Question

The augmented matrix of a system of linear equations is given in reduced row-echelon form. Find the solution of the system.

1. $\left[\begin{array}{llll}1& 0& 0& 2\\ 0& 1& 0& 1\\ 0& 0& 1& 3\end{array}\right]$$\begin{array}{l}x=\_\_\_\\ y=\_\_\_\\ z=\_\_\_\end{array}$
   
2. $\left[\begin{array}{llll}1& 0& 1& 2\\ 0& 1& 1& 1\\ 0& 0& 0& 0\end{array}\right]$$\begin{array}{l}x=\_\_\_\\ y=\_\_\_\\ z=\_\_\_\end{array}$
   
3. $\left[\begin{array}{llll}1& 0& 0& 2\\ 0& 1& 0& 1\\ 0& 0& 0& 3\end{array}\right]$$\begin{array}{l}x=\_\_\_\\ y=\_\_\_\\ z=\_\_\_\end{array}$
   

Short answer

1. Thus, the augmented matrix of system of equation is $x=2,y=1,z=3$
2. $x=2-t,y=1-t,z=t$
3. No solution

Step-by-step solution

a.Step 1. Given information

A matrix is given as-

$\left[\begin{array}{llll}1& 0& 0& 2\\ 0& 1& 0& 1\\ 0& 0& 1& 3\end{array}\right]$

Step 2. Concept used

The matrix that represents the system of linear equations is referred to as an augmented matrix of the system, it contains the same information as the system but in a simpler form as only the coefficients and constants that appear in the equations are used.

Using the given augmented matrix, we will find the result.

Step 3. Calculation

Let the given matrix be

$A=\left[\begin{array}{llll}1& 0& 0& 2\\ 0& 1& 0& 1\\ 0& 0& 1& 3\end{array}\right]$

From $\$A\$$the system of equations are,

$\begin{array}{l}x+0y+0z=2\\ 0x+y+0z=1\\ 0x+0y+z=3\end{array}$

So,

$\begin{array}{l}x=2\\ y=1\\ z=3\end{array}$

b.Step 1. Given information

A matrix is given as-

$\left[\begin{array}{llll}1& 0& 0& 2\\ 0& 1& 0& 1\\ 0& 0& 0& 0\end{array}\right]$

Step 2. Concept used

The matrix that represents the system of linear equations is referred to as an augmented matrix of the system, it contains the same information as the system but in a simpler form as only the coefficients and constants that appear in the equations are used.

Using the given augmented matrix, we will find the result

Step 3. Calculation

The given augmented matrix of a system of linear equation is

$\left[\begin{array}{llll}1& 0& 0& 2\\ 0& 1& 0& 1\\ 0& 0& 0& 0\end{array}\right]$

Here, the last row represents the equation $0=0$. This equation is always true, no matter what values are used for x, y and z.

So, the last matrix corresponds to the system

$\left\{\begin{array}{l}x=2\\ y=1\end{array}\right.$

Now we solve for the leading variables x and y in terms of the non leading variable z

To obtain the complete solution, we let z be any real number t, and we express x, y and z in terms of t.

$\begin{array}{l}x=2-t\\ y=1-t\\ z=t\end{array}$

c.Step 1. Given information

A matrix is given as-

$\left[\begin{array}{llll}1& 0& 0& 2\\ 0& 1& 0& 1\\ 0& 0& 0& 3\end{array}\right]$

Step 2. Concept used

The matrix that represents the system of linear equations is referred to as an augmented matrix of the system, it contains the same information as the system but in a simpler form as only the coefficients and constants that appear in the equations are used.

Using the given augmented matrix, we will find the result.

Step 3. Calculation

Here, the augmented matrix is in a row reduced echelon form so to solve this we use the back substitution method.

$System\left\{\begin{array}{l}x=2\\ y=1\\ 0=3\end{array}\right.$

Here, we saw that the third equation is False.