+0  
 
0
326
2
avatar+266 

 

 

 

?

Paypay  Jan 15, 2016

Best Answer 

 #1
avatar+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

 

 

 

cool cool cool

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

2+0 Answers

 #1
avatar+78618 
+20
Best Answer

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

 

 

 

cool cool cool

CPhill  Jan 15, 2016
edited by CPhill  Jan 15, 2016
 #2
avatar+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.

 

 

laugh

heureka  Jan 15, 2016

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