Question

Help with matrices?

Answer 1

you can multiply by the inverse to solve that

Answer 2

ok can you help me find the inverse?

Answer 3

@AutumnRoseT did you get the answer.

Answer 4

@Haseeb96 no, I'm not sure how to get the inverse

Answer 5

For inverse we use this formula
A^-1(call A inverse) = adj A/ I A I

Answer 6

I'm still not sure how to do that...

Answer 7

have you hear the words adjoint matrix and discriminant matrix?

Answer 8

I know of discriminant matrix but not adjoint...

Answer 9

its called determinant, not discriminant :)

Answer 10

okay

Answer 11

there is a nice formula for the inverse of a 2x2 matrix

Answer 12

Ya I thought it sounded funny thank you perl

Answer 13

$$\large
\mathbf{A}^{-1} = \begin{bmatrix}
a & b \\ c & d \\
\end{bmatrix}^{-1} =
\frac{1}{\det(\mathbf{A})} \begin{bmatrix}
\,\,\,d & \!\!-b \\ -c & \,a \\
\end{bmatrix} =
\frac{1}{ad - bc} \begin{bmatrix}
\,\,\,d & \!\!-b \\ -c & \,a \\
\end{bmatrix}

$$

Answer 14

I don't suppose someone could walk me through the equation with this problem?

Answer 15

i can

Answer 16

I would really appreciate it so I can understand it better

Answer 17

$$
\mathbf{A} = \begin{bmatrix}
5 & -2 \\ 2 & -1 \\
\end{bmatrix}


\\~\\
\mathbf{A}^{-1} = \begin{bmatrix}
5 & -2 \\ 2 & -1 \\
\end{bmatrix}^{-1} =
\frac{1}{\det(\mathbf{A})} \begin{bmatrix}
\ -1 & \ 2 \\ -2 & 5 \\
\end{bmatrix} =
\frac{1}{ad - bc} \begin{bmatrix}
\,\,\,d & \!\!-b \\ -c & \,a \\
\end{bmatrix}

$$

Answer 18

notice how i am using the formula above

Answer 19

yes ok I can kinda see how that works...

Answer 20

$$
\mathbf{A} = \begin{bmatrix}
a & b \\ c & d \\
\end{bmatrix}= \begin{bmatrix}
5 & -2 \\ 2 & -1 \\
\end{bmatrix}

\\~\\ \mathbf{A}^{-1} =
\frac{1}{\det(\mathbf{A})} \begin{bmatrix}
\,\,\,d & \!\!-b \\ -c & \,a \\
\end{bmatrix} =\frac{1}{5(-1) - (-2)(2)}
\begin{bmatrix}
\ -1 & \ 2 \\ -2 & 5 \\
\end{bmatrix}

$$

Answer 21

ok this will take a minute for me to look at and understand...

Answer 22

the positions a,d switch, and b,c both become negative

Answer 23

oh ok I can see it now

Answer 24

$$ \large{
\mathbf{A} = \begin{bmatrix}
a & b \\ c & d \\
\end{bmatrix}= \begin{bmatrix}
5 & -2 \\ 2 & -1 \\
\end{bmatrix}

\\~\\ \mathbf{A}^{-1} =
\frac{1}{ad - bc} \begin{bmatrix}
\,\,\,d & \!\!-b \\ -c & \,a \\
\end{bmatrix} =\frac{1}{5(-1) - (-2)(2)}
\begin{bmatrix}
\ -1 & \ 2 \\ -2 & 5 \\
\end{bmatrix}
\\~\\ \mathbf{A}^{-1} =\frac{1}{-1}\begin{bmatrix}
\ -1 & \ 2 \\ -2 & 5 \\
\end{bmatrix}
\\~\\ \mathbf{A}^{-1} =\begin{bmatrix}
\ 1 & -2 \\ 2 & -5 \\
\end{bmatrix}

}
$$

Answer 25

oooooh I see! and then multiply that by the other matrix in the original problem?

Answer 26

$$ \Large{
A~\bf {x} = b
\\~\\
A^{-1}A~\bf {x} = A^{-1}b
\\~\\
I~\bf {x} = A^{-1}b
\\~\\ \bf {x} = A^{-1}b
\\~\\ \bf {x} = \begin{bmatrix}
\ 1 & -2 \\ 2 & -5 \\
\end{bmatrix} \cdot \begin{bmatrix} 2 \\4\end{bmatrix}




}
$$

Answer 27

sooo.... I've never done multiplication with different dimensions before...

Answer 28

@perl can you help me multiply it?

Answer 29

$$
\Large \rm {
A~\mathbf {x} = \mathbf{b }
\\~\\
A^{-1}A~\mathbf {x} = A^{-1}\mathbf{b }
\\~\\
I~\mathbf {x} = A^{-1}\mathbf{b }
\\~\\ \mathbf {x} = A^{-1}\mathbf{b }
\\~\\
\mathbf {x} = \begin{bmatrix}
\ 1 & -2 \\ 2 & -5 \\
\end{bmatrix} \cdot \begin{bmatrix} 2 \\-4\end{bmatrix}
\\~\\
\mathbf {x} = \begin{bmatrix}
\ 1\cdot 2 + (-2)\cdot (-4) \\ 2\cdot 2 + (-5)\cdot (-4) \\
\end{bmatrix}
\\~\\
\mathbf {x} = \begin{bmatrix}
10\\24 \\
\end{bmatrix}

}
$$

Answer 30

There is also another way to solve this matrix equation, might be faster.
The matrix equation given is equivalent to a system of equations as follows.
$$
\Large{

\begin{bmatrix}
5 & -2 \\ 2 & -1 \\
\end{bmatrix}
\cdot
\begin{bmatrix}
x \\ y \end{bmatrix}
= \begin{bmatrix}
2 \\ 4 \end{bmatrix}

\iff \begin{cases} 5x -2y = 2\\ 2x -1y = 4 \end{cases}
}
$$

Answer 31

Thank you so much!