MATHEMATICS HIGH SCHOOL

Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.

Answers

Answer 1

Answer:

Answer:

Option (a) is correct.

The solution is (1, -1 , -4)

Step-by-step explanation:

Given:

A system of equation having 3 equations,

$2x+y+z=-3\n\n 3x-5y+3z=-4\n\n 5x-y+2z=-2$

We have to solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.

Consider the given system

$2x+y+z=-3\n\n 3x-5y+3z=-4\n\n 5x-y+2z=-2$

Write in matrix form as

$\begin{pmatrix}2&1&1\n \:3&-5&3\n \:5&-1&2\end{pmatrix}\begin{pmatrix}x\n \:y\n \:z\end{pmatrix}=\begin{pmatrix}-3\n \:-4\n \:-2\end{pmatrix}$

⇒ AX = b

Writing in Augmented matrix form , [A | b]

$\begin{pmatrix}2&1&1&-3\n 3&-5&3&-4\n 5&-1&2&-2\end{pmatrix}$

Apply row operations to make A an identity matrix.

$R_1\:\leftrightarrow \:R_3$

$=\begin{pmatrix}5&-1&2&-2\n 3&-5&3&-4\n 2&1&1&-3\end{pmatrix}$

$R_2\:\leftarrow \:R_2-(3)/(5)\cdot \:R_1$

$=\begin{pmatrix}5&-1&2&-2\n 0&-(22)/(5)&(9)/(5)&-(14)/(5)\n 2&1&1&-3\end{pmatrix}$

$R_3\:\leftarrow \:R_3-(2)/(5)\cdot \:R_1$

$=\begin{pmatrix}5&-1&2&-2\n 0&-(22)/(5)&(9)/(5)&-(14)/(5)\n 0&(7)/(5)&(1)/(5)&-(11)/(5)\end{pmatrix}$

$R_3\:\leftarrow \:R_3+(7)/(22)\cdot \:R_2$

$=\begin{pmatrix}5&-1&2&-2\n 0&-(22)/(5)&(9)/(5)&-(14)/(5)\n 0&0&(17)/(22)&-(34)/(11)\end{pmatrix}$

$R_3\:\leftarrow (22)/(17)\cdot \:R_3$

$=\begin{pmatrix}5&-1&2&-2\n 0&-(22)/(5)&(9)/(5)&-(14)/(5)\n 0&0&1&-4\end{pmatrix}$

$R_2\:\leftarrow \:R_2-(9)/(5)\cdot \:R_3$

$=\begin{pmatrix}5&-1&2&-2\n 0&-(22)/(5)&0&(22)/(5)\n 0&0&1&-4\end{pmatrix}$

$R_1\:\leftarrow \:R_1-2\cdot \:R_3$

$=\begin{pmatrix}5&-1&0&6\n 0&-(22)/(5)&0&(22)/(5)\n 0&0&1&-4\end{pmatrix}$

$R_2\:\leftarrow \:-(5)/(22)\cdot \:R_2$

$=\begin{pmatrix}5&-1&0&6\n 0&1&0&-1\n 0&0&1&-4\end{pmatrix}$

$R_1\:\leftarrow \:R_1+1\cdot \:R_2$

$=\begin{pmatrix}5&0&0&5\n 0&1&0&-1\n 0&0&1&-4\end{pmatrix}$

$R_1\:\leftarrow (1)/(5)\cdot \:R_1$

$=\begin{pmatrix}1&0&0&1\n 0&1&0&-1\n 0&0&1&-4\end{pmatrix}$

Thus, We obtained an identity matrix

Thus, The solution is (1, -1 , -4)

Answer 2

Answer:

This involves quite a lot of arithmetic to do manually.

The first thing you do is to make the first number in row 2 = to 0.

This is done by R2 = -3/2 R1 + R2

so the matrix becomes

( 2 1 1) ( -3 )

( 0 -13/2 3/2) (1/2 )

(5 -1 2) (-2)

Next step is to make the 5 in row 5 = 0

then the -1 must become zero

You aim for the form

( 1 0 0) (x)

(0 1 0) (y)

(0 0 1) ( z)

x , y and z will be the required solutions.

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Answers

Answer:

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Answers

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Answers

Answer:

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Answers

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Answers

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