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# ​ Matrix. pt.3

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Paypay  Jan 15, 2016

#1
+78618
+20

A =    0  - 1   1            B =   1    2

-5   -1   0                     0   1

-2   -1

AB  will produce  a  (   [2] x 3 ) x ( 3 x [2] )  =    2 x 2 matrix

This is because if the columns of the first matrix = the rows of the second matrix, we can multiply them.....and the resultant matrix will have the number of rows of the first matrix and the number of columns of the second

element 1,1  = [ 0 * 1  +   -1* 0   + 1 * -2  ]   = -2

element 1,2  = [ 0*2 +  -1 * 1   + 1* -1]   =  -2

We can see now that H is the answer, but let's finish it

element 2, 1  =  [ -5 * 1 + -1 * 0  + 0 * -2] =   -5

element 2, 2 = [ -5 * 2 + -1 * 1 + 0 * -1 ]  =  -11

And the resulting matrix is   H  =

-2   -2

-5  -11

CPhill  Jan 15, 2016
edited by CPhill  Jan 15, 2016
Sort:

#1
+78618
+20

A =    0  - 1   1            B =   1    2

-5   -1   0                     0   1

-2   -1

AB  will produce  a  (   [2] x 3 ) x ( 3 x [2] )  =    2 x 2 matrix

This is because if the columns of the first matrix = the rows of the second matrix, we can multiply them.....and the resultant matrix will have the number of rows of the first matrix and the number of columns of the second

element 1,1  = [ 0 * 1  +   -1* 0   + 1 * -2  ]   = -2

element 1,2  = [ 0*2 +  -1 * 1   + 1* -1]   =  -2

We can see now that H is the answer, but let's finish it

element 2, 1  =  [ -5 * 1 + -1 * 0  + 0 * -2] =   -5

element 2, 2 = [ -5 * 2 + -1 * 1 + 0 * -1 ]  =  -11

And the resulting matrix is   H  =

-2   -2

-5  -11

CPhill  Jan 15, 2016
edited by CPhill  Jan 15, 2016
#2
+18712
+20

$$\begin{array}{rcl} \begin{bmatrix} 0&-1&1\\ -5&-1&0 \end{bmatrix}\cdot \begin{bmatrix} 1&2&\\ 0&1\\ -2&-1 \end{bmatrix} &=& \begin{bmatrix} \begin{pmatrix}0\\-1\\1\end{pmatrix}\cdot \begin{pmatrix}1\\0\\-2\end{pmatrix} & \begin{pmatrix}0\\-1\\1\end{pmatrix}\cdot \begin{pmatrix}2\\1\\-1\end{pmatrix}\\ \begin{pmatrix}-5\\-1\\-0\end{pmatrix}\cdot \begin{pmatrix}1\\0\\-2\end{pmatrix} & \begin{pmatrix}-5\\-1\\-0\end{pmatrix}\cdot \begin{pmatrix}2\\1\\-1\end{pmatrix}\\ \end{bmatrix}\\\\ &=& \begin{bmatrix} 0\cdot 1 +(-1)\cdot 0 + 1\cdot(-2) & 0\cdot 2 + (-1)\cdot 1 + 1\cdot (-1)\\ (-5)\cdot 1 +(-1)\cdot 0 + 0\cdot(-2) & (-5)\cdot 2 + (-1)\cdot 1 + 0\cdot (-1)\\ \end{bmatrix}\\\\ &=& \begin{bmatrix} 0+0-2 & 0-1-1\\ -5+ 0 + 0 & -10 -1 +0\\ \end{bmatrix}\\\\ &=& \begin{bmatrix} -2 & -2\\ -5 & -11\\ \end{bmatrix} \end{array}$$

Matrix H is the result.

heureka  Jan 15, 2016

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