Chapter 10.3, Problem 3E

(a)

To determine

To fill: The property of an augmented matrix of a system of linear equation.

Answer to Problem 3E

The leading variables in the augmented matrix of a system of linear equation with variables x, y and z are x and y.

Explanation of Solution

Given:

Augmented matrix

$\left(\begin{array}{cccc}1& 0& -1& 3\\ 0& 1& 2& 5\\ 0& 0& 0& 0\end{array}\right)$

The augmented matrix of a system is written by the use of only coefficient and constants that appears in the equation.

For example,

$\{\begin{array}{l}3x-2y+z=3\\ x+4y+2z=0\\ -x+0y+4z=11\end{array}$

Write the matrix in augmented form

$\left[\begin{array}{cccc}3& -2& 1& 3\\ 1& 4& 2& 0\\ -1& 0& 4& 11\end{array}\right]$

The augmented form of this matrix is

$\begin{array}{l}x-z=3\\ y+2z=5\\ z=0\end{array}$

The variable z can assigned any value. It is not a leading variable in the system.

The only leading variables in the system are x and y.

Thus, the leading variables in the system is x and y.

(b)

To determine

To check: Whether the system is inconsistent or dependent.

Answer to Problem 3E

The given system of augmented matrix is dependent in nature.

Explanation of Solution

Given:

Augmented matrix

$\left(\begin{array}{cccc}1& 0& -1& 3\\ 0& 1& 2& 5\\ 0& 0& 0& 0\end{array}\right)$

The augmented matrix of a system is written by the use of only coefficient and constants that appears in the equation.

For example,

$\{\begin{array}{l}3x-2y+z=3\\ x+4y+2z=0\\ -x+0y+4z=11\end{array}$

Write the matrix in augmented form

$\left[\begin{array}{cccc}3& -2& 1& 3\\ 1& 4& 2& 0\\ -1& 0& 4& 11\end{array}\right]$

A system which has infinitely many solutions, and if it is not inconsistent, the variables in row-echelon form are not leading in nature is called as dependent variable.

If the row-echelon form contains a row that represent $0=c$, where c is not zero, then the system has no solution. Such type of system is called as inconsistent system.

Thus, the given system of augmented matrix with variables x,y and z is dependent in nature.

(c)

To determine

The solution for the system.

Answer to Problem 3E

The solution for the given system is $x=3+k$, $y=5-2k$ and $z=k$

Explanation of Solution

Given:

Augmented matrix

$\left(\begin{array}{cccc}1& 0& -1& 3\\ 0& 1& 2& 5\\ 0& 0& 0& 0\end{array}\right)$

For (I) row

$x-z=3$ (I)

For (II) row

$y+2z=5$ (II)

As z is arbitrary it can take any value.

Take $z=k$ (III)

Substitute $z=k$ from (III) in (I) and (II),

$x=3+k$

$y=5-2k$

$z=k$

Thus, the equation for the above augmented matrix is $x=3+k$ $y=5-2k$ and $z=k$.