The account charged $35 for the first hour of work and $23 for each hour after that.

Solve the system of linear equations using the Gauss-Jordan elimination method. 2x + 3y 6212 3x + (x, y. z)

Gekata [30.6K]

Answer:

The solution of the system of linear equations is $x=3, y=4, z=1$

Step-by-step explanation:

We have the system of linear equations:

$2x+3y-6z=12\\x-2y+3z=-2\\3x+y=13$

Gauss-Jordan elimination method is the process of performing row operations to transform any matrix into reduced row-echelon form.

The first step is to transform the system of linear equations into the matrix form. A system of linear equations can be represented in matrix form (Ax=b) using a coefficient matrix (A), a variable matrix (x), and a constant matrix(b).

From the system of linear equations that we have, the coefficient matrix is

$\left[\begin{array}{ccc}2&3&-6\\1&-2&3\\3&1&0\end{array}\right]$

the variable matrix is

$\left[\begin{array}{c}x&y&z\end{array}\right]$

and the constant matrix is

$\left[\begin{array}{c}12&-2&13\end{array}\right]$

We also need the augmented matrix, this matrix is the result of joining the columns of the coefficient matrix and the constant matrix divided by a vertical bar, so

$\left[\begin{array}{ccc|c}2&3&-6&12\\1&-2&3&-2\\3&1&0&13\end{array}\right]$

To transform the augmented matrix to reduced row-echelon form we need to follow these row operations:

multiply the 1st row by 1/2

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\1&-2&3&-2\\3&1&0&13\end{array}\right]$

add -1 times the 1st row to the 2nd row

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&-7/2&6&-8\\3&1&0&13\end{array}\right]$

add -3 times the 1st row to the 3rd row

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&-7/2&6&-8\\0&-7/2&9&-5\end{array}\right]$

multiply the 2nd row by -2/7

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&-7/2&9&-5\end{array}\right]$

add 7/2 times the 2nd row to the 3rd row

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&0&3&3\end{array}\right]$

multiply the 3rd row by 1/3

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&0&1&1\end{array}\right]$

add 12/7 times the 3rd row to the 2nd row

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&0&4\\0&0&1&1\end{array}\right]$

add 3 times the 3rd row to the 1st row

$\left[\begin{array}{ccc|c}1&3/2&0&9\\0&1&0&4\\0&0&1&1\end{array}\right]$

add -3/2 times the 2nd row to the 1st row

$\left[\begin{array}{ccc|c}1&0&0&3\\0&1&0&4\\0&0&1&1\end{array}\right]$

From the reduced row echelon form we have that

$x=3\\y=4\\z=1$

Since every column in the coefficient part of the matrix has a leading entry that means our system has a unique solution.

7 0

3 years ago

The sphere is centered in the cube. Find an expression for the volume of the cube outside the sphere

Nikolay [14]

Solution -

The sphere is centered in the cube, so diameter of the sphere is equal to the side of the cube. Hence ,the radius of the sphere is half of the length of the side.

If the length of side of the cibe = a , then radius of the sphere is $\frac{a}{2}$

Volume of the cube = a³, volume of the sphere = $\frac{4}{3}$ π(radius)³ = $\frac{4}{3}$ π( $\frac{a}{2}$ )³

volume of the cube outside the sphere = volume of the cube - volume of the sphere

= $a^3 - \frac{4}{3}$ π( $\frac{a}{2}$ )³

= $a^3 -\frac{1}{6} \pi a^{3}$

=a³(1 - $\frac{\pi }{6}$ )

7 0

3 years ago

How do you find the midpoint of a line segment

egoroff_w [7]

Answer:

Add the two x-coordinates and divide by 2 and the same thing for y-coordinates....takes two coordinates must be the ones that are the end points of the line

3 0

3 years ago

Read 2 more answers

Pls solve this...need help thanks :)

WINSTONCH [101]

Step-by-step explanation:

2² * 3 * 5 = 4 * 3 * 5 = 60.

3² * 5² = 9 * 25 = 225.

7 0

3 years ago

Read 2 more answers

Tom rented a truck for one day. There was a base fee of $16.99 , and there was an additional charge of 90 cents for each mile dr