• Oct 10th 2006, 01:46 PM
Greg
I am really lost with this could anyway show me with working out how to solve

A=(2 1 3)
(1 0-6)
(4 2 7)

B=(3 2 3 )
( 4 6 5)

C=(1 -1 4)
(2 5 -3)
(-3 4 0)

D=(4 0 -3)
( 3-3 4)

E= (3 7)
(1-6)
(0 4)

Evaluate where possible.

1) 3A
2) 2D+3B
3)3A-2C
4)AB

I know it is probably very easy but if I can see the working out I am sure I can understand it.
• Oct 10th 2006, 01:59 PM
topsquark
Quote:

Originally Posted by Greg
I am really lost with this could anyway show me with working out how to solve

A=(2 1 3)
(1 0-6)
(4 2 7)

B=(3 2 3 )
( 4 6 5)

C=(1 -1 4)
(2 5 -3)
(-3 4 0)

D=(4 0 -3)
( 3-3 4)

E= (3 7)
(1-6)
(0 4)

Evaluate where possible.

1) 3A
2) 2D+3B
3)3A-2C
4)AB

I know it is probably very easy but if I can see the working out I am sure I can understand it.

1 - 3 are fairly straightforward to see how to do.

For the first, simply multiply each component of the matrix by 3.

For the second, calculate 2D and 3B. Now, when you add two matrices, you simply add each component. Ex:

(8 0 -6) = 2D
(6 -6 8)

(9 6 9 )
(12 18 15) = 3B

(8+9 0+6 -6+9)
(6+12 -6+18 8+15) = 2D + 3B

Problem 3 is done the same way.

-Dan
• Oct 10th 2006, 02:06 PM
CaptainBlack
Quote:

Originally Posted by Greg
I am really lost with this could anyway show me with working out how to solve

A=(2 1 3)
(1 0-6)
(4 2 7)

B=(3 2 3 )
( 4 6 5)

C=(1 -1 4)
(2 5 -3)
(-3 4 0)

D=(4 0 -3)
( 3-3 4)

E= (3 7)
(1-6)
(0 4)

Evaluate where possible.

1) 3A

This is just the matrix A with all of its elements multiplied by 3

Quote:

2) 2D+3B
As D and B have the same dimensions (2x3) this can be done and
each element of the resulting matrix is 2 times the corresponding element
of D plus 3 times the corresponding element of B

Quote:

3)3A-2C
Yes can be done as A and C have the same dimensions (3x3), and
each element of the resulting matrix is 3 times the corresponding element
of A minus 2 times the corresponding element of C

Quote:

4)AB
Can't be done as A and B are non-comformant. That is the number of rows
of B is not the same as the number of columns of A, which is a necessary
condition for the matrix product to be defined.

RonL
• Oct 10th 2006, 02:11 PM
topsquark
Quote:

Originally Posted by Greg
A=(2 1 3)
(1 0-6)
(4 2 7)

B=(3 2 3 )
( 4 6 5)

E= (3 7)
(1-6)
(0 4)

Find AB

This problem cannot be done. Let me demonstrate how to do the problem by calculating AE.

The general rule can be described as follows: we are going to multiply the rows of matrix A by the columns of matrix E in the following manner:

The (1,1) component of AE is going to be
(2 1 3) x (3 1 0) = 2*3 + 1*1 + 3*0 = 7

The (1,2) component of AE is going to be
(2 1 3) x (7 -6 4) = 2*7 + 1*-6 + 3*4 = 20

Now, E has no more rows to use so we are now done with the first row of AE.

The (2,1) component of AE is going to be
(1 0-6) x (3 1 0) = 1*3 + 0*1 + -6*0 = 3

etc. until we run out of rows in A. We finally get:

(7 20)
(3 -17) = AE
(14 44)

Note that, in order to make a sensible answer at the end that the number of columns of A has to be equal to the number of rows of E. This is why we can't do AB: A has 3 columns and B only has 2 rows.

-Dan
• Oct 10th 2006, 02:13 PM
AfterShock
A is a 3 x 3 matrix; B is a 2 x 3 matrix.

In order to be able to multiply matrices together, the columns of the first matrix HAS to equal the rows of the second matrix.

Since 3 does not equal 2, it can not be multiplied. Note, too, that AB does not equal BA.

One more useful tool is if the columns of the first does equal the rows of the second, that is, you can multiply the matrices, the answer will the number of rows the first matrix has by the number of columns the second matrix has.
• Oct 10th 2006, 08:31 PM
nak
I think It is done like this
Matrices can be multiplied only where the number of rows=number of columns in any two given matrices. If this is not the case then two matrices cannot be multiplied. e.g. u say evaluate 3A where

A=(2 1 3)
(1 0-6)
(4 2 7)

Just multiply each element of A by 3 hence 3A=(6 3 9)
(3 0 -18)
(12 6 21)