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 method of inversion

$5 x - y + 4 z = 5, 2 x + 3 y + 5 z = 2$ and $5 x - 2 y + 6 z = - 1$

Accepted Answer

Given equations are $5 x - y + 4 z = 5, 2 x + 3 y + 5 z = 2$ and $5 x - 2 y + 6 z = - 1$

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

$[\begin{matrix} 5 & - 1 & 4 \\ 2 & 3 & 5 \\ 5 & - 2 & 6 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 5 \\ 2 \\ - 1 \end{matrix}]$

Here, $[\begin{matrix} 5 & - 1 & 4 \\ 2 & 3 & 5 \\ 5 & - 2 & 6 \end{matrix}] = A$ , $[\begin{matrix} x \\ y \\ z \end{matrix}] = X$ and $[\begin{matrix} 5 \\ 2 \\ - 1 \end{matrix}] = B$

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 $|A|$

$A = [\begin{matrix} 5 & - 1 & 4 \\ 2 & 3 & 5 \\ 5 & - 2 & 6 \end{matrix}]$

$|A| = 5 (18 + 10) + 1 (12 - 25) + 4 (- 4 - 15) = 51$

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

$A_{11} = {(- 1)}^{1 + 1} |\begin{matrix} 3 & 5 \\ - 2 & 6 \end{matrix}| = 1 (18 + 10) = 28$

$A_{12} = {(- 1)}^{1 + 2} |\begin{matrix} 2 & 5 \\ 5 & 6 \end{matrix}| = - 1 [12 - 25] = 13$

$A_{13} = {(- 1)}^{1 + 3} |\begin{matrix} 2 & 3 \\ 5 & - 2 \end{matrix}| = 1 [- 4 - 15] = - 19$

$A_{21} = {(- 1)}^{2 + 1} |\begin{matrix} - 1 & 4 \\ - 2 & 6 \end{matrix}| = - 1 [- 6 + 8] = - 2$

$A_{22} = {(- 1)}^{2 + 2} |\begin{matrix} 5 & 4 \\ 5 & 6 \end{matrix}| = 1 [30 - 20] = 10$

$A_{23} = {(- 1)}^{2 + 3} |\begin{matrix} 5 & - 1 \\ 5 & - 2 \end{matrix}| = - 1 [- 10 + 5] = 5$

$A_{31} = {(- 1)}^{3 + 1} |\begin{matrix} - 1 & 4 \\ 3 & 5 \end{matrix}| = 1 [- 5 - 12] = - 17$

$A_{32} = {(- 1)}^{3 + 2} |\begin{matrix} 5 & 4 \\ 2 & 5 \end{matrix}| = - 1 [25 - 8] = - 17$

$A_{33} = {(- 1)}^{3 + 3} |\begin{matrix} 5 & - 1 \\ 2 & 3 \end{matrix}| = 1 [15 + 2] = 17$

$∴$ $adj A = [\begin{matrix} 28 & - 2 & - 17 \\ 13 & 10 & - 17 \\ - 19 & 5 & 17 \end{matrix}]$

Therefore, by using the formula for $A^{- 1}$

$A^{- 1} = \frac{1}{|A|} (adj A)$

$∴ A^{- 1} = \frac{1}{51} [\begin{matrix} 28 & - 2 & - 17 \\ 13 & 10 & - 17 \\ - 19 & 5 & 17 \end{matrix}]$

Use equation (i) to get

$X = \frac{1}{51} [\begin{matrix} 28 & - 2 & - 17 \\ 13 & 10 & - 17 \\ - 19 & 5 & 17 \end{matrix}] [\begin{matrix} 5 \\ 2 \\ - 1 \end{matrix}]$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{51} [\begin{matrix} 28 & - 2 & - 17 \\ 13 & 10 & - 17 \\ - 19 & 5 & 17 \end{matrix}] [\begin{matrix} 5 \\ 2 \\ - 1 \end{matrix}]$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{51} [\begin{matrix} 153 \\ 102 \\ - 102 \end{matrix}]$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 3 \\ 2 \\ - 2 \end{matrix}]$

Hence, by equality of matrices we get $x = 3$ , $y = 2$ and $z = - 2$ .

Solve the following equations by the method of inversion

$5 x - y + 4 z = 5, 2 x + 3 y + 5 z = 2$ and $5 x - 2 y + 6 z = - 1$

Important Questions on Matrices

Learn and Prepare for any exam you want !

Solve the following equations by the method of inversion 5x-y+4z=5, 2x+3y+5z=2 and 5x-2y+6z=-1

Important Questions on Matrices

Learn and Prepare for any exam you want !

Solve the following equations by the method of inversion

$5 x - y + 4 z = 5, 2 x + 3 y + 5 z = 2$ and $5 x - 2 y + 6 z = - 1$