Chapter 16.3, Problem 6MRE

To determine

To find: the real solution of the system and value of the variable $a,b\text{ and }c$ .

Answer to Problem 6MRE

  $a=-1,b=1andc=-2$ .

Explanation of Solution

Given:

  $\begin{array}{l}a+b-c=-2\\ 2a-b+c=5\\ a+3b+2c=2\end{array}$

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,

  $\begin{array}{l}a+b-c=-2\\ 2a-b+c=5\\ a+3b+2c=2\end{array}$

It can be written on the basis of linear equations in matrix form $\left[\begin{array}{ccc}1& 1& -1\\ 2& -1& 1\\ 1& 3& 2\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}-2\\ 5\\ 2\end{array}\right]$ .

Let $A=\left[\begin{array}{ccc}1& 1& -1\\ 2& -1& 1\\ 1& 3& 2\end{array}\right]$ , $\text{X=}\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$ and $\text{B=}\left[\begin{array}{c}-2\\ 5\\ 2\end{array}\right]$ . $\text{adjA=}{\left[\begin{array}{ccc}-5& -3& 7\\ -5& 3& -2\\ 0& -3& -3\end{array}\right]}^T=\left[\begin{array}{ccc}-5& -5& 0\\ -3& 3& -3\\ 7& -2& -3\end{array}\right].$

Since, $\text{AX=B}$

  ${\text{X=A}}^{\text{-1}}\text{B}$

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

  $\begin{array}{l}{A}^{-1}=\frac{adjA}{\left|A\right|}\\ \left|A\right|=\left|1\left(-2-3\right)-1\left(4-1\right)-1\left(6+1\right)\right|=\left|-5-3-7\right|=\left|-15\right|=15.\\ A^{-1}=\frac{\left[\begin{array}{ccc}-5& -5& 0\\ -3& 3& -3\\ 7& -2& -3\end{array}\right]}{15}=\left[\begin{array}{ccc}\frac{-1}{3}& \frac{-1}{3}& 0\\ \frac{-1}{5}& \frac{1}{5}& \frac{-1}{5}\\ \frac{7}{15}& \frac{-2}{15}& \frac{-1}{5}\end{array}\right].\\ A^{-1}=\left[\begin{array}{ccc}\frac{-1}{3}& \frac{-1}{3}& 0\\ \frac{-1}{5}& \frac{1}{5}& \frac{-1}{5}\\ \frac{7}{15}& \frac{-2}{15}& \frac{-1}{5}\end{array}\right].\end{array}$

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

  $\begin{array}{l}X=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{ccc}\frac{-1}{3}& \frac{-1}{3}& 0\\ \frac{-1}{5}& \frac{1}{5}& \frac{-1}{5}\\ \frac{7}{15}& \frac{-2}{15}& \frac{-1}{5}\end{array}\right]\left[\begin{array}{c}-2\\ 5\\ 2\end{array}\right].\\ X=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}\frac{2}{3}-\frac{5}{3}+0\\ \frac{2}{5}+1-\frac{2}{5}\\ \frac{-14}{15}-\frac{10}{15}-\frac{2}{5}\end{array}\right]\\ \left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}-1\\ 1\\ -2\end{array}\right].\end{array}$

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

  $a=-1,b=1andc=-2$ .