# matrix multiplication

• Feb 15th 2009, 05:55 AM
karimwahab
matrix multiplication
{ 3 2 } x { a b } = { 1 0 }
{ 1 -1 } { c d } { 0 1 }

So, what is:
{ a b } = { 1 0 } x Inv. { 3 2 }
{ c d } { 0 1 } { 1 -1 }

can any help me figure it out this matrix
its do to with "inverse"

• Feb 15th 2009, 08:07 PM
CaptainBlack
Quote:

Originally Posted by karimwahab
(3 2) + (a b) = (1 0)
(1 -1) (c d) (0 1)

can any help me figure it out this matrix
its do to with "inverse"

Why does this show a "+" sign?

CB
• Feb 16th 2009, 08:21 AM
Inverse of a 2x2 matrix
Hello karimwahab
Quote:

Originally Posted by karimwahab
{ 3 2 } x { a b } = { 1 0 }
{ 1 -1 } { c d } { 0 1 }

So, what is:

Quote:

Originally Posted by karimwahab
{ a b } = { 1 0 } x Inv. { 3 2 }
{ c d } { 0 1 } { 1 -1 }

can any help me figure it out this matrix
its do to with "inverse"

I'm not entirely sure what you mean by the part in red, but the first part is clear enough. It is:

$\displaystyle \begin{pmatrix}3&2\\1&-1 \end{pmatrix}\begin{pmatrix}a&b\\c&d \end{pmatrix}=\begin{pmatrix}1&0\\0&1 \end{pmatrix}$

Now the matrix $\displaystyle \begin{pmatrix}1&0\\0&1 \end{pmatrix}$ is called the Identity Matrix, usually denoted by $\displaystyle I$. This matrix has the special property that when any other 2x2 matrix is multiplied by it (either on the left or the right) then that matrix is unchanged - it keeps its identity. In other words, for any matrix $\displaystyle A$:

$\displaystyle A\times I = I\times A = A$

You then need to know that, if $\displaystyle A$ and $\displaystyle B$ are 2x2 matrices and $\displaystyle A\times B = I$, then $\displaystyle A$ and $\displaystyle B$ are inverses of each other. This is written $\displaystyle B= A^{-1}$ and $\displaystyle A = B^{-1}$.

So the matrix $\displaystyle \begin{pmatrix}a&b\\c&d \end{pmatrix}$ is the inverse of the matrix $\displaystyle \begin{pmatrix}3&2\\1&-1 \end{pmatrix}$. There's a formula for the inverse of a 2x2 matrix here: The inverse of a 2x2 matrix - mathcentre

If you use that formula, you'll find that:

$\displaystyle \begin{pmatrix}a&b\\c&d \end{pmatrix} = \frac{1}{5} \begin{pmatrix}1&2\\1&-3 \end{pmatrix}$

I hope that answers the question.