Chapter 4, Problem 27SGR

To determine

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

Answer to Problem 27SGR

  $\text{x=2, y=-3 and z=6}$ .

Explanation of Solution

Given:

  $\begin{array}{l}\text{5x+2y=4}\\ \text{3x+4y+2z=6}\\ \text{7x+3y+4z=29}\end{array}$

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,

  $\begin{array}{l}\text{5x+2y=4}\\ \text{3x+4y+2z=6}\\ \text{7x+3y+4z=29}\end{array}$

It can be written on the basis of linear equations in matrix form $\left[\begin{array}{ccc}5& 2& 0\\ \text{3}& 4& 2\\ 7& 3& 4\end{array}\right]\left[\begin{array}{c}\text{x}\\ \text{y}\\ \text{z}\end{array}\right]\text{=}\left[\begin{array}{c}4\\ 6\\ 29\end{array}\right]$ .

Let $\text{A=}\left[\begin{array}{ccc}5& 2& 0\\ \text{3}& 4& 2\\ 7& 3& 4\end{array}\right]$ , $\text{X=}\left[\begin{array}{c}\text{x}\\ \text{y}\\ \text{z}\end{array}\right]$ and $\text{B=}\left[\begin{array}{c}4\\ 6\\ 29\end{array}\right]$ . $\text{adjA=}{\left[\begin{array}{ccc}\text{16-6}& \text{-}\left(\text{12-14}\right)& \text{9-28}\\ \text{-}\left(\text{8-0}\right)& \text{20-0}& \text{-}\left(\text{15-14}\right)\\ \text{4-0}& \text{-}\left(\text{10-0}\right)& \text{20-6}\end{array}\right]}^{\text{T}}\text{=}{\left[\begin{array}{ccc}\text{10}& \text{2}& \text{-19}\\ \text{-8}& \text{20}& \text{ -1}\\ \text{4}& \text{-10}& \text{14}\end{array}\right]}^{\text{T}}=\left[\begin{array}{ccc}10& -8& 4\\ 2& 20& -10\\ -19& -1& 14\end{array}\right]$ .

Since, $\text{AX=B}$

  ${\text{X=A}}^{\text{-1}}\text{B}$

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

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

  $\left|\text{A}\right|\text{=}\left|5\left(10\right)-2\left(-2\right)+0\right|=54.$

  ${\text{A}}^{\text{-1}}\text{=}\frac{\left[\begin{array}{ccc}10& -8& 4\\ 2& 20& -10\\ -19& -1& 14\end{array}\right]}{54}=\left[\begin{array}{ccc}\frac{10}{54}& \frac{-8}{54}& \frac{4}{54}\\ \frac{2}{54}& \frac{20}{54}& \frac{-10}{54}\\ \frac{-19}{54}& \frac{-1}{54}& \frac{14}{54}\end{array}\right]$ .

  ${\text{A}}^{\text{-1}}=\left[\begin{array}{ccc}\frac{5}{27}& \frac{-4}{27}& \frac{2}{27}\\ \frac{\text{1}}{27}& \frac{10}{27}& \frac{-5}{27}\\ \frac{-19}{54}& \frac{-1}{54}& \frac{7}{27}\end{array}\right]$ .

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

  $\text{X=}\left[\begin{array}{c}\text{x}\\ \text{y}\\ \text{z}\end{array}\right]=\left[\begin{array}{ccc}\frac{5}{27}& \frac{-4}{27}& \frac{2}{27}\\ \frac{\text{1}}{27}& \frac{10}{27}& \frac{-5}{27}\\ \frac{-19}{54}& \frac{-1}{54}& \frac{7}{27}\end{array}\right]\left[\begin{array}{c}4\\ 6\\ 29\end{array}\right]$ .

  $\text{X=}\left[\begin{array}{c}\text{x}\\ \text{y}\\ \text{z}\end{array}\right]\text{=}\left[\begin{array}{c}\frac{\text{20-24+58}}{27}\\ \frac{4+60-145}{27}\\ \frac{-76-6+406}{54}\end{array}\right]\text{=}\left[\begin{array}{c}2\\ -3\\ 6\end{array}\right]$ .

  $\left[\begin{array}{c}\text{x}\\ \text{y}\\ \text{z}\end{array}\right]=\left[\begin{array}{c}2\\ -3\\ 6\end{array}\right]$ .

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

  $\text{x=2, y=-3 and z=6}$ .