Chapter 3.4, Problem 9PPS

To determine

To solve the following system of linear equations

  $\{\begin{array}{l}4a+5b-6c=2\\ -3a-2b+7c=-15\\ -a+4b+2c=-13\end{array}$

Answer to Problem 9PPS

  $a=-3,b=-2,c=-4.$

Explanation of Solution

Given:

  $\{\begin{array}{l}4a+5b-6c=2\\ -3a-2b+7c=-15\\ -a+4b+2c=-13\end{array}$

Concept used:

System of linear equations can be solved using inverse matrix method.

To find inverse of a square matrix A, following steps should be followed.

1. Augment given matrix with the identity matrix
2. Perform row operations to make the identity matrix to the left and the right matrix becomes inverse matrix.

In this way, inverse matrix of a matrix can be obtained.

Formula used:

  $AX=B\iff X=A^{-1}B$

Calculation:

Consider given system of equations,

  $\{\begin{array}{l}4a+5b-6c=2\\ -3a-2b+7c=-15\\ -a+4b+2c=-13\end{array}$

Writing above system of linear equations in matrix form $(AX=B)$ ,

  $\left[\begin{array}{ccc}4& 5& -6\\ -3& -2& 7\\ -1& 4& 2\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}2\\ -15\\ -13\end{array}\right]$

Here,

  $\begin{array}{l}A=\left[\begin{array}{ccc}4& 5& -6\\ -3& -2& 7\\ -1& 4& 2\end{array}\right]\\ X=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]\\ B=\left[\begin{array}{c}2\\ -15\\ -13\end{array}\right]\end{array}$

Now, find the inverse matrix of matrix A.

Step 1: Augmenting matrix A with the identity matrix

  $[\begin{array}{ccc}4& 5& -6\\ -3& -2& 7\\ -1& 4& 2\end{array}\begin{array}{c}\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\end{array}]$

Step 2: Performing row operations to make the identity matrix to the left.

  $[\begin{array}{ccc}4& 5& -6\\ -3& -2& 7\\ -1& 4& 2\end{array}\begin{array}{c}\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\end{array}]$

  $\begin{array}{l}\text{Performing}{R}_1=\frac{R_1}{4}\\ \\ \left[\begin{array}{ccc}1& \frac{5}{4}& -\frac{3}{2}\\ -3& -2& 7\\ -1& 4& 2\end{array}|\begin{array}{ccc}\frac{1}{4}& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ \\ \text{Performing}{R}_2=R_2+3R_1\\ \\ \left[\begin{array}{ccc}1& \frac{5}{4}& -\frac{3}{2}\\ 0& \frac{7}{4}& \frac{5}{2}\\ -1& 4& 2\end{array}|\begin{array}{ccc}\frac{1}{4}& 0& 0\\ \frac{3}{4}& 1& 0\\ 0& 0& 1\end{array}\right]\\ \\ \text{Performing}{R}_3=R_3+R_1\\ \\ \left[\begin{array}{ccc}1& \frac{5}{4}& -\frac{3}{2}\\ 0& \frac{7}{4}& \frac{5}{2}\\ 0& \frac{21}{4}& \frac{1}{2}\end{array}|\begin{array}{ccc}\frac{1}{4}& 0& 0\\ \frac{3}{4}& 1& 0\\ \frac{1}{4}& 0& 1\end{array}\right]\\ \\ \end{array}$

  $\begin{array}{l}\text{Performing}{R}_1=R_1+\frac{23}{7}{R}_3\\ \\ \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& \frac{10}{7}\\ 0& 0& 1\end{array}|\begin{array}{ccc}\frac{32}{49}& \frac{34}{49}& -\frac{23}{49}\\ \frac{3}{7}& \frac{4}{7}& 0\\ \frac{2}{7}& \frac{3}{7}& -\frac{1}{7}\end{array}\right]\\ \\ \text{Performing}{R}_2=R_2-\frac{10}{7}{R}_3\\ \\ \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{ccc}\frac{32}{49}& \frac{34}{49}& -\frac{23}{49}\\ \frac{1}{49}& -\frac{2}{49}& \frac{10}{49}\\ \frac{2}{7}& \frac{3}{7}& -\frac{1}{7}\end{array}\right]\end{array}$

So, inverse matrix is,

  $A^{-1}=\left[\begin{array}{ccc}\frac{32}{49}& \frac{34}{49}& -\frac{23}{49}\\ \frac{1}{49}& -\frac{2}{49}& \frac{10}{49}\\ \frac{2}{7}& \frac{3}{7}& -\frac{1}{7}\end{array}\right]$

Now, applying the formula $AX=B\iff X=A^{-1}B$ ,$\left[\begin{array}{ccc}4& 5& -6\\ -3& -2& 7\\ -1& 4& 3\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}2\\ -15\\ -13\end{array}\right]\iff \left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{ccc}\frac{32}{49}& \frac{34}{49}& -\frac{23}{49}\\ \frac{1}{49}& -\frac{2}{49}& \frac{10}{49}\\ \frac{2}{7}& \frac{3}{7}& -\frac{1}{7}\end{array}\right]\left[\begin{array}{c}2\\ -15\\ -13\end{array}\right]$

Finding the solutions,

  $\begin{array}{l}\Rightarrow \left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{ccc}\frac{32}{49}& \frac{34}{49}& -\frac{23}{49}\\ \frac{1}{49}& -\frac{2}{49}& \frac{10}{49}\\ \frac{2}{7}& \frac{3}{7}& -\frac{1}{7}\end{array}\right]\left[\begin{array}{c}2\\ -15\\ -13\end{array}\right]\\ \Rightarrow \left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}\frac{32}{49}\times 2+\left(\frac{34}{49}\right)\times -15+\left(-\frac{23}{49}\right)\times -13\\ \frac{1}{49}\times 2+\left(-\frac{2}{49}\right)\times -15+\left(\frac{10}{49}\right)\times -13\\ \frac{2}{7}\times 2+\left(\frac{3}{7}\right)\times -15+\left(-\frac{1}{7}\right)\times -13\end{array}\right]\\ \Rightarrow \left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}-3\\ -2\\ -4\end{array}\right]\end{array}$

Conclusion:

Therefore, the solution to given system of equations is $a=-3,b=-2,c=-4.$