac{15}{11}$ and $y = \frac{12}{11}$ .

Solve the following equations by the methods of inversion $2 x - y = - 2, x + 5 y = 5$

Important Questions on Matrices

Learn and Prepare for any exam you want !

Solve the following equations by the methods of inversion 2x-y=-2, x+5y=5

Important Questions on Matrices

Learn and Prepare for any exam you want !

Solve the following equations by the methods of inversion $2 x - y = - 2, x + 5 y = 5$