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Please show all your work. I need step by step. How did you solve? Please help me both part or both question

5. Use the Gauss-Jordan method to solve the given system of equations. 2 +9 + 2 + 300 = 1, 2 + 2 + 4 + 5 = = 3, (a) -ہے .5- =

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Answer #1

a) The gauss jordan method is as follows

Step 1

The augemented matrix is

\begin{bmatrix} 1 & 1 & 1 & 3 & 1 \\ 1&2 &4 & 5 & 3 \\ 1& 0 & -2 &1 & -5 \end{bmatrix}

Step 2

Perform row elementary operarion on the augemented matrix

R_{2}-R_{1}\rightarrow R_{2} and R_{3}-R_{1}\rightarrow R_{3}

\begin{bmatrix} 1 & 1 & 1 & 3 & 1 \\ 0 &1 &3 & 2 & 2 \\ 0& -1 & -3 &-2 & -6 \end{bmatrix}

Step 3:

Perform row elementary operarion on the augemented matrix

R_{3}+R_{2}\rightarrow R_{3}

The matrix in reduced row echelon form is

\begin{bmatrix} 1 & 1 & 1 & 3 & 1 \\ 0 &1 &3 & 2 & 2 \\ 0& 0 & 0 &0 & -4 \end{bmatrix}

Step 4:

The linear system corresponding to the matrix in reduced row echelon form is

x+y+z+3w=1

y+3z+2w=2

0=-4

Since 0 ≠ -4, there is no solution for given system of equation.

b) The gauss jordan method is as follows

Step 1

The augemented matrix is

\begin{bmatrix} 1 & 2 & -1 & 1\\ 0 &3 & 1 &-3 \\ 3& 3 & -4 & 6 \end{bmatrix}

Step 2

Perform row elementary operarion on the augemented matrix

R_{3}-3R_{1}\rightarrow R_{3}

\begin{bmatrix} 1 & 2 & -1 & 1\\ 0 &3 & 1 &-3 \\ 0& -3 & -1 & 3 \end{bmatrix}

\frac{R_{2}}{2}\rightarrow R_{2}

\begin{bmatrix} 1 & 2 & -1 & 1\\ 0 &1 & \frac{1}{3} &-1 \\ 0& -3 & -1 & 3 \end{bmatrix}

Step 3:

Perform row elementary operarion on the augemented matrix

R_{3}+3R_{2}\rightarrow R_{3}

The matrix in reduced row echelon form is

\begin{bmatrix} 1 & 2 & -1 & 1\\ 0 &1 & \frac{1}{3} &-1 \\ 0& 0 & 0 & 0 \end{bmatrix}

Step 4: The linear system corresponding to the matrix in reduced row echelon form is

x+2y-z=1

y+\frac{z}{3}=-1

The solutions are

x=3+\frac{5}{3}t

y=-1-\frac{1}{3}t

z=t, t\epsilon R

there is Infinite number of solutions for given system of equation.

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