HELP PLEASE- Linear Algebra

[seksy]

I have two questions!
1) So the question is about matrices (A^-1 --> inverse of a matrix) and it is as follows:

Use the method of inverses, if possible, to solve the system x+2y = 3
2x-y = 1

What I thought was that you would write out 1 2 3 | 1 0 1
2-1 1| 0 1 0

and then row-reduce the matrix until the left side is in row-reduced form and then whatever answers you have on the right side are x and y? Can someone please explain this and show me how they got the answer? Thanks.

2) This question is about ranks.. I don't understand how you figure that out of a matrix.. I'm really confused, please help!

Find the rank of the matrix A = 1 0 2
1-2 2
3 1 6

Can you please show me how to do this and what the solution is?


- Thank you so much!

 

[galactus]

I have two questions!
1) So the question is about matrices (A^-1 --> inverse of a matrix) and it is as follows:

Use the method of inverses, if possible, to solve the system x+2y = 3
2x-y = 1

What I thought was that you would write out 1 2 3 | 1 0 1
2-1 1| 0 1 0

and then row-reduce the matrix until the left side is in row-reduced form and then whatever answers you have on the right side are x and y? Can someone please explain this and show me how they got the answer? Thanks.

\(\displaystyle \begin{bmatrix}1&2\\2&-1\end{bmatrix}\cdot \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\1\end{bmatrix}\)

The solutions are \(\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1&2\\2&-1\end{bmatrix}^{-1}\cdot\begin{bmatrix}3\\1\end{bmatrix}\)

To find the inverse of a 2-by-2 matrix use \(\displaystyle \begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\)

Use this easy formula to find the inverse. Multiply it by \(\displaystyle \begin{bmatrix}3\\1\end{bmatrix}\) and you have your solutions.

2) This question is about ranks.. I don't understand how you figure that out of a matrix.. I'm really confused, please help!

Find the rank of the matrix A = 1 0 2
1-2 2
3 1 6

Can you please show me how to do this and what the solution is?

Find the RREF of the given matrix. The number of leading variables is the rank.

 

[soroban]

Hello, seksy!

\(\displaystyle \text{(1) Use the method of inverses to solve the system: }\;\begin{array}{ccc}x+2y &=& 3 \\ 2x-y &=& 1\end{array}\)

Your game plan had a good start . . .

\(\displaystyle \text{We have: }\:Ax \:=\:B \Longleftrightarrow\quad \begin{pmatrix}1 & 2 \\ 2 & \text{-}1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} \:=\:\begin{pmatrix}3 \\ 1 \end{pmatrix}\) .[1]
. . \(\displaystyle \text{and we want }A^{\text{-}1}\)

\(\displaystyle \text{We have: }\: \left|\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 2 & \text{-}1 & 0 & 1 \end{array}\right|\quad\)

\(\displaystyle \begin{array}{c} \\ R_2 - 2R_1\end{array}\left|\begin{array}{cc|cc}1 & 2 & 1 & 0 \\ 0 & \text{-}5 & \text{-}2 & 1 \end{array}\right|\)

. . . \(\displaystyle \begin{array}{c} \\ \text{-}\frac{1}{5}R_2\end{array}\left|\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & 1 & \frac{2}{5} & \text{-}\frac{1}{5}\end{array}\right|\)

\(\displaystyle \begin{array}{c}R_1-2R_2 \\ \\ \end{array}\left|\begin{array}{cc|cc} 1 & 0 & \frac{1}{5} & \frac{2}{5} \\ 0 & 1 & \frac{2}{5} & \text{-}\frac{1}{5}\end{array}\right| \text{ }\)

\(\displaystyle \text{Hence: }\:A^{\text{-}1} \;=\;\begin{pmatrix}\frac{1}{5} & \frac{2}{5} \\ \\[-3mm]\frac{2}{5} & \text{-}\frac{1}{5}\end{pmatrix}\)


Left-multiply [1] by \(\displaystyle A^{\text{-}1}\)

. . \(\displaystyle \begin{pmatrix}\frac{1}{5} & \frac{2}{5} \\ \\[-3mm] \frac{2}{5} & \text{-}\frac{1}{5}\end{pmatrix}\begin{pmatrix}1 & 2 \\ 2 & \text{-}1 \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix} \;=\;\begin{pmatrix}\frac{1}{5} & \frac{2}{5} \\ \\[-3mm] \frac{2}{5} & \text{-}\frac{1}{5}\end{pmatrix}\begin{pmatrix}3 \\ 1 \end{pmatrix}\)

. . . . . . . . . \(\displaystyle \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix} \;=\;\begin{pmatrix}\frac{3}{5} + \frac{2}{5} \\ \\[-3mm] \frac{6}{5} - \frac{1}{5}\end{pmatrix}\)

. . . . . . . . . . . . . . .\(\displaystyle \begin{pmatrix}x \\ y\end{pmatrix} \;=\;\begin{pmatrix}1 \\ 1\end{pmatrix}\)

\(\displaystyle \text{Therefore: }\:\begin{Bmatrix} x \:=\: 1 \\ y \:=\: 1 \end{Bmatrix}\)


Edit: Ah, galactus did it with the formula . . . good!
.

 

[seksy]

Thank you! I did the question as suggested and I got the same answer

- I have a question for the second question I posted it up, I worked it out and got a rank of 2, is that correct?

My final answer was
1 0 0
0 1 0
0 0 0

 

[galactus]

-

I have a question for the second question I posted it up, I worked it out and got a rank of 2, is that correct?

My final answer was
1 0 0
0 1 0
0 0 0

Yep.