Math / Algebra

QuestionCompleting Gauss-Jordan elimination with a $3 \times 3$ matrix $\begin{aligned} 5 x+y & =-7 \\ -2 x-y+2 z & =11 \\ -4 y+5 z & =13 \end{aligned}$
Solve the system by completing the steps below to produce a reduced row-echelon form. $R_{1}, R_{2}$ , and $R_{3}$ denote the first, second, and third rows, respectively. The arrow notation $(\rightarrow)$ stands for "replaces," where the expression on the left of the arrow replaces the expression on
Here is the augmented matrix: $\left[\begin{array}{ccc:c} 5 & 1 & 0 & -7 \\ -2 & -1 & 2 & 11 \\ 0 & -4 & 5 & 13 \end{array}\right]$
Enter the missing coefficients for the row operations. (1) $\square$ $\cdot R_{1} \rightarrow R_{1}:$ $\left[\begin{array}{ccc:c} 1 & \frac{1}{5} & 0 & -\frac{7}{5} \\ -2 & -1 & 2 & 11 \\ 0 & -4 & 5 & 13 \end{array}\right]$

Studdy Solution

STEP 1

What is this asking? We're asked to perform a series of row operations on a matrix to get it into reduced row-echelon form, which will help us solve a system of equations.
Specifically, we need to figure out what to multiply the first row by to get a leading 1. Watch out! Don't forget that whatever we do to one row, we have to do to every element in that row!
Also, remember that our goal is to get a 1 in the top left corner and 0s below it.

STEP 2

1. Find the multiplier for $R_1$ .
2. Perform the row operation.

STEP 3

Our current $R_1$ is $\begin{bmatrix} 5 & 1 & 0 & -7 \end{bmatrix}$ .
We want the first element to be 1.

STEP 4

To turn the 5 into a 1, we need to multiply it by $\frac{1}{5}$ .
Remember, dividing by $5$ is the same as multiplying by $\frac{1}{5}$ !

STEP 5

$5 \cdot \frac{1}{5} = 1$ $1 \cdot \frac{1}{5} = \frac{1}{5}$ $0 \cdot \frac{1}{5} = 0$ $-7 \cdot \frac{1}{5} = -\frac{7}{5}$

STEP 6

So, our new $R_1$ is $\begin{bmatrix} 1 & \frac{1}{5} & 0 & -\frac{7}{5} \end{bmatrix}$ .

STEP 7

The matrix now looks like this: $\begin{bmatrix} 1 & \frac{1}{5} & 0 & -\frac{7}{5} \\ -2 & -1 & 2 & 11 \\ 0 & -4 & 5 & 13 \end{bmatrix}$

STEP 8

The missing coefficient for the row operation is $\frac{1}{5}$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the multiplier for $R_1$ .
2. Perform the row operation.

STEP 3

Our current $R_1$ is $\begin{bmatrix} 5 & 1 & 0 & -7 \end{bmatrix}$ .
We want the first element to be 1.

STEP 4

To turn the 5 into a 1, we need to multiply it by $\frac{1}{5}$ .
Remember, dividing by $5$ is the same as multiplying by $\frac{1}{5}$ !

STEP 5

$5 \cdot \frac{1}{5} = 1$ $1 \cdot \frac{1}{5} = \frac{1}{5}$ $0 \cdot \frac{1}{5} = 0$ $-7 \cdot \frac{1}{5} = -\frac{7}{5}$

STEP 6

So, our new $R_1$ is $\begin{bmatrix} 1 & \frac{1}{5} & 0 & -\frac{7}{5} \end{bmatrix}$ .

STEP 7

The matrix now looks like this: $\begin{bmatrix} 1 & \frac{1}{5} & 0 & -\frac{7}{5} \\ -2 & -1 & 2 & 11 \\ 0 & -4 & 5 & 13 \end{bmatrix}$

STEP 8

The missing coefficient for the row operation is $\frac{1}{5}$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the multiplier for R1R_1R1​.2. Perform the row operation.

STEP 3

Our current R1R_1R1​ is [510−7]\begin{bmatrix} 5 & 1 & 0 & -7 \end{bmatrix}[5​1​0​−7​].We want the first element to be **1**.

STEP 4

To turn the **5** into a **1**, we need to multiply it by 15\frac{1}{5}51​.Remember, dividing by 555 is the same as multiplying by 15\frac{1}{5}51​!

STEP 5

5⋅15=1 5 \cdot \frac{1}{5} = 1 5⋅51​=1 1⋅15=15 1 \cdot \frac{1}{5} = \frac{1}{5} 1⋅51​=51​ 0⋅15=0 0 \cdot \frac{1}{5} = 0 0⋅51​=0−7⋅15=−75 -7 \cdot \frac{1}{5} = -\frac{7}{5} −7⋅51​=−57​

STEP 6

So, our new R1R_1R1​ is [1150−75]\begin{bmatrix} 1 & \frac{1}{5} & 0 & -\frac{7}{5} \end{bmatrix}[1​51​​0​−57​​].

STEP 7

The matrix now looks like this: [1150−75−2−12110−4513]\begin{bmatrix} 1 & \frac{1}{5} & 0 & -\frac{7}{5} \\ -2 & -1 & 2 & 11 \\ 0 & -4 & 5 & 13 \end{bmatrix}⎣⎡​1−20​51​−1−4​025​−57​1113​⎦⎤​

STEP 8

The missing coefficient for the row operation is 15\frac{1}{5}51​.

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1. Find the multiplier for $R_1$ .
2. Perform the row operation.

Our current $R_1$ is $\begin{bmatrix} 5 & 1 & 0 & -7 \end{bmatrix}$ .
We want the first element to be 1.

To turn the 5 into a 1, we need to multiply it by $\frac{1}{5}$ .
Remember, dividing by $5$ is the same as multiplying by $\frac{1}{5}$ !

$5 \cdot \frac{1}{5} = 1$ $1 \cdot \frac{1}{5} = \frac{1}{5}$ $0 \cdot \frac{1}{5} = 0$ $-7 \cdot \frac{1}{5} = -\frac{7}{5}$

So, our new $R_1$ is $\begin{bmatrix} 1 & \frac{1}{5} & 0 & -\frac{7}{5} \end{bmatrix}$ .

The matrix now looks like this: $\begin{bmatrix} 1 & \frac{1}{5} & 0 & -\frac{7}{5} \\ -2 & -1 & 2 & 11 \\ 0 & -4 & 5 & 13 \end{bmatrix}$

The missing coefficient for the row operation is $\frac{1}{5}$ .