Chapter 4, Problem 19PT

To determine

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

Answer to Problem 19PT

  $\text{x=-2 and y=5}$ .

Explanation of Solution

Given:

  $\begin{array}{l}\text{2x-y=-9}\\ \text{x+2y=8}\end{array}$

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

  $\begin{array}{l}\text{2x-y=-9}\\ \text{x+2y=8}\end{array}$ .

It can be written on the basis of linear equations in matrix form $\left[\begin{array}{cc}\text{2}& \text{-1}\\ \text{1}& \text{2}\end{array}\right]\left[\begin{array}{c}\text{x}\\ \text{y}\end{array}\right]\text{=}\left[\begin{array}{c}\text{-9}\\ \text{8}\end{array}\right]$

Let $\text{A=}\left[\begin{array}{cc}\text{2}& \text{-1}\\ \text{1}& \text{2}\end{array}\right]$ , $\text{X=}\left[\begin{array}{c}\text{x}\\ \text{y}\end{array}\right]$ and $\text{B=}\left[\begin{array}{c}\text{-9}\\ \text{8}\end{array}\right]$ .

Since, $\text{AX=B}$

  ${\text{X=A}}^{\text{-1}}\text{B}$

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

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

  $\text{adjA=}{\left[\begin{array}{cc}2& -1\\ 1& 2\end{array}\right]}^{\text{T}}=\left[\begin{array}{cc}2& 1\\ -1& 2\end{array}\right]$ .

  $\left|\text{A}\right|\text{=}\left|\text{4+1}\right|\text{=5}\text{.}$

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

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

  $\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}\frac{\text{2}}{\text{5}}& \frac{\text{1}}{\text{5}}\\ \frac{\text{-1}}{\text{5}}& \frac{\text{2}}{\text{5}}\end{array}\right]\text{.}\left[\begin{array}{c}\text{-9}\\ \text{8}\end{array}\right]$

  $\left[\begin{array}{c}\text{x}\\ \text{y}\end{array}\right]\text{=}\left[\begin{array}{c}\frac{\text{-18}}{\text{5}}\text{+}\frac{\text{8}}{\text{5}}\\ \frac{\text{9}}{\text{5}}\text{+}\frac{\text{16}}{\text{5}}\end{array}\right]$

  $\left[\begin{array}{c}\text{x}\\ \text{y}\end{array}\right]\text{=}\left[\begin{array}{c}\text{-2}\\ \text{5}\end{array}\right]$

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

  $\text{x=-2 and y=5}$ .