Find the Product of Two Matrices - Pre-Calculus

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

We know that to be able to have the product of the 2 matrices, the size of the column of A must equal the size of the row of B. This gives :

.

Solving for n, we find

Since n is a natural number is the only possible solution.

Note that every entry of the product matrix is the sum of ( times) .

This gives as every entry of the product of the two matrices.

Note that when we multiply the first row by the first colum we get: ( times), this gives the value of .

All other rows are zeros, and therefore we have zeros in the other entries.

Note that multiplying every row of by the first column of gives .

Mutiplying every row of by the second column of gives .

Now the remaining columns are columns of zeros, and therefore this product gives zero in every row-column product.

Knowing these three aspects we get the resulting matrix.

The first matrix is (4x1) and the second matrix is (1x3). We can perform the matrix multiplication in this case. The resulting matrix is (4x3).

The first entry in the formed matrix is on the first row and the first column.

It is coming from the product of the first row of A and the first column of B.

This gives .We continue in this fashion.

The entry (4,3) is coming from the 4th row of A and the 3rd column of B.

This gives . To obtain the whole matrix we need to remember that any entry on AB say(i,j) is coming from the product of the rom i from A and the column j of B.

After doing all these calculations we obtain:

We note first that A is 4x4 , B is 4x1.

To be able to do BA the number of columns of B must equal the number of rows

of A.

Since the number of columns of B is 1 and the number of rows of A is 4, we do not have equality and therefore we can't have the product BA.

Since we are assuming that the two matrices 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 from A to the entry from B:

Since the entries from A are the same and given by ln(2) and the entries from B are the same and given by ln(3), we add these two to obtain :

ln(2)+ln(3) and by the properties of the logarithm we have ln(2)+ln(3)=ln(2x3)=ln(6).

Therefore our matrix is given by:

The number of columns of A is equal to the number of rows of B. Therefore we can perform this operation.

Any entry of the matrix product is the result of the sum of the product of the elements of the row of A with the colum of of B. To obtain the first entry of the matrix product, we use the the first row of A and the first column of B, multiplying componentwise and adding. Doing this operation for each entry,

we obtain our matrix:

We can treat i as a scalar. To do this multiplication all we need to do is to multply each entry of the matrix by i.

We see that when we multiply each entry of the matrix by i, we obtain and we know that .

This means that all the entries are equal to same scalar . Now placing all these scalar in the matrix entries we obtain:

To find the product, line up the rows of the left matrix individually with the one column in the right matrix:

The size of every matrix can be written in the form rows x cols. The following matrix is of the size 2 x 1 because it has 2 rows and 1 column.

For two matrices to be able to be multiplied, their sizes must line up that the number of columns in the first matrix is equal to the number of rows of the first matrix. For example:

So these two matrices can be multiplied. However, if the case were such that:

Here, the # of columns in the first matrix does not line up with the # of rows in the second matrix, so the two matrices cannot be multiplied.

First, note that the order of the matrix multiplication is important . Multiplication of two matrices is possible only if the number of columns of the first matrix is equal to the number of rows of the second matrix . Both and are matrices (2 rows and 2 columns, respectively). Thus, is possible since the number of columns of (2) equals to the number of rows of (2). Furthermore, the size of is equal to the number of rows of and the number of columns of .

To avoid confusion, I will use the notation , , and to denote the constituents of matrices , , and , respectively. For example, refers to the constituent in that is in row 1 and column 2. The general version of the three matrices are shown below:

Using the rules of multiplying two matrices, the definition of is shown below:

Thus,