Help........ Again and thanks

Mathematics

1 answer:

Hunter-Best [27]3 years ago

5 0

Answer:

x=3, y=1

Step-by-step explanation:

2x+y =7

3x-4y =5

Multiply the first equation by 4 so we can eliminate y

4(2x+y =7)

8x +4y =28

Add this to the second equation

8x +4y =28

3x-4y =5

----------------

11x = 33

Divide by 11

11x/11=33/11

x=3

Now we still need to find y

2x+y=7

2(3)+y =7

6+y=7

Subtract 6 from each side

6-6+y=7-6

y= 1

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Is X=[−14−32] the inverse of A=[15−25310−110]?

Elis [28]

The product of the matrices is an identity matrix. Therefore, X and A are inverse of each other.

The matrices are given as:

$X = \left[\begin{array}{cc}-1&-3\\4&2\end{array}\right]$

$A = \left[\begin{array}{cc}\frac 15&\frac 3{10}\\-\frac 25 &-\frac 1{10}\end{array}\right]$

To check the matrices are inverse, we calculate their products.

$A \times X = \left[\begin{array}{cc}\frac 15&\frac 3{10}\\-\frac 25 &-\frac 1{10}\end{array}\right] \times \left[\begin{array}{cc}-1&-3\\4&2\end{array}\right]$

Multiply the rows of A by the column of X.

This gives

$A \times X = \left[\begin{array}{cc}\frac 15 \times -1 + \frac{3}{10} \times 4&\frac 15 \times -3 + \frac{3}{10} \times 2\\ -\frac 25 \times -1 - \frac{1}{10} \times 4&-\frac 25 \times -3 + -\frac{1}{10} \times 2\end{array}\right]$