Chapter 3.4, Problem 5CYU

To determine

To solve the following system of linear equations

  $\{\begin{array}{l}3x+5y-z=12\\ -2x-3y+5z=14\\ 4x+7y+3z=38\end{array}$

Answer to Problem 5CYU

No Solution

Explanation of Solution

Given:

  $\{\begin{array}{l}3x+5y-z=12\\ -2x-3y+5z=14\\ 4x+7y+3z=38\end{array}$

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,

  $\{\begin{array}{l}3x+5y-z=12\\ -2x-3y+5z=14\\ 4x+7y+3z=38\end{array}$

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

  $\left[\begin{array}{ccc}3& 5& -1\\ -2& -3& 5\\ 4& 7& 3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}12\\ 14\\ 28\end{array}\right]$

Here,

  $\begin{array}{l}A=\left[\begin{array}{ccc}3& 5& -1\\ -2& -3& 5\\ 4& 7& 3\end{array}\right]\\ X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\\ B=\left[\begin{array}{c}12\\ 14\\ 28\end{array}\right]\end{array}$

Now, find the inverse matrix of matrix A.

Step 1: Augmenting matrix A with the identity matrix

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

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

  $\begin{array}{l}\text{Performing}{R}_1=R_1-5R_2\\ \\ [\begin{array}{ccc}1& 0& -22\\ 0& \frac{1}{3}& \frac{13}{3}\\ 0& \frac{1}{3}& \frac{13}{3}\end{array}\begin{array}{c}\begin{array}{ccc}-3& -5& 0\\ \frac{2}{3}& 1& 0\\ -\frac{4}{3}& 0& 1\end{array}\end{array}]\\ \\ \text{Performing}{R}_3=R_3-R_2\\ \\ [\begin{array}{ccc}1& 0& -22\\ 0& \frac{1}{3}& \frac{13}{3}\\ 0& 0& 0\end{array}\begin{array}{c}\begin{array}{ccc}-3& -5& 0\\ \frac{2}{3}& 1& 0\\ -2& -1& 1\end{array}\end{array}]\\ \end{array}$

Here, row 3 contains only zeros; hence the determinant of the matrix is zero.

So, matrix is not invertible.

Since the determinant of matrix is zero there is no possible inverse matrix for the given matrix.

Hence, there is no possible solution for the given system of equations.

Conclusion:

Therefore, there is no solution to given system of equations.