How do I do 3 matrices? ?

Question

A= 0 1 0 B= 0 0 0

0 0 1 1 0 0

1 0 0 0 1 0

I have to find the products of AB and BA and determine whether B is the multiplicative inverse of A. I now know how to do 2 matrix but not 3. Someone just showed me how to do the 2 but I am having a problem doing the 3. I don't do which steps to follow.

Answer 1

Here's a big tip.  If A and B are inverses of each other, than AB = BA = I where I is the identity matrix.  This is like saying a * 1/a = 1/a * a = 1.  Do you know how to multiply 2 X 2 matrices?  Well 3 X 3 is just an extension.  Word of advice, when asking these questions; make sure your problem is readable.  To the untrained eye, this looks like a bunch of random numbers.  And this is super easy because of all the zeroes in your matrices.  If you did it right and the two matrices are in fact inverses, then your answer will be....

I = 1  0  0

     0  1  0

     0  0  1

Answer 2

A * B

[0..1..0][0..0..0]

[0..0..1][1..0..0]

[1..0..0][0..1..0]

To get the new matrix's first row first column, follow the first row of A multiplied like a dot product with the first column of B

[0..1..0][0]

...........[1]

...........[0]

0 * 0 + 1 * 1 + 0* 0 = 1

To get the new matrix's first row second column, follow the first row of A multiplied like a dot product with the second column of B

[0..1..0][0]

...........[0]

...........[1]

0 * 0 + 1 * 0 + 0* 1 = 0

To get the new matrix's first row third column, follow the first row of A multiplied like a dot product with the third column of B

[0..1..0][0]

...........[0]

...........[0]

0 * 0 + 1 * 0 + 0* 0 = 0

Do the same things for the second row of A and each of the columns of B to get the second row of the answer matrix

Do the same things for the third row of A and each of the columns of B to get the third row of the answer matrix

The Answer matrix for A*B should be

[1..0..0] ← This is the row I worked out in detail!

[0..1..0]

[0..0..0]

Now do the reverse.

B * A

[0..0..0][0..1..0]

[1..0..0][0..0..1]

[0..1..0][1..0..0]

The answer is

[0..0..0]

[0..1..0]

[0..0..1]

So no B is not the multiplicative inverse of A