Question

How do u solve matrices

Answer 1

see explanation

Explanation:

multiplying the matrices on the left side

`\left[\begin{array}{ccc}3x+4y\\6x+y\\\end{array}\right]` =
`\left[\begin{array}{ccc}6\\-9\\\end{array}\right]`

Equating corresponding components on both sides gives

3x + 4y = 6 → (1)

6x + y = - 9 → (2)

multiply (2) by - 4

- 24x - 4y = 36 → (3)

Add (1) and (3) term by term

- 21x = 42 ( divide both sides by - 21

x = - 2

substitute x = - 2 into either (1) or (2)

(1) : - 6 + 4y = 6 ( add 6 to both sides )

4y = 12 ( divide both sides by 4 )

y = 3

solution is (- 2, 3 )

Answer 2

Answer: y = 3

Explanation:

Multiply the matrices on the left and set each row equal to the right side. This will result in a system of equations.

Use the following format for multiplying:

`\left[\begin{array}{cc}a_(11)&a_(12)\\a_(21)&a_(22)\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] =\left[\begin{array}{c}c\\d\end{array}\right]\\ \\\\\rightarrow \left\{\begin{array}{cc}a_(11)(x) + a_(12)(y)&=c\\a_(21)(x)+a_(22)(y)&=d\end{array}\right\}`

`\left[\begin{array}{cc}3&4\\6&1\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] =\left[\begin{array}{c}\ \ 6\\-9\end{array}\right]\\ \\\\\rightarrow \left\{\begin{array}{cl}3x+4y&=\ \ 6\\6x+1y&=-9\end{array}\right\}`

Now solve the system. Since you are looking for "y", eliminate "x" by multiplying Row 1 by -2:

`\begin{array}{clll}3x+4y&=\ \ 6&\rightarrow -2(3x + 4y = 6)&\rightarrow -6x-8y=-12\\6x+1y&=-9&\rightarrow 1(6x + 1y = -9)&\rightarrow \underline{\ \ 6x + 1y =\  -9}\\& & &\ \quad \qquad -7y=-21\\& & &\ \ \qquad \qquad y=3\\\end{array}`