Chapter 4, Problem 32SGR

To determine

To find: the solution of the equation using Cramer’s rule.

Answer to Problem 32SGR

  $\text{x=8}$ and $\text{y=}-12.$

Explanation of Solution

Given:

  $\left[\begin{array}{cc}\text{5}& \text{3}\\ \text{3}& \text{2}\end{array}\right]\text{.}\left[\begin{array}{c}\text{x}\\ \text{y}\end{array}\right]\text{=}\left[\begin{array}{c}\text{4}\\ \text{0}\end{array}\right]$

Concept used:

To multiply the two matrices column of the first matrices must equals to the raw of the second Matrices.

The product will be defined. The dimension of the product of two matrices can be known taking Raw of the first matrix and taking the column of the second matrix.

Using substitution, elimination and addition can solve the two un-known variables.

Calculation:

  $\left[\begin{array}{cc}\text{5}& \text{3}\\ \text{3}& \text{2}\end{array}\right]\text{.}\left[\begin{array}{c}\text{x}\\ \text{y}\end{array}\right]\text{=}\left[\begin{array}{c}\text{4}\\ \text{0}\end{array}\right]$

Solving by matrix multiplication.

  $\Rightarrow \left[\begin{array}{c}\text{5x+3y}\\ \text{3x+2y}\end{array}\right]\text{=}\left[\begin{array}{c}\text{4}\\ \text{0}\end{array}\right]$ .

  $\text{5x+3y}$ =4 ………… $\left(\text{i}\right)$

  $\text{3x+2y}$

  $\text{=0}$ …………. $\left(\text{ii}\right)$

By multiply the equation $\left(\text{i}\right)$ by $\text{2}$ and $\left(\text{ii}\right)$ by $\text{3}$ :

  $\text{10x+6y=8}$ ……. $\left(\text{iii}\right)$

  $\text{9x+6y=0}$ ……… $\left(\text{iv}\right)$

By using eliminating method in equation $\left(\text{iii}\right)$and $\left(\text{iv}\right)$ .

  $\text{x=8}$ .

By putting value of $\text{x=8}$ in equation $\left(\text{iv}\right)$ :

  $\text{9}\left(\text{8}\right)\text{+6y=0}$ .

  $\text{y=-}\frac{\text{72}}{6}=-12.$

  $\text{y=}-12.$

Hence $\text{x=8}$ and $\text{y=}-12.$