Question

PLEASE HELP QWQ AsAp with these 4 questions

Answer 1

Explanation:

I can't believe I'm doing this for 5 points, but ok!

For the first 3, we are going to multiply to find the value of that 3 x 3 matrix by picking up the first 2 columns and plopping them down at the end and then multiplying through using the rules for multiplying matrices:

`\left[\begin{array}{ccccc}7&4&6&7&4\\-4&8&9&-4&8\\1&8&7&1&8\end{array}\right]` and from there find the sum of the products of the main axes minus the sum of the products of the minor axes, as follows (I'm not going to state the process in the next 2 problems, so make sure you follow it here. This is called the determinate. The determinate is what you get when you evaluate or find the value of a matrix. Just so you know):

`(7*8*7)+(4*9*1)+(6*-4*8)-[(1*8*6)+(8*9*7)+(7*-4*4)]` which gives us:

392 + 36 - 192 - [48 + 504 - 112] which simplifies to

236 - 440 which is -204

On to the second one:

`\left[\begin{array}{ccccc}-8&-4&-1&-8&-4\\1&7&-3&1&7\\8&9&9&8&9\end{array}\right]` and multiplying gives us

`(-8*7*9)+(-4*-3*8)+(-1*1*9)-[(8*7*-1)+(9*-3*-8)+(9*1*-4)]` which gives us:

-504 + 96 - 9 - [-56 + 216 - 36] which simplifies to

-417 - 124 which is -541, choice c.

Now for the third one:

`\left[\begin{array}{ccccc}-2&-2&-5&-2&-2\\2&7&-3&2&7\\8&9&9&8&9\end{array}\right]` and multiplying gives us

`(-2*7*9)+(-2*-3*8)+(-5*2*9)-[(8*7*-5)+(9*-3*-2)+(9*2*-2)]` which gives us:

`-126+48-90-[-280+54-36]` which simplifies to

-168 - (-262) which is 94, choice c again.

Now for the last one. I'll show you the set up for the matrix equation; I solved it using the inverse matrix. So I'll also show you the inverse and how I found it.

`\left[\begin{array}{cc}-4&-5&\\-6&-8\\\end{array}\right]`
`\left[\begin{array}{c}x\\y\\\end{array}\right]` =
`\left[\begin{array}{c}-5\\-2\\\end{array}\right]` and I found the inverse of the 2 x 2 matrix on the left.

Find the inverse by:

* finding the determinate

* putting the determinate under a 1

* multiply that by the "mixed up matrix (you'll see...)

First things first, the determinate:

|A| = (-4*-8) - (-6*-5) which simplifies to

|A| = 32 - 30 so

|A| = 2; now put that under a 1 and multiply it by the mixed up matrix. The mixed up matrix is shown in the next step:

`(1)/(2)\left[\begin{array}{cc}-8&5\\6&-4\end{array}\right]` (to get the mixed up matrix, swap the positions of the numbers on the main axis and then change the signs of the numbers on the minor axis). Now we multiply in the 1/2 to get the inverse:

`\left[\begin{array}{cc}-4&(5)/(2)\\3&-2\\\end{array}\right]` Multiply that inverse by both sides of the equation. This inverse "undoes" the matrix that's already there (like dividing the matrix that's already there by itself) which leaves us with just the matrix of x and y. Multiply the inverse matrix by the solution matrix:

`\left[\begin{array}{c}x&y\end{array}\right] =\left[\begin{array}{cc}-4&(5)/(2) \\3&-2\end{array}\right] *\left[\begin{array}{c}-5&-2\\\end{array}\right]` and that right side multiplies out to

x = 20 - 5 which is

x = 15 and

y = -15 + 4 which is

y = -11

(It works, I checked it)