Chapter 4, Problem 26SGR

To determine

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

Answer to Problem 26SGR

  $\text{x=2, y=6}$ .

Explanation of Solution

Given:

  $\begin{array}{l}\text{3x-y=0}\\ \text{5x+2y=22}\end{array}$

Concept used:

Steps to solve the inverse of matrix by

Step1: calculating the matrix of minors.

Step2: turn that into the matrix of cofactors.

Step3: then ad jugate by transposing the cofactor.

Step4: last multiply by the $\frac{1}{\text{determinant}}$ .

Formula for inverse of matrix A is:

  ${\text{A}}^{\text{-1}}\text{=}\frac{\text{adjA}}{\left|\text{A}\right|}$ .

  $\text{AX=B}$

  ${\text{X=A}}^{\text{-1}}\text{B}$ .

Calculation:

Here,

  $\begin{array}{l}\text{3x-y=0}\\ \text{5x+2y=22}\end{array}$

It can be written on the basis of linear equations in matrix form $\left[\begin{array}{cc}\text{3}& \text{-1}\\ \text{5}& \text{2}\end{array}\right]\left[\begin{array}{c}\text{x}\\ \text{y}\end{array}\right]\text{=}\left[\begin{array}{c}\text{0}\\ \text{22}\end{array}\right]$ .

Let $\text{A=}\left[\begin{array}{cc}\text{3}& \text{-1}\\ \text{5}& \text{2}\end{array}\right]$ , $\text{X=}\left[\begin{array}{c}\text{x}\\ \text{y}\end{array}\right]$ and $\text{B=}\left[\begin{array}{c}\text{0}\\ \text{22}\end{array}\right]$ . $\text{adjA=}{\left[\begin{array}{cc}2& -5\\ 1& 3\end{array}\right]}^{\text{T}}=\left[\begin{array}{cc}2& 1\\ -5& 3\end{array}\right]$ .

Since, $\text{AX=B}$

  ${\text{X=A}}^{\text{-1}}\text{B}$

Finding ${\text{A}}^{-1}$

  ${\text{A}}^{\text{-1}}\text{=}\frac{\text{adjA}}{\left|\text{A}\right|}$ .

  $\left|\text{A}\right|\text{=}\left|6+5\right|=11.$

  ${\text{A}}^{\text{-1}}\text{=}\frac{\left[\begin{array}{cc}2& 1\\ -5& 3\end{array}\right]}{11}=\left[\begin{array}{cc}\frac{2}{11}& \frac{1}{11}\\ \frac{-5}{11}& \frac{3}{11}\end{array}\right]$ .

  ${\text{A}}^{\text{-1}}=\left[\begin{array}{cc}\frac{2}{11}& \frac{1}{11}\\ \frac{-5}{11}& \frac{3}{11}\end{array}\right]$ .

Since ${\text{X=A}}^{\text{-1}}\text{B}$

  $\text{X=}\left[\begin{array}{c}\text{x}\\ \text{y}\end{array}\right]=\left[\begin{array}{cc}\frac{2}{11}& \frac{1}{11}\\ \frac{-5}{11}& \frac{3}{11}\end{array}\right]\left[\begin{array}{c}\text{0}\\ \text{22}\end{array}\right]$ .

  $\text{X=}\left[\begin{array}{c}\text{x}\\ \text{y}\end{array}\right]\text{=}\left[\begin{array}{c}0+2\\ 0+6\end{array}\right]\text{=}\left[\begin{array}{c}2\\ 6\end{array}\right]$ .

  $\left[\begin{array}{c}\text{x}\\ \text{y}\end{array}\right]=\left[\begin{array}{c}2\\ 6\end{array}\right]$

Hence comparing matrix of L.H.S and R.H.S

  $\text{x=2, y=6}$ .