Determining dimensions

If you want to multiply matrices, the number of columns in the first matrix has to equal the number of rows in the second matrix. After you multiply matrices, you get a whole new matrix that features the number of rows the first matrix had and the number of columns from the original second matrix. The process is sort of like cross-pollinating white and red petunias and getting pink.

In algebraic terms, if matrix A has the dimension and matrix B has the dimension , to multiply , n must equal p. The dimension of the resulting matrix will be , the number of rows in the first matrix and the number of columns in the second.

For example, if you multiply a matrix times a matrix, you get a matrix. However, you can’t multiply a matrix by a matrix. Further emphasizing that the order in which you multiply the matrices does indeed matter, you can multiply a matrix by a matrix and get a matrix.

Defining the process

Multiplying matrices is no easy matter, but it isn’t all that difficult — if you can multiply and add correctly. When you multiply two matrices, you compute the elements with the following rule:

If you find the element after multiplying matrix A times matrix B, is the sum of the products of the elements in the ith row of matrix A and the jth column of matrix B.

Okay, so that rule may sound like a lot of hocus pocus. How about I give you something a bit more concrete? Here’s a matrix A that has the dimension and a matrix B with the dimension . According to the rules that govern multiplying matrices, you can multiply A times B because the number of columns in matrix A is two, and the number of rows in matrix B is two. The matrix that you create when multiplying A and B has the dimension .

You find the element by multiplying the elements in the first row of A times the elements in the first column of B and then adding the products: . You find by multiplying the second row of matrix A times the third column of matrix B and adding: .

Although the last example was more concrete, it didn’t provide you with numbers. With the following matrices, you can multiply matrix J times matrix K because the number of columns in matrix J matches the number of rows in matrix K. You can also see the computations needed to find the resulting matrix.

You get a matrix when you multiply because multiplying a matrix times a matrix leaves you with two rows and two columns. The multiplication process may seem a bit complicated, but after you get the hang of it, you can do the multiplication and addition in your head.

Determining how many of each item was sold

The first question is: How many TVs, smartphones, and computers did the store sell during those three months? Because all the matrices J, F, and M have the same dimensions, you can add them together (see the section “Adding and subtracting matrices”). I next show you how you find the sales for the first three months of the year by adding the matrices (the totals of the individual salespeople).

The way to use matrices to find the total sales for each type of electronic, you multiply the sum matrix (the matrix you just found) by a row matrix, . You’re multiplying a matrix with the dimension times a matrix with the dimension , so your result is a matrix with the dimension ; the totals of each electronic appear in order from left to right. Think of the row matrix T with the salespersons’ names across the top. When you multiply this matrix by the sum matrix, you match each of the columns in T with each of the rows in the sum. The columns in T and the rows in the sum are the salespersons, so they line up. By multiplying all the numbers in the second matrix by 1’s, you essentially multiply by 100 percent — you add everything up. The resulting matrix is a row matrix with the numbers for the electronics in the respective columns. First, the matrix T with its labels for each element is given by

And the following shows the computation and result.

You’re wondering why you can’t just add the numbers in each column, aren’t you? Go right ahead! That works, too. I show you this process with a small, reasonable number of items so you can see how you can instruct a computer to do it with hundreds or thousands of entries. You may also want to “weight” some entries — make some items worth more than others — if you’re keeping track of points. The row matrix can have different entries that align with the different items.

Determining sales by salesperson

Here’s another question you can answer with the matrix sum : How much money did each salesperson bring in? For instance, assume that the average costs of a TV, smartphone, and computer are $1,500, $400, and $2,000, respectively. You can construct a column matrix containing these dollar amounts, call it the dollar matrix, and multiply it by the sum matrix .

The following matrix multiplication shows the multiplication of the sum matrix times the dollar matrix, with the resulting amount of money brought in by each salesperson.

Here are the results of multiplying the entries in the row of the first matrix by the columns of the second:

• A:
• B:
• C:
• D:

Determining how to increase sales

Here’s one last question that deals with a percentage: How many electronic devices in each category must each salesperson sell if they have to increase sales to 125 percent during the next quarter?

You identify this problem as a scalar multiplication problem (see the section “Multiplying matrices by scalars”) because you multiply each entry in the sum matrix by 125 percent to get the sales target. The value 1.25 represents 125 percent — 25 percent more than the last quarter. This next computation shows the result of the scalar multiplication and a second matrix with the numbers rounded up. (You can’t sell half a computer, and rounding up gives the salesperson the number he or she needs to make or exceed the goal [not come in slightly below].)

Identifying an inverse for any size square matrix

The general method you use to find a matrix’s inverse involves writing down the matrix, inserting an identity matrix, and then changing the original matrix into an identity matrix.

To solve for the inverse of a matrix, follow these steps:

1. Create a double matrix — consisting of the target matrix and the identity matrix of the same size — with the identity matrix to the right of the original.

2. Perform row operations until the elements on the left become an identity matrix (see the “Defining Row Operations” section earlier in the chapter).

   The elements created in the new left-hand matrix need to feature a diagonal of ones, with zeros above and below the ones. Upon completion of this step, the elements on the right become the elements of the inverse matrix.

For example, say you want to solve for the inverse of matrix M in the following figure. You first put the identity matrix to the right of the elements in matrix M. The goal is to make the elements on the left look like an identity matrix by using matrix row operations.

The identity matrix has a diagonal slash of ones, and zeros lie above and below the ones. The first thing you do to create your identity matrix is to get the zeros below the one in the upper left-hand corner of the original identity. Here are the row operations you use for this example:

1. Multiply Row 1 times 3 and add the result to Row 2, making the result the new Row 2.
2. Multiply Row 1 by   and add the result to Row 3, putting the resulting answer in Row 3.

The following shows what the matrix looks like after you perform the row operations. The notation on the arrow between the two matrices describes the row operations you use.

You see a one in the second column and second row of the resulting matrix — along the diagonal of ones that you want. Getting the one in this position is a nice coincidence; it doesn’t always work out as well. If the one hadn’t fallen in that position, you’d have to divide the whole row by whatever number makes that element a one.

Now, you want zeros above and below the 1 in the middle, so you follow these steps:

1. Multiply Row 2 by and add the result to Row 1.
2. Multiply Row 2 by 7 and add the result to Row 3.

You now need to turn the element in the third row, third column of the matrix into a 1, so you multiply the row through by ; this multiplication is the same as dividing through by .

You have one last set of row operations to perform to get zeros above the last one on the diagonal:

1. Multiply Row 3 by 8 and add the elements in the row to Row 1.
2. Multiply Row 3 by  and add the result to Row 2.

The following shows the final two steps and the result.

The operations gave you a identity matrix on the left and a new matrix on the right. The elements in the matrix on the right form the inverse matrix, . The product of M and its inverse, , is the identity matrix.

The following checks your work — showing that is the identity matrix:

How can you tell when a matrix doesn’t have an inverse? You can’t get ones along the diagonal by using the row operations. You usually get a whole row of zeros as a result of your row operations. A row of zeros is the warning that no inverse is possible.

Using a quick-and-slick rule for 2 × 2 matrices

You have a special rule at your disposal to find inverses of matrices. To implement the rule for a matrix, you have to switch two elements, negate two elements, and divide all the elements by the difference of the cross products of the elements. This may sound complicated, but the math is really neat and sweet, and the process is much quicker than the general method (see the previous section).

Here is the general formula for the matrix rule.

As you can see from this special rule, the upper-left and lower-right corner elements are switched; the upper-right corner and lower-left corner elements are negated (changed to the opposite sign); and all the elements are divided by the result of doing two cross products and subtracting.

To find the inverse of matrix Z using the method, you switch the 5 and the 11, you change the 6 to and the 9 to , and you divide by the difference of the cross products: . Watch the order in which you do the subtraction — the order does matter. Dividing each element by 1 doesn’t change the elements, as you can see.