Ask question

Ask question

All categories

3 years ago

5

Use row reduction to find the inverse of the given matrix if it exists, and check your answer by multiplication. hint [see examp

le 2.] (if the inverse doesn't exist, enter dne in all answer boxes.) 1 6 4 0 1 2 0 0 1

1 answer:

Marianna [84]3 years ago

3 0

The given matrix is
| 1 6 4 |
| 0 1 2 |
| 0 0 1 |

Create the augmented matrix.
| 1 6 4 | 1 0 0 | R1
| 0 1 2 | 0 1 0 | R2
| 0 0 1 | 0 0 1 | R3

R1 -> R1 - 6*R2
| 1 0 -8 | 1 -6 0 | R1
| 0 1 2 | 0 1 0 | R2
| 0 0 1 | 0 0 1 | R3

R1 -> R1 + 8*R3
| 1 0 0 | 1 -6 8 | R1
| 0 1 2 | 0 1 0 | R2
| 0 0 1 | 0 0 1 | R3

R2 -> R2 - 2*R3
| 1 0 0 | 1 -6 8 |
| 0 1 0 | 0 1 -2 |
| 0 0 1 | 0 0 1 |

The inverse is
| 1 -6 8 |
| 0 1 -2 |
| 0 0 1 |

Verification:
| 1 6 4 | | 1 -6 8 | | 1 0 0 |
| 0 1 2 | | 0 1 -2 | = | 0 1 0 |
| 0 0 1 | | 0 0 1 | | 0 0 1 |

You might be interested in

Hannah needs 350 colored pencils for the craft kits for summer camp.

sashaice [31]

Hannah would need 4 boxes

8x12 to find out how many colored pencils in each box you would get 96 then divided that by 350 to see how many boxes you would need you would get 3.645... and so on you would need to round up since Hannah needs 350 she can’t have less but she can have more therefore Hannah needs 4 boxes of pencils

5 0

3 years ago

Read 2 more answers

[tex]cos {}^{4} α+sin {}^{4} α= \frac{1}{4} (3+cos4α)<br>Prove:<br>

asambeis [7]

Given:

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

To prove:

The given statement.

Proof:

We have,

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

$LHS=\cos^4 \alpha+\sin^4\alpha$

$LHS=(\cos^2 \alpha)^2+(\sin^2 \alpha)^2$

$LHS=(\cos^2 \alpha+\sin^2\alpha)^2-2\sin ^2\alpha\cos^2 \alpha$ $[\because a^2+b^2=(a+b)^2-2ab]$

$LHS=(1)^2-2(1-\cos^2 \alpha)\cos^2 \alpha$ $[\because \cos^2 \alpha+\sin^2\alpha=1]$

$LHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

Now,

$RHS=\dfrac{1}{4}(3+\cos 4 \alpha)$

$RHS=\dfrac{1}{4}[3+(2\cos^2 2\alpha-1)]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2\cos^2 2\alpha]$

$RHS=\dfrac{1}{4}[2+2(2\cos^2 \alpha-1)^2]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2(4\cos^4 \alpha-4\cos \alpha+1)]$ $[\because (a-b)^2=a^2-2ab+b^2]$

$RHS=\dfrac{1}{4}[2+8\cos^4 \alpha-8\cos \alpha+2]$

$RHS=\dfrac{1}{4}[4+8\cos^4 \alpha-8\cos \alpha]$

$RHS=1+2\cos^4 \alpha-2\cos \alpha$

$RHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

$LHS=RHS$

Hence proved.

8 0

3 years ago

Which of the following is the equation of a circle whose center is at the origin and whose radius is 4?

Gennadij [26K]

$\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{0}{ h},\stackrel{0}{ k})\qquad \qquad 
radius=\stackrel{4}{ r}
\\\\\\
(x-0)^2+(y-0)^2=4^2\implies x^2+y^2=16$

4 0

3 years ago

Help !!!!!! math pleaseee

IRINA_888 [86]

Answer:

(f + g)(2) = 10, (h – g)(2) = –9, (f × h)(2) = –12, (h / g)(2) = –0.5

Step-by-step explanation:

4 0

3 years ago

Help ASAP thanks :)))

barxatty [35]

a raise to m ∙ a raise to n = a raise to (m + n)

so 7 raise to (3+4)=7 raise to &

7 0

3 years ago