Chapter 7.4, Problem 52E

To determine

The linear equations associated with given augmented matrix and hence find the values of the variables.

Answer to Problem 52E

The required system of equations are $x+5y=0$ ; $y=6$ and the values of the variables are $x=-30$ and $y=6$ .

Explanation of Solution

Given: The given augmented matrix is $\left[\begin{array}{cc}1& 5\\ 0& 1\end{array}|\begin{array}{c}0\\ 6\end{array}\right]$ .

Concept Used: An augmented matrix is a matrix representation of a standard system of simultaneous linear equations. It is a matrix which consists of two parts with a line separating both parts. On the left hand side of the line is the coefficient matrix (generally denoted byA ) of the system of equations and on the right hand side of the line are the constant terms (generally denoted by B ), written in order. Multiplying the coefficient matrix part to the variable column matrix (generally denoted by X) gives the algebraic expressions which are then equated to the constant terms to get the system of equations. Thus, the system of equations can be found using the matrix equation-

  $AX=B$

Calculations: The given augmented matrix is $\left[\begin{array}{cc}1& 5\\ 0& 1\end{array}|\begin{array}{c}0\\ 6\end{array}\right]$ .

Consider the variables of the system of equations to be x and y.

The coefficient matrix and the constant matrix are-

  $A=\left[\begin{array}{cc}1& 5\\ 0& 1\end{array}\right]$ and $B=\left[\begin{array}{c}0\\ 6\end{array}\right]$ respectively.

The variable column matrix is-

  $X=\left[\begin{array}{c}x\\ y\end{array}\right]$

Multiplying the coefficient matrix with the variable column matrix and equating it to the constant matrix gives-

  $AX=B$

  $\Rightarrow \left[\begin{array}{cc}1& 5\\ 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}0\\ 6\end{array}\right]$

  $\Rightarrow \left[\begin{array}{c}x+5y\\ y\end{array}\right]=\left[\begin{array}{c}0\\ 6\end{array}\right]$

  $\Rightarrow x+5y=0$ . . . . . . . ( i )

  $y=6$ . . . . . . . . . ( ii )

Using back substitution method, using $y=6$ in equation ( i ) gives-

  $x+5(6)=0$

  $\Rightarrow x+30=0$

  $\Rightarrow x=-30$

Thus, the values of the variables are $x=-30$ and $y=6$ .