2. Consider the following matrices, determine which of the...

2. Consider the following matrices, determine which of the following products can be calculated, and if so, find their product. $\begin{pmatrix} 2 & -3 & 6 \\ -1 & 2 & 4 \end{pmatrix}$ $A =$ $\vec{x} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}$ $B = \begin{pmatrix} 2 & -1 \\ 3 & -2 \end{pmatrix}$ (a) $A\vec{x}$ (c) $A^T\vec{x}$ (e) $BA$ (b) $B\vec{x}$ (d) $AB$ (f) $\vec{x}\vec{x}^T$ (g) $\vec{x}^T\vec{x}$ (h) $A^TB$ (i) $A^TB^T$ (j) $B^TA^T$

Transcript

00:01 Hey there.

00:02 So we have been given six multiplication of matrices a, c and b in six orders.

00:09 So we have to determine whether any of it is performable or not.

00:13 And then if it is, then we will perform it.

00:15 Okay.

00:16 So the process is to identify all the orders and then as our rule, we will see whether we can multiply it or not.

00:27 Okay.

00:27 So, a is of order 2 by 4, then b is of order 1 by 4, and then c is of order 4 by 1.

00:47 You can perform, you cannot perform a .b, so there is no way we could perform this whole multiplication if you go in this way.

00:59 But you can perform bc if you go in the opposite way, but the resultant cannot be multiplied with a.

01:08 Okay as because the matrix which will come after b c multiplication will be a one by one matrix and that matrix cannot be multiplied with a 2 by 4 matrix okay so this is not possible now for ac b again it is 2 by 4 c is 4 by 1 and b is 1 by 4 you can easily see that ac is possible as because these 2 are equal then cb is also possible because they're row and column and row are equal so this is possible now for b a c we have 4 by 1 sorry that would be 4 by 1 is for c is 4 by 1 b is 1 by 4 okay b is 1 by 4 so we cannot perform b a we cannot perform b we cannot perform because they are column and row doesn't match and then we cannot if we perform a and c we cannot perform it with b so in either way you cannot perform this entire operation for b c a let us see b is of the order 1 by 4 c is 4 by 1 and a is 2 by so here you can perform bc but the resultant matrix would be a one by one matrix.

03:03 That matrix could not be multiplied with this a because it is a 2 by 4 matrix.

03:10 So this bca is also not possible.

03:13 Now let us come to c ab.

03:16 So the c is 4 by 1, a is 2 by 4 and b is 1 by 4.

03:23 So there is no way you can multiply any of this because they are not equal.

03:29 So c -a -b is not also possible.

03:33 Now for c -b -a, c is 4x1, b is 1 by 4, and a is 2x.

03:45 You can perform c -b as because the column of c is equal to the row of b...