Chapter 10.3, Problem 4E

(a)

To determine

The solution of the augmented matrix.

Answer to Problem 4E

The solution for the augmented matrix for a system of linear equation is

$x=2$

$y=1$

$z=3$

Explanation of Solution

Given:

Augmented matrix

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

Calculation:

A vector in a plane is a line segment with an assigned direction.

The starting point of a vector is known as initial point of the vector and the end point of the vector is known as terminal point of the vector.

A vector u can be represented as $u=\overrightarrow{AB}$.

For a matrix to be in row-echelon form,

The first non zero number in each row is 1. This entry is called as leading entry.

The leading entry in each row is to the right of the leading entry in the row immediately above it.

All rows consist of entirely of zeros are at the bottom of the matrix.

Every number above and below each leading entry is 0.

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

The matrix is in row echelon form. Back substitution method is applied to get the solution for the variables x, y and z.

The first row gives the solution for x

$x=2$

The second row gives the solution for y

$y=1$

The third row gives the solution for z

$z=3$

Thus, the solution for the given augmented matrix is $x=2$, $y=1$ and $z=3$.

(b)

To determine

The solution of the augmented matrix.

Answer to Problem 4E

The solution for the augmented matrix for a system of linear equation is

$x=2-k$

$y=1-k$

$z=k$

Explanation of Solution

Given:

Augmented matrix

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

Calculation:

A vector in a plane is a line segment with an assigned direction.

The starting point of a vector is known as initial point of the vector and the end point of the vector is known as terminal point of the vector.

A vector u can be represented as $u=\overrightarrow{AB}$.

For a matrix to be in row-echelon form,

The first non zero number in each row is 1. This entry is called as leading entry.

The leading entry in each row is to the right of the leading entry in the row immediately above it.

All rows consist of entirely of zeros are at the bottom of the matrix.

Every number above and below each leading entry is 0.

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

The matrix is in row echelon form. Back substitution method is applied to get the solution for the variables x, y and z.

The third row contains all zeros. So, the variable z can assume any value.

$z=k$ (1)

The first row gives the solution for x

$\begin{array}{c}x+z=2\\ x=2-z\end{array}$ (2)

Substitute $k$ for $z$ from (1) in (2)

$x=2-k$

The second row gives the solution for y

$\begin{array}{c}y+z=1\\ y=1-z\end{array}$ (3)

$y=1-k$

Thus, the solution for the given augmented matrix is $x=2-k$ $y=1-k$ and $z=k$.

(c)

To determine

The solution of the augmented matrix.

Answer to Problem 4E

The augmented matrix has no solution

Explanation of Solution

Given:

Augmented matrix

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

Calculation:

A vector in a plane is a line segment with an assigned direction.

The starting point of a vector is known as initial point of the vector and the end point of the vector is known as terminal point of the vector.

A vector u can be represented as $u=\overrightarrow{AB}$.

For a matrix to be in row-echelon form,

The first non zero number in each row is 1. This entry is called as leading entry.

The leading entry in each row is to the right of the leading entry in the row immediately above it.

All rows consist of entirely of zeros are at the bottom of the matrix.

Every number above and below each leading entry is 0.

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

For the first row $x=2$

For the second row $y=1$.

For the third row there is no particular value of $z$ assigned.

The third row states that $0=3$.

It is sense less. There are no particular values for $z$

Thus, the system of given augmented matrix has no solution.