# what am i doing wrong? finding the inverse of the matrix

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• Feb 21st 2013, 01:11 PM
MariaBlood777
what am i doing wrong? finding the inverse of the matrix
the question is find the inverse the matrix.

[3 1][1 0]
[4 2][0 1]

in Calc Chat (step by step) they multiplied 1/2(R2) first then did -R2 + R1 and so forth.

I thought I was suppose to multiply 1/3(R1) first to get 1 on R1 and so forth. I am so confused. I know how to set it up but I still dont get why I could have just multiplied 1/3(R1). ??? In Gaus-elimination are you suppose to make sure the coefficent is 1 on the first row??(Worried)

thank you
• Feb 21st 2013, 05:02 PM
Soroban
Re: what am i doing wrong? finding the inverse of the matrix
Hello, MariaBlood777!

Quote:

$\displaystyle \text{Find the inverse of: }\:A \;=\;\begin{bmatrix}3&1\\4&2\end{bmatrix}$

$\displaystyle \left[\begin{array}{cc|cc}3&1&1&0 \\4&2&0&1\end{array}\right]$

in Calc Chat (step by step) they multiplied 1/2(R2) first then did -R2 + R1 and so forth.
I can see why they did that, but it's a very strange opening move.

I thought I was suppose to multiply 1/3(R1) first to get 1 on R1 and so forth.
That is correct, but there are other ways to get 1 on R1.

We can do it "your way", but it will introduce fractions.
If you can handle them, great!

$\displaystyle \text{We have: }\:\left[\begin{array}{cc|cc}3&1&1&0 \\ 4&2&0&1 \end{array}\right]$

. . . . $\displaystyle \begin{array}{c}\frac{1}{3}R_1 \\ \frac{1}{4}R_2 \end{array}\:\left[\begin{array}{cc|cc} 1&\frac{1}{3} & \frac{1}{3}&0 \\ 1 & \frac{1}{2} & 0 & \frac{1}{4}\end{array}\right]$

. $\displaystyle \begin{array}{c}\\ R_2-R_1 \end{array}\:\left[\begin{array}{cc|cc} 1 & \frac{1}{3} & \frac{1}{3} & 0 \\ 0 & \frac{1}{6} & \text{-}\frac{1}{3} & \frac{1}{4} \end{array}\right]$

. . . . $\displaystyle \begin{array}{c}\\6R_2 \end{array}\:\left[\begin{array}{cc|cc}1 & \frac{1}{3} & \frac{1}{3} & 0 \\ 0 & 1 & \text{-}2 & \frac{3}{2}\end{array}\right]$

$\displaystyle \begin{array}{c}R_1-\frac{1}{3}R_2 \\ \\ \end{array}\:\left[\begin{array}{cc|cc} 1&0 & 1 & \text{-}\frac{1}{2} \\ 0&1&\text{-}2 & \frac{3}{2}\end{array}\right]$

Therefore: .$\displaystyle A^{-1} \;=\;\begin{bmatrix}1 & \text{-}\frac{1}{2} \\ \text{-}2 & \frac{3}{2} \end{bmatrix}$