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Use matrices to solve the system of equations if possible. Use Gaussian elimination with back substitution or gauss Jordan elimi

CaHeK987 [17]

In matrix form, the system is given by

$\begin{bmatrix} -1 & 1 & -1 \\ 2 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -20 \\ 29 \\ 29 \end{bmatrix}$

I'll use G-J elimination. Consider the augmented matrix

$\left[ \begin{array}{ccc|c} -1 & 1 & -1 & -20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]$

• Multiply through row 1 by -1.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]$

• Eliminate the entries in the first column of the second and third rows. Combine -2 (row 1) with row 2, and -3 (row 1) with row 3.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 5 & -2 & -31 \end{array} \right]$

• Eliminate the entry in the second column of the third row. Combine -5 (row 2) with row 3.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 3 & 24 \end{array} \right]$

• Multiply row 3 by 1/3.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]$

• Eliminate the entry in the third column of the second row. Combine row 2 with row 3.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]$

• Eliminate the entries in the second and third columns of the first row. Combine row 1 with row 2 and -1 (row 3).

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

Then the solution to the system is

$\boxed{x=9, y=-3, z=8}$

If you want to use G elimination and substitution, you'd stop at the step with the augmented matrix

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]$

The third row tells us that $z=8$ . Then in the second row,

$y-z = -11 \implies y=-11 + 8 = -3$

and in the first row,

$x-y+z=20 \implies x=20 + (-3) - 8 = 9$

5 0

2 years ago

A right triangle is ____ a scalene triangle.<br> Always, sometimes, or never

konstantin123 [22]

Answer: Sometimes

Step-by-step explanation:

This is a very tricky question. We know that in an isosceles triangle, two sides are the same. Because of this, we know that the right angle is always 90 degrees, and the other sides would be 45 degrees each because all triangles add up to 180 degrees. A right triangle could also be scalene, meaning none of the sides are the same, for example a scalene triangle could have sides of 44, 46, and 90 degrees. Rest assured, a right triangle will never be equalateral. Hope this helps :) Merry Christmas

4 0

3 years ago

Please hurry

tester [92]

Answer:

D

Step-by-step explanation:

divide 10 times starting with 100.

The answer is 25/256 or 0.09765625

5 0

3 years ago

Lines m and n are parallel lines cut by a transversal l.

lord [1]

The second option. Angles 2 and 6 are corresponding angles, and 6 and 7 make a linear pair, so 2 and 7 are supplementary.

The first option doesn’t even mention angle 7, the third option doesn’t mention angle 2, and the last option doesn’t mention 7 either.

4 0

3 years ago

I don’t know how to solve for Q.

Naya [18.7K]

Answer:

∠Q = 75°

Step-by-step explanation:

Start by recognizing that the triangle is isosceles (the long sides are marked as being equal-length). That means angles Q and R have the same measure.

Next, you use the fact that the sum of angles is 180° to write an equation.

∠R +∠P +∠Q = 180°

(2x +15)° +x° +(2x +15)° = 180° . . . . substitute the known values

5x +30 = 180 . . . . . . . . . . . . . . . . divide by °, collect terms

5x = 150 . . . . . . . . subtract 30

x = 30 . . . . . . . divide by 5

Then angle Q is ...

∠Q = (2x +15)° = (2×30 +15)°

∠Q = 75°

8 0

3 years ago