What is 6 2/3 times 4 1/5. Estimate the product

We cannot subtract matrices of different sizes. Matrix B is 2x2 while C is 3x2. Both matrices must have the same number of rows, and they must also have the same number of columns. The matrices don't have to be square.

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b)

$A+B = \begin{bmatrix}-5 & 6\\7 & 4\end{bmatrix}$

You add the corresponding elements. For instance, in the top left corner we have -1+(-4) = -5. The other entries are treated in a similar manner.

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c)

$-2E = \begin{bmatrix}6 & -4 & 8\\-12 & 14 & -16\\-10 & -18 & 20\end{bmatrix}$

You get this from multiplying each entry in matrix E by -2. Eg: top left corner has -2*(-3) = 6

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d)

$CD = \begin{bmatrix}4 & 7 & 8\\ 1 & 3 & 4\\14 & 17 & 16\end{bmatrix}$

Matrix C has 3 rows and D has 3 columns. The final answer will be size 3x3

To generate each value in the answer matrix, you'll highlight rows of C to pair with columns of D. Then you'll multiply out the corresponding values, after which you add those products. This is done for every entry in the answer shown above.

For example, the first row of C is highlighted and the second column of D is highlighted. Those values pair up and multiply getting -1*(-1) + 2*3 = 1+6 = 7, which goes in the first row and second column of the answer matrix. The other entries are handled in a similar fashion.

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e)

$\det(B) = -46$

The 2x2 matrix determinant formula is $\begin{vmatrix}a & b\\c & d\end{vmatrix} = a*d - b*c$

In this case, a = -4, b = 6, c = 5, d = 4.

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f)

$B^{-1} = \begin{bmatrix}-2/23 & 3/23\\ 5/46 & 2/23\end{bmatrix}$

Swap the top left and bottom right corners of matrix B. Change the sign of the other two corner values. Then multiply each entry by 1/d where d is the determinant found back in part (e) above. Be sure to reduce any fraction as much as possible.

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Problem 4

Answers: x = 2 and y = -10

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Work Shown:

$\begin{bmatrix}2x & 4\\-5 & -2\end{bmatrix}+\begin{bmatrix}3 & -7\\11 & y-1\end{bmatrix} = \begin{bmatrix}7 & -3\\6 & -13\end{bmatrix}\\\\\\\begin{bmatrix}2x+3 & 4+(-7)\\-5+11 & -2+(y-1)\end{bmatrix} = \begin{bmatrix}7 & -3\\6 & -13\end{bmatrix}\\\\\\\begin{bmatrix}2x+3 & -3\\6 & y-3\end{bmatrix} = \begin{bmatrix}7 & -3\\6 & -13\end{bmatrix}\\\\\\$

In the top left corners of each matrix, in line 3, we have 2x+3 = 7 which solves to...

2x+3 = 7

2x = 7-3

2x = 4

x = 4/2

x = 2

In the bottom right corners, we have y-3 = -13 which solves to....

y-3 = -13

y = -13+3

y = -10

4 0

3 years ago

You will spend no less than $80 on a gift. You have already spent $50 on part of the gift. Let s be the amount of money you will

GalinKa [24]

Assuming you have to create an equation...
s is what you have to figure out so it is the subject of the rule
s=
you now need to find out what to do with the other side
you are taking 50 FROM 80 so
s=80-50
s=30
your answer is 30

4 0

4 years ago

A truck can be rented from Company A for $60 a day plus $0.60 per mile . Company B charges $30 a day plus $0.90 per mile to rent

Mamont248 [21]

x = the number of miles

y = the total cost

Company A:

0.60x + 60 = y [Company A charges $60 plus $0.60 per mile(x)]

Company B:

0.90x + 30 = y [Company B charges $30 plus $0.90 per mile(x)]

To find the number of miles where the costs for both companies are the same, you can set the equations equal to each other as the costs(y) are the same:

y = y Substitute the equations into "y" (substitute (0.60x + 60) and (0.90x + 30) into "y" since y = 0.60x + 60 and y = 0.90x + 30)

0.60x + 60 = 0.90x + 30 To find x, isolate/get the variable "x" by itself. Subtract 30 on both sides

0.60x + 60 - 30 = 0.90x + 30 - 30

0.60x + 30 = 0.90x Subtract 0.60x on both sides to get "x" on one side of the equation

0.60x - 0.60x + 30 = 0.90x - 0.60x

30 = 0.30x Divide 0.30 on both sides to get "x" by itself

100 = x 100 miles

(if you need to find out the cost where both companies cost the same, you can substitute/plug in the value of x into one of the equations.)

0.60x + 60 = y Plug in 100 into "x" since x = 100

0.60(100) + 60 = y

120 = y At 100 miles, both companies cost $120

8 0

4 years ago