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${:[x_(1)+2x_(2)+3x_(3)],[x_(1)+(2)(2)+3(3)],[x_(1)+5=6]:} An m × n matnix A is quadx_(1)+5=6 In recuad row ecchelon form if it satisfies quadx_(1)=6 the follouinn annorties: {:[x_(1)+x_(2)^(1)+2x_(3)=-1],[x_(1)-2x_(2)+x_(3)=-5],[3x_(1)+x_(2)+x_(3)=3]:} a) Find all solutions by unang the Gausion elimination? Gause-Jordan Reduction$

$\begin{array}{r} x_{1}+2 x_{2}+3 x_{3} \\ x_{1}+(2)(2)+3(3) \\ x_{1}+5=6 \end{array}$ $\newline$ An $m \times n$ matnix $A$ is $\quad x_{1}+5=6$ $\newline$ In recuad row ecchelon form if it satisfies $\quad x_{1}=6$ the follouinn annorties: $\newline$ $\begin{array}{l} x_{1}+x_{2}^{1}+2 x_{3}=-1 \\ x_{1}-2 x_{2}+x_{3}=-5 \\ 3 x_{1}+x_{2}+x_{3}=3 \end{array}$ $\newline$ a) Find all solutions by unang the Gausion elimination? Gause-Jordan Reduction

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Math Problems

Grade 8

Write an equation word problem

Full solution

\begin{array}{r} x_{1}+2 x_{2}+3 x_{3} \\ x_{1}+(2)(2)+3(3) \\ x_{1}+5=6 \end{array}

\newline

m \times n

matnix

A

\quad x_{1}+5=6

\newline

In recuad row ecchelon form if it satisfies

\quad x_{1}=6

the follouinn annorties:

\newline

\begin{array}{l} x_{1}+x_{2}^{1}+2 x_{3}=-1 \\ x_{1}-2 x_{2}+x_{3}=-5 \\ 3 x_{1}+x_{2}+x_{3}=3 \end{array}

\newline

a) Find all solutions by unang the Gausion elimination? Gause-Jordan Reduction

Identify Equations: Identify the system of equations to solve: $\newline$ $\begin{align*} x_1 + 2x_2 + 3x_3 &= -1 \\ x_1 - 2x_2 + x_3 &= -5 \\ 3x_1 + x_2 + x_3 &= 3 \end{align*}$
Write Augmented Matrix: Write the augmented matrix for the system: $\newline$ $\begin{bmatrix} 1 & 2 & 3 & |-1 \\ 1 & -2 & 1 & |-5 \\ 3 & 1 & 1 & |3 \end{bmatrix}$
Perform Row Operations: Perform row operations to get the matrix in echelon form. Start by eliminating $x_1$ from the second and third rows: $\newline$ Subtract row $1$ from row $2$ : $\newline$ $\begin{bmatrix} 1 & 2 & 3 & |-1 \\ 0 & -4 & -2 & |-4 \\ 3 & 1 & 1 & |3 \end{bmatrix}$ $\newline$ Subtract $3$ times row $1$ from row $3$ : $\newline$ $\begin{bmatrix} 1 & 2 & 3 & |-1 \\ 0 & -4 & -2 & |-4 \\ 0 & -5 & -8 & |6 \end{bmatrix}$
Simplify Matrix: Simplify row $2$ by dividing by $-4$ : $\newline$ $\begin{bmatrix} 1 & 2 & 3 & |-1 \\ 0 & 1 & 0.5 & |1 \\ 0 & -5 & -8 & |6 \end{bmatrix}$ $\newline$ Add $5$ times row $2$ to row $3$ : $\newline$ $\begin{bmatrix} 1 & 2 & 3 & |-1 \\ 0 & 1 & 0.5 & |1 \\ 0 & 0 & -5.5 & |11 \end{bmatrix}$
Back Substitution: Simplify row $3$ by dividing by $-5$ . $5$ : $\newline$ $\begin{bmatrix} 1 & 2 & 3 & |-1 \\ 0 & 1 & 0.5 & |1 \\ 0 & 0 & 1 & |-2 \end{bmatrix}$ $\newline$ Back substitute to find $x_2$ and $x_1$ : $\newline$ Subtract $0$ . $5$ times row $3$ from row $2$ : $\newline$ $\begin{bmatrix} 1 & 2 & 3 & |-1 \\ 0 & 1 & 0 & |2 \\ 0 & 0 & 1 & |-2 \end{bmatrix}$ $\newline$ Subtract $3$ times row $3$ from row $1$ : $\newline$ $\begin{bmatrix} 1 & 2 & 0 & |5 \\ 0 & 1 & 0 & |2 \\ 0 & 0 & 1 & |-2 \end{bmatrix}$ $\newline$ Subtract $2$ times row $2$ from row $1$ : $\newline$ $\begin{bmatrix} 1 & 0 & 0 & |1 \\ 0 & 1 & 0 & |2 \\ 0 & 0 & 1 & |-2 \end{bmatrix}$