# Math Help - Invert Matrix

1. ## Invert Matrix

Been a while since I had to do this... totally forgot how to <_<

the only thing I can remember is $A^{-1} = \frac{1}{ad-bc} * \left(\begin{array}{cc}d&-b\\-c&a\end{array}\right)$ or something...

But this leaves me absolutely clueless on how to solve

$A^{-1} = \left(\begin{array}{ccc}1&2&3\\2&3&4\\3&4&6\end{ar ray}\right)$

2. Hi there.

That's the method for finding the inverse of a 2x2 matrix. Are you looking for a method to find the inverse of a 3x3 matrix? If so, have a look here: How to Inverse a 3X3 Matrix - wikiHow

3. Thanks

(Stupid question: why is this in university math? I first saw this in the 5th grade of high school? (just couldn't remember the formula xd))

4. Here is what I consider a simpler method of inverting a matrix using "row reduction".

First write the matrix and the identity matrix side by side:
$\begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 6\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$

Now use "row-reduction" to reduce the given matrix to the identity matrix while applying the same reductions to the identity matrix. When you are done, you will have the identity matrix now where the given matrix was and the inverse where the identity matrix was.

"Row operations" basically correspond to the operations we can do to entire equations (each being a row of the matrix):
1) Multiply or divide every number in one row of the matrix by a number.
2) Add (or subtract) a multiple of one row to (or from) another.
3) Swap two row.

Work one column at a time from left to right- your objective is to get a "1" on the diagonal (at the "pivot" position) which you can do by dividing the entire row by the number in that spot, and then get "0" at every position below or above the diagon which you can do by subtracting the "pivot" row multiplied by the number in the row above or below the "pivot" from that row.

Here, I see that the number on the diagonal in the first row, the "pivot" number, is 1 so I don't have to divide that row by anything. Directly below it, in the second row, is a "2" so I subtract 2 times the first row from the second row. In the third row I have a "3" so I subtract 3 times the first row from the third row. And of course, I extend those same row operations to the identity matrix on the right.

That gives:
$\begin{bmatrix}1 & 2 & 3 \\ 2-2(1) & 3-2(2) & 4-2(3) \\ 3- 3(1) & 4- 3(2) & 6- 3(3)\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0-2(1) & 1- 2(0) & 0- 2(0) \\ 0- 3(1) & 0- 3(0) & 1- 3(0)\end{bmatrix}$
$\begin{bmatrix}1 & 2 & 3 \\ 0 & -1 & -2 \\ 0 & -2 & -3\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ -2 & 1 & 0 \\ -3 & 0 & 1\end{bmatrix}$

Now, the "pivot" is the second number in the second row and it is "-1". I can make that "1" by multiplying the entire row by -1. Above the "pivot" is "2" and I can make that "0" by subtracting 2 times the (new) second row from the first row. Below the "pivot" is -2 and I can make that "0" by adding 2 times the (new) second row to it (or "subtracting -2 times...).

That gives:
$\begin{bmatrix}1- 2(0) & 2- 2(1) & 3- 2(2) \\ 0 & 1 & 2 \\ 0+ 2(0) & -2+ 2(1) & -3+ 2(2)\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & -1 & 0\ -3+ 2(2) & 0- 2(-1) & 1+ 2(0)\end{bmatrix}$
$\begin{bmatrix}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}-3 & 2 & 0 \\ 2 & -1 & 0 \\ 1 & -2 & 1\end{bmatrix}$

Notice that because of the "0" s in the first column, the first column is left as it was- already correct.

Now I see that the "pivot", last row, last column, is "1" so I don't have to multiply of divide the last row by anything. There is a "-1} in the first row of the last column so to get "0" in the first row above that, I need to add the third row to the first row. There is a "2" in the second row of the last column so to get a "0" there, I need to subtract 2 times the third row from the second row.

That gives:
$\begin{bmatrix}1+ 1(0) & 0+ 1(0) & -1+ 1(1) \\0- 1(0) & 1- 1(0) & 2-2(1) \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}-3+ 1(1) & 2+ 1(-2) & 0+ 1(1) \\ 2- 2(1) & -1-2(=2) & 0- 2(1)\end{bmatrix}$
$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} -2 & 0 & 1 \\ 0 & 3 & -2 \\ 1 & -2 & 1\end{bmatrix}$.

That is, the inverse matrix to
$\begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 6\end{bmatrix}$
is
$\begin{bmatrix} -2 & 0 & 1 \\ 0 & 3 & -2 \\ 1 & -2 & 1\end{bmatrix}$.

5. Originally Posted by shinhidora
Thanks

(Stupid question: why is this in university math? I first saw this in the 5th grade of high school? (just couldn't remember the formula xd))
You may well have learned the inverse of a 2 by 2 matrix in the fifth grade. I doubt you learned to find the inverse of a 3 by 3 matrix in the fifth grade. That may be why you "forgot" the formula!

6. Great Explanation! Thank you very much!

7. Hello, shinhidora!

HallsofIvy gave an excellent explanation.
Here it is without words . . .

Solve: . $A^{-1} \:=\: \left(\begin{array}{ccc}1&2&3\\2&3&4\\3&4&6\end{ar ray}\right)$

We have: . $\left[\begin{array}{ccc|ccc}
1&2&3 &1&0&0 \\ 2&3&4 & 0&1&0 \\ 3&4&6 & 0&0&1\end{array}\right]$

$\begin{array}{c} \\ R_2 - 2R_1 \\ R_3 - 3R_1\end{array} \left[\begin{array}{ccc|ccc}1&2&3 & 1&0&0 \\ 0&\text{-}1&\text{-}2 & \text{-}2 &1&0 \\ 0&\text{-}2&\text{-}3 & \text{-}3&0& 1 \end{array}\right]$

. . $\begin{array}{c}\\ \text{-}1\!\cdot\! R_2 \\ \\ \end{array} \left[\begin{array}{ccc|ccc} 1&2&3 & 1&0&0 \\ 0&1&2 & 2&\text{-}1&0 \\ 0&\text{-}2&\text{-}3 & \text{-}3&0&1 \end{array}\right]$

$\begin{array}{c}R_1-2R_2 \\ \\ R_3+2R_2 \end{array} \left[\begin{array}{ccc|ccc}1&0&\text{-}1 & \text{-}3&2&0 \\ 0&1&2 & 2&\text{-}1&0 \\ 0&0&1 & 1&\text{-}2&1 \end{array}\right]$

$\begin{array}{c}R_1+R_3 \\ R_2-2R_3 \\ \\ \end{array} \left[\begin{array}{ccc|ccc}1&0&0 & \text{-}2&0&1 \\ 0&1&0 & 0&3&\text{-}2 \\ 0&0&1 & 1&\text{-}2&1 \end{array}\right]$

Therefore: . $A^{-1} \;=\;\left[\begin{array}{ccc}\text{-}2&0&1 \\ 0&3&\text{-}2 \\ 1&\text{-}2&1\end{array}\right]$

8. Said it before but, I so need you people to replace my math teacher