Solve a 3x3 matrix times a 3x3 matrix?

Question

0 1 0 ||||||| -1 2 0
2 - 1 1 ||||||| 4 6 0
0 2 -1 ||||||| 1 0 1
Ty if you do it

Answer 1

Let's say you are multiplying two matrix A and B... condition for matrix multiplication is number of columns of A must be equal to number of rows of B... if it is true, then only you can doo multiplication operation and the resulting matrix will be of thiss dimension... number of rows of A and number of columns of B...
...............c1a..c2a..c3a
.......r1a.....0......1......0
A =..r2a.....2.....-1.....1
.......r3a.....0......2.....-1
...............c1b..c2b..c3b
.......r1b.....-1....2.......0
B = r2b......4....6........0
.......r3b......1....0.......1

A is a 3x3 matrix... 3 rows and 3 columns
and B is also a 3x3 matrix... [a matrix that has same number of rows and columns is called a square matrix]...

Since number of columns of A is equal to number of rows of B = 3, we can perform the matrix multiplication operation... and we also know the resulting matrix will be a 3x3 matrix because number of rows of A is 3 and so is number of columns of B...
____________c1ab____c2ab____c3ab
.............r1ab [ r1a*c1b....r1a*c2b....r1a*c3b]
A x B = r2ab [ r2a*c1b....r2a*c2b....r2a*c2b]
.............r3ab [ r3a*c1b....r3a*c2b....r3a*c3b]
note: x and asterik represents multiplication operation...

r1a*c1b = [0*-1 + 1*4 + 0*1] = 4
r1a*c2b = [0*2 + 1*6 + 0*0] = 6
r1a*c3b = [0*0 + 1*0 +0*1] = 0
r2a*c1b = [2*-1 + -1*4 + 1*1] = -5
r2a*c2b = [2*2 + -1*6 + 1*0] = -2
r2a*c3b = [2*0 + -1*0 + 1*1] = 1
r3a*c1b = [0*-1 + 2*4 + -1*1] = 7
r3a*c2b = [0*2 + 2*6 + -1*0] = 12
r3a*c3b = [0*0 + 2*0 + -1*1] = -1

.................[ 4....6....0 ]
A x B = ....[-5...-2....1 ]
.................[ 7...12...-1]

in the A x B matrix the one before the above one, where I have tried to name the columns as c1ab, c2ab... the third column is c3ab...

Answer 2

Let x[m,n] be the element in matrix x, row m, column n.
Call the three matrices a, b, and c

To work out c[1,1] you need to do (a[1,1]*b[1,1]) + (a[1,2]*b[2,1]) + ....

So in your case, c[1,1] is (0 * -1) + (1 * 4) + (0 * 1) = 4
c[2,1] is (2 * -1) + (-1 * 4) + (1 * 1) = -5

4 6 0
-5 -2 1
7 12 -1

note: matrix multiplication is only possible if the number of COLUMNS in the first matrix is equal to the number of ROWS in the second matrix. The answer will be a matrix with the same number of ROWS as the first matrix, and the same number of COLUMNS as the second matrix.

So you can multiply a 3 x 2 matrix by a 2 x 4 matrix which will give an answer which is a 3 x 4 matrix, but you can't multiply a 2 x 4 matrix by a 3 x 2 matrix

Answer 3

[0 1 0] [-1 2 0]
[2 -1 1] [4 6 0]
[0 2 -1] [1 0 1]

Using matrix multiplication:

[4 6 0]
[-5 -2 1]
[7 12 -1]

Answer 4

You know the rules then even you can do it.

4 6 0
-5 -2 1
7 12 -1

This is the answer I guess.

Answer 5

a[i,j] = row[i] * column[j]

Answer 6

Do it.