Use a system of equations to find the specified equation that passes through the points (-2,4) (0,6) and (2,0). Solve the system using matrices.

Find AB where A= 8 2 and B= -3 -2
-3 -3 -8 0

Sove the system of linear equations using an inverse matrix.

7x+4y=5

14x-20y=-15

Given matrix A= 5 21 Find a^-1 the inverse matrix.
-10 7

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Given matrix A= 5 21 Find a^-1 the inverse matrix.
-10 7


The inverse of a 2X2 matrix is easy.

A=\begin{bmatrix}5&21\\-10&7\end{bmatrix}

If we have a matrix A=\begin{bmatrix}a&b\\c&d\end{bmatrix}

A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}=\begin{bmatrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\ \frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{bmatrix}

Use this on your matrix and you have it.

Find AB where A= \(\ \ 8\ \ \ 2\\-3\ -3 \) and B= \(-3\ -2\\ -8\ \ \ 0\)

Just place them side by side like this:

AB= \(\ \ 8\ \ \ 2\\-3\ -3 \) \(-3\ -2\\ -8\ \ \ 0\)

Then, multiply the numbers.
1. 1st row, multiplied by first column. Here, first of the row multiplied by first or the column, second of the row multiplied by second in the column. Then, add the two results and put them in the first position of a 2x2 matrix.

2. 1st row multiplied by second column. Here again, first of the row multiplied by first or the column, second of the row multiplied by second in the column. Then, add again the two results and put them in the second horizontal position of a 2x2 matrix.

3. 2nd row multiplied by first column. Here again, first of the row multiplied by first or the column, second of the row multiplied by second in the column. Then, add again the two results and put them in the first position of the second row of a 2x2 matrix.

4. 2nd row multiplied by second column. Here again, first of the row multiplied by first or the column, second of the row multiplied by second in the column. Then, add again the two results and put them in the second position of the second row of a 2x2 matrix.

Like this:

A = \(a\ b \\ c \ d\) \ and\ B = \(w\ x \\ y\ z\)

AB = \(a\ b \\ c \ d\)\(w\ x \\ y\ z\) = \(aw+by \ ax+bz \\ cw+dy \ cx + dz\)

Sove the system of linear equations using an inverse matrix.

7x+4y=5

14x-20y=-15

Look at those equations like matrices:

\(7\ \ \ \ 4\\ 14 \ -20 \)\(x\\ y \) = \(\ \ 5\\ -15\)

Now to have x and y alone, you need to remove the previous matrix. This is done by multiplying both sides by the inverse matrix of the (7,4 | 14,-20) matrix.

Use what you have learned now to continue.