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3241004551 [841]
3 years ago
10

Find the inverse of the given​ matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3

rd Column negative 2 2nd Row 1st Column 4 2nd Column 1 3rd Column 3 3rd Row 1st Column negative 4 2nd Column 2 3rd Column 3 EndMatrix right bracket 1 0 −2 4 1 3 −4 2 3
Mathematics
1 answer:
BabaBlast [244]3 years ago
6 0

Answer:

A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]

Step-by-step explanation:

We want to find the inverse of A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

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

  • Subtract row 1 multiplied by 4 from row 2

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]

  • Add row 1 multiplied by 4 to row 3

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]

  • Subtract row 2 multiplied by 2 from row 3

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]

  • Divide row 3 by −27

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

  • Add row 3 multiplied by 2 to row 1

\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

  • Subtract row 3 multiplied by 11 from row 2

\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

As can be seen, we have obtained the identity matrix to the left. So, we are done.

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Berto has $12 to put gas in his car. If gas costs $3.75 per gallon, which ordered pair relating number of gallons of gas, x, to
Salsk061 [2.6K]

Answer:

If gas costs $3.75 per gallon and Berto has $12, then he can purchase 12/3.75 gallons. This is approximately 3.2 gallons. So the coordinate on this line would be (3.2, 12).

Actually... It's from web

3 0
3 years ago
Read 2 more answers
Help me solve these please
Burka [1]
<span>______________________________________________________</span>
7y-1=-1+9(y-9)

Add 1 on both sides.
7y=9(y-9)
Distribute.
7y=9y-81
Subtract 7y and add 81 on both sides.
2y=81
y=40.5
______________________________________________________

<span>4c+3=-12+c-1
</span>Subtract 3 on both sides.
4c=-12-1-3+c
Subtract c on both sides.
3c=-16
c=-16/3
______________________________________________________
<span>-(k+1)-10=-5k+3
</span>Add 10 on both sides.
-(k+1)=-5k+13
Distribute.
-k-1=-5k+13
Add 1 and add 5k on both sides.
4k=14
k=4/14
______________________________________________________
-w-10=-9-2(w-8)
Distribute.
-w-10=-9-2w+16
Simplify.
-w-10=-2w+7
Add 2w and 10 on both sides.
w=17
______________________________________________________
-2(n-3)-5=-5n+1
Distribute.
-2n+6-5=-5n+1
Simplify.
-2n+1=-5n+1
Add 5n and subtract 1 on both sides.
3n=0
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______________________________________________________

To make sure you got the right answer, you can check your work by plugging the result back into the question and seeing if both sides become equal. If both sides are equal, then the result is correct.
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4 0
3 years ago
Use the ordinary division of polynomials to find the quotient and remainder when the first polynomial is divided by the second.
Vanyuwa [196]

The quotient and remainder when the first polynomial is divided by the second are -4w^2 - 7w - 21 and -71 respectively

<h3>How to determine the quotient and remainder when the first polynomial is divided by the second?</h3>

The polynomials are given as:

-4w^3 + 5w^2 - 8, w - 3

Set the divisor to 0.

So, we have

w - 3 = 0

Add 3 to both sides

w = 3

Substitute w = 3 in -4w^3 + 5w^2 - 8 to determine the remainder

-4(3)^3 + 5(3)^2 - 8

Evaluate the expression

-71

This means that the remainder when -4w^3 + 5w^2 - 8 is divided by w - 3 is -71

The quotient (Q) is calculated as follows:

Q = [-4w^3 + 5w^2 - 8]/[w - 3]

The numerator can be expressed as follows:

Numerator = -4w^3 + 5w^2 - 8

Subtract the remainder.

So, we have:

Numerator = -4w^3 + 5w^2 - 8 + 71

This gives

Numerator = -4w^3 + 5w^2 + 61

So, the quotient becomes

Q = [-4w^3 + 5w^2 + 61]/[w - 3]

Expand

Q = [(w - 3)(-4w^2 - 7w - 21)]/[w - 3]

Evaluate

Q = -4w^2 - 7w - 21

Hence, the quotient and remainder when the first polynomial is divided by the second are -4w^2 - 7w - 21 and -71 respectively

Read more about polynomials at:

brainly.com/question/4142886

#SPJ1

6 0
2 years ago
ky has 3 times more books than grant and grant has 6 fewer books than Jaime if the total combined number of books is 176 how man
sergiy2304 [10]
Number of books each of them have:

Jaime: x
Grant: x - 6
Ky: 3 (x - 6)

There are 176 books in total, so we can write it down as:

176 = x + x - 6 + 3 (x - 6)
176 = x + x - 6 + 3 * x + 3 * (-6)
176 = x + x - 6 + 3x - 18
176 = 2x - 6 + 3x - 18
176 = 5x - 24                        / + 24 (both sides)
5x = 200                              / ÷ 5 (both sides)
x = 40

Doublecheck:

Jamie: 40
Grant: 40 - 6 = 34
Ky: 3 (40 - 6) = 3 * 34 = 102

102 + 34 + 40 = 176, so it's correct :)
5 0
3 years ago
Read 2 more answers
Q1W7 Learning Task 1 (Introduction)
Tatiana [17]

Answer:

Column A                                            Column B

1. x² + 6x + 8                                        x-3,x+2

2. x³ - 7x + 6                                       x+1, x+2, x+3

3. x³ - 2x² - 5x + 6                              x-1, x+2, x-3

Step-by-step explanation:

Column A                                            Column B

1. x² + 6x + 8                                        x-3,x+2

2. x³ - 7x + 6                                       x+1, x+2, x+3

3. x³ - 2x² - 5x + 6                              x-1, x+2, x-3

Using Factor theorem we put values of x = ±1,±2,±3 in each of the polynomials unless we get a zero.

1. x² + 6x + 8      

= 1+6(1) +8= 15

1. x² + 6x + 8

  4+ 12+8 = 24

1. x² + 6x + 8

 (-1)² + 6(-1)+ 8

= 1-6+8= 3

1. x² + 6x + 8

 (-2)² + 6(-2)+ 8

= 4-12+8= 0

1. x² + 6x + 8

(3)²+ 6(3) +8

= 9+18+8 ≠ 0

1. x² + 6x + 8

(-3)²+ 6(-3) +8

= 9-18+8 =-1

For this polynomial we have x+2= 0 or x=-2, x-3= 0 , x=3

2. x³ - 7x + 6

1-7+6= 0

2. x³ - 7x + 6

(-1)³-7(-1) +6

= 13-1≠0

2. x³ - 7x + 6

(2)³-7(2) +6

= 8-14+6= 0

2. x³ - 7x + 6

(-2)³-7(-2) +6

= -8 +14+6

2. x³ - 7x + 6

(-3)³-7(-3) +6

= -27+21+6 = 0

For this polynomial we have x+1= 0 , x+2 = 0  and x+3= 0, or x=-1,-2,-3

3. x³ - 2x² - 5x + 6

(1)³-2(1)²-5(1)+6

= 0

3. x³ - 2x² - 5x + 6

(-1)³-2(-1)²-5(-1)+6

= -1 -2 +5+6

=8

3. x³ - 2x² - 5x + 6

(2)³-2(2)²-5(2)+6

= 8-8-10+6

=-4

3. x³ - 2x² - 5x + 6

(-2)³-2(-2)²-5(-2)+6

= -8-8+10+6

=0

3. x³ - 2x² - 5x + 6

(3)³-2(3)²-5(3)+6

= 27-18-15+6

=0

3. x³ - 2x² - 5x + 6

(-3)³-2(-3)²-5(-3)+6

= -27-18+15+6

=-14

For this polynomial we have x-1= 0 ,x+2=0, x-3= 0or x=1,-2,3

7 0
3 years ago
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