Multiplication of Matrices - Pre-Calculus

When we multiply a scalar (regular number) by a matrix, all we need to do is mulitply it to every entry inside the matrix:

When we multiply a scalar (regular number) by a matrix, all we need to do is mulitply it through to every entry inside the matrix:

When we multiply a scalar (regular number) by a matrix, all we need to do is mulitply it through to every entry inside the matrix:

We note k is simply a scalar. To do this multiplication all we need to do is to multply each entry of the matrix by k.

we see that when we multiply we have :

this gives the entry of the matrix kA.

Therefore the resulting matrix is :

Since we are adding the matrix to itself, we have the same size, we can perform the matrices addition.

We know that when adding matrices, we add them componenwise. Let (i,j) be any entry of the addition matrix. We add the entry form A to the entry from B which is the same as A. This means that to add A+A we simply add each entry of A to itself.

Since the entries from A are the same and given by 1 and the entries from B=A are the same and given by 1, we add these two to obtain:

1+1 and this means that each entry of A+A is 2. We continue in this fashion by additing the entries of A each one to itself n times to obtain that the entries of A+A+....A( n times ) are given by:

We note n is simply a scalar. To do this multiplication all we need to do is to multply each entry of the matrix by n.

We see that when we multiply we have : .

This means that each entry of the resulting matrix is .

This gives the nA which is :

This problem involves a scalar multiplication with a matrix. Simply distribute the negative three and multiply this value with every number in the 2 by 3 matrix. The rows and columns will not change.

A scalar that multiplies a one by two matrix will result in a one by two matrix.

Multiply the scalar value with each value in the matrix.

Scalar multiplication and addition of matrices are both very easy. Just like regular scalar values, you do multiplication first:

The addition of matrices is very easy. You merely need to add them directly together, correlating the spaces directly.

You can begin by treating this equation just like it was:

That is, you can divide both sides by :

Now, for scalar multiplication of matrices, you merely need to multiply the scalar by each component:

Then, simplify:

Therefore,

Begin by distributing the fraction through the matrix on the left side of the equation. This will simplify the contents, given that they are factors of :

Now, this means that your equation looks like:

This simply means:

and

or

Therefore,

To multiply a scalar and a matrix, simly multiply each number in the matrix by the scalar. Thus,

To find the product of 2 matrices, first line up the first row of the left matrix with the first column of the right matrix. Multiply the first, second, and third entries and then add them together.

Next, line up the second row of the left matrix with the second column of the right matrix. Then the first row and the second column, and finally the second row and the first column:

The dimensions of A and B are as follows: A= 3x3, B= 3x1

When we mulitply two matrices, we need to keep in mind their dimensions (in this case 3x_3 and 3_x1).

The two inner numbers need to be the same. Otherwise, we cannot multiply them. The product's dimensions will be the two outer numbers: 3x1.

The dimensions of both A and B are 2x2. Therefore, the matrix that results from their product will have the same dimensions.

Thus plugging in our values for this particular problem we get the following:

The dimensions of A and B are as follows: A=1x3, B= 3x1.

Because the two inner numbers are the same, we can find the product.

The two outer numbers will tell us the dimensions of the product: 1x1.

Therefore, plugging in our values for this problem we get the following:

The dimensions of A and B are as follows: A= 3x1, B= 2x3

In order to be able to multiply matrices, the inner numbers need to be the same. In this case, they are 1 and 2. As such, we cannot find their product.

The answer is No Solution.

To have the above equality we need to have and .

means that , or . Trying all different values of , we see that no can satisfy both matrices.

Therefore there is no that satisfies the above equality.

We will use an induction proof to show this result.

We first note the above result holds for n=1. This means

We suppose that and we need to show that:

By definition . By inductive hypothesis, we have:

Therefore,

This shows that the result is true for n+1. By the principle of mathematical induction we have the result.

Note that:

Since .

This means that