Multiplication of Matrix

Adding two matrices is simple: you add the entries that sit in the same place. Multiplying two matrices works differently. Each ROW of the first matrix is combined with each COLUMN of the second matrix. You multiply the matching entries in pairs, first with first and second with second, and then add the products. That single number becomes one entry of the answer.

Take the matrices \( A = \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & -3 \\ -2 & -5 \end{bmatrix} \). Each has 2 rows and 2 columns, so each is of order \( 2 \times 2 \). The product of A and B is written AB, with the two matrices placed side by side:

\[ AB = \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 \\ -2 & -5 \end{bmatrix} \]

The answer will also be a \( 2 \times 2 \) matrix, so there are four entries to find. Each entry comes from one row of A and one column of B.

Step 1: First row of A times first column of B. The first row of A is 3, 4 and the first column of B is 1, \( -2 \). Multiplying first with first and second with second, then adding:

\[ 3 \times 1 + 4 \times (-2) = 3 - 8 = -5 \]

Step 2: First row of A times second column of B, which is \( -3 \), \( -5 \). This entry goes in row 1, column 2:

\[ 3 \times (-3) + 4 \times (-5) = -9 - 20 = -29 \]

Step 3: Second row of A, which is 5, 6, times first column of B. This entry goes in row 2, column 1:

\[ 5 \times 1 + 6 \times (-2) = 5 - 12 = -7 \]

Step 4: Second row of A times second column of B. This entry goes in row 2, column 2:

\[ 5 \times (-3) + 6 \times (-5) = -15 - 30 = -45 \]

Step 5: Placing the four entries in their positions:

\[ AB = \begin{bmatrix} -5 & -29 \\ -7 & -45 \end{bmatrix}_{2 \times 2} \]

Now reverse the order and find BA. This time the rows of B are combined with the columns of A.

Step 1: Row 1 of B is 1, \( -3 \). Multiplying it by the first column of A, which is 3, 5:

\[ 1 \times 3 + (-3) \times 5 = 3 - 15 = -12 \]

Step 2: Row 1 of B times the second column of A, which is 4, 6:

\[ 1 \times 4 + (-3) \times 6 = 4 - 18 = -14 \]

Step 3: Row 2 of B is \( -2 \), \( -5 \). Multiplying it by the first column of A:

\[ (-2) \times 3 + (-5) \times 5 = -6 - 25 = -31 \]

Step 4: Row 2 of B times the second column of A:

\[ (-2) \times 4 + (-5) \times 6 = -8 - 30 = -38 \]

Step 5: Collecting the four entries:

\[ BA = \begin{bmatrix} -12 & -14 \\ -31 & -38 \end{bmatrix} \]

Compare this with \( AB = \begin{bmatrix} -5 & -29 \\ -7 & -45 \end{bmatrix} \). The two products are different. With ordinary numbers, \( 3 \times 4 \) and \( 4 \times 3 \) are always equal, but with matrices the order of multiplication can change the answer completely. So whenever you multiply matrices, keep careful track of which matrix comes first.

In AB, every row of A must be paired, entry by entry, with a column of B. A row of A has as many entries as A has columns. A column of B has as many entries as B has rows. If these two numbers are not the same, some entries are left without a partner and the product cannot be made.

When the condition holds, the order of the product is easy to predict. If A is of order \( m \times n \) and B is of order \( n \times p \), then AB is of order \( m \times p \). In symbols:

\[ A_{m \times n} \times B_{n \times p} = (AB)_{m \times p} \]

A quick way to use this rule: write the two orders side by side. The two numbers in the middle must be equal, and the two numbers on the outside give the order of the answer.

Suppose A is \( m \times n \) and B is \( n \times p \), so AB can be found. Can BA be found as well? In BA the matrix B comes first. B has p columns and A has m rows, so BA exists only when \( p = m \). Even then, BA has order \( n \times n \), which may be different from the order \( m \times m \) of AB. So knowing that AB exists tells you nothing certain about BA.

Ordinary numbers obey several laws of multiplication. Three of these laws carry over to matrices, and one of them does not. Each law below is checked on actual matrices, so you can see it working instead of taking it on trust.

You have already seen this happen. With \( A = \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & -3 \\ -2 & -5 \end{bmatrix} \), AB came out as \( \begin{bmatrix} -5 & -29 \\ -7 & -45 \end{bmatrix} \) while BA came out as \( \begin{bmatrix} -12 & -14 \\ -31 & -38 \end{bmatrix} \). A few special pairs, such as a square matrix and the identity matrix, do give AB equal to BA, but you can never assume it.

Check it with \( A = \begin{bmatrix} 2 & 0 \\ 3 & 1 \end{bmatrix} \), \( B = \begin{bmatrix} 4 & -1 \\ -2 & 5 \end{bmatrix} \) and \( C = \begin{bmatrix} 1 & 3 \\ 6 & 0 \end{bmatrix} \). Each matrix below shows the products already worked out and waiting to be added. For example, the top left entry of AB is \( 2 \times 4 + 0 \times (-2) = 8 + 0 \).

Step 1: Multiplying A by B, row by column:

\[ \begin{aligned} AB &= \begin{bmatrix} 8 + 0 & -2 + 0 \\ 12 - 2 & -3 + 5 \end{bmatrix} \\ &= \begin{bmatrix} 8 & -2 \\ 10 & 2 \end{bmatrix} \end{aligned} \]

Step 2: Multiplying the result AB by C:

\[ \begin{aligned} (AB)C &= \begin{bmatrix} 8 - 12 & 24 + 0 \\ 10 + 12 & 30 + 0 \end{bmatrix} \\ &= \begin{bmatrix} -4 & 24 \\ 22 & 30 \end{bmatrix} \end{aligned} \]

Step 3: Now the other grouping. Multiplying B by C first:

\[ \begin{aligned} BC &= \begin{bmatrix} 4 - 6 & 12 - 0 \\ -2 + 30 & -6 + 0 \end{bmatrix} \\ &= \begin{bmatrix} -2 & 12 \\ 28 & -6 \end{bmatrix} \end{aligned} \]

Step 4: Multiplying A by the result BC:

\[ \begin{aligned} A(BC) &= \begin{bmatrix} -4 + 0 & 24 + 0 \\ -6 + 28 & 36 - 6 \end{bmatrix} \\ &= \begin{bmatrix} -4 & 24 \\ 22 & 30 \end{bmatrix} \end{aligned} \]

Steps 2 and 4 give the same matrix, so \( (AB)C = A(BC) \).

Check it with \( A = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 4 \\ 2 & 5 \end{bmatrix} \) and \( C = \begin{bmatrix} -1 & -3 \\ 6 & 0 \end{bmatrix} \).

Step 1: Adding B and C, entry by entry:

\[ \begin{aligned} B + C &= \begin{bmatrix} 1 - 1 & 4 - 3 \\ 2 + 6 & 5 + 0 \end{bmatrix} \\ &= \begin{bmatrix} 0 & 1 \\ 8 & 5 \end{bmatrix} \end{aligned} \]

Step 2: Multiplying A by (B + C). The top left entry, for example, is \( 1 \times 0 + 2 \times 8 = 0 + 16 \):

\[ \begin{aligned} A(B + C) &= \begin{bmatrix} 0 + 16 & 1 + 10 \\ 0 + 24 & 0 + 15 \end{bmatrix} \\ &= \begin{bmatrix} 16 & 11 \\ 24 & 15 \end{bmatrix} \end{aligned} \]

Step 3: Multiplying A by B:

\[ \begin{aligned} AB &= \begin{bmatrix} 1 + 4 & 4 + 10 \\ 0 + 6 & 0 + 15 \end{bmatrix} \\ &= \begin{bmatrix} 5 & 14 \\ 6 & 15 \end{bmatrix} \end{aligned} \]

Step 4: Multiplying A by C:

\[ \begin{aligned} AC &= \begin{bmatrix} -1 + 12 & -3 + 0 \\ 0 + 18 & 0 + 0 \end{bmatrix} \\ &= \begin{bmatrix} 11 & -3 \\ 18 & 0 \end{bmatrix} \end{aligned} \]

Step 5: Adding AB and AC:

\[ \begin{aligned} AB + AC &= \begin{bmatrix} 5 + 11 & 14 - 3 \\ 6 + 18 & 15 + 0 \end{bmatrix} \\ &= \begin{bmatrix} 16 & 11 \\ 24 & 15 \end{bmatrix} \end{aligned} \]

Steps 2 and 5 give the same matrix, so \( A(B + C) = AB + AC \).

The \( 2 \times 2 \) identity matrix is \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), with 1 on the leading diagonal and 0 everywhere else. Take \( A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \).

Step 1: Multiplying A by I. The top left entry is \( 2 \times 1 + 3 \times 0 = 2 + 0 \):

\[ \begin{aligned} AI &= \begin{bmatrix} 2 + 0 & 0 + 3 \\ 1 + 0 & 0 + 4 \end{bmatrix} \\ &= \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} = A \end{aligned} \]

Step 2: Multiplying I by A. The top left entry is \( 1 \times 2 + 0 \times 1 = 2 + 0 \):

\[ \begin{aligned} IA &= \begin{bmatrix} 2 + 0 & 3 + 0 \\ 0 + 1 & 0 + 4 \end{bmatrix} \\ &= \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} = A \end{aligned} \]

Both products give back A, so \( AI = IA = A \). This is one of the special cases in which changing the order does not change the answer. Here are the four properties side by side:

Let \( A = \begin{bmatrix} 2 & 4 & 0 \\ 3 & 5 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} -1 & 4 \\ 8 & 2 \\ 3 & 7 \end{bmatrix} \). Decide whether AB and BA are defined, giving a reason, and find each one that is.

A has 2 rows and 3 columns, so its order is \( 2 \times 3 \). B has 3 rows and 2 columns, so its order is \( 3 \times 2 \).

For AB: the number of columns of A is 3 and the number of rows of B is 3. They are equal, so AB is defined, and its order will be \( 2 \times 2 \).

\[ AB = \begin{bmatrix} 2 & 4 & 0 \\ 3 & 5 & 1 \end{bmatrix} \begin{bmatrix} -1 & 4 \\ 8 & 2 \\ 3 & 7 \end{bmatrix} \]

Step 1: Row 1 of A times column 1 of B. There are now three pairs to multiply before adding:

\[ \begin{aligned} & 2 \times (-1) + 4 \times 8 + 0 \times 3 \\ &= -2 + 32 + 0 = 30 \end{aligned} \]

Step 2: Row 1 of A times column 2 of B:

\[ \begin{aligned} & 2 \times 4 + 4 \times 2 + 0 \times 7 \\ &= 8 + 8 + 0 = 16 \end{aligned} \]

Step 3: Row 2 of A times column 1 of B:

\[ \begin{aligned} & 3 \times (-1) + 5 \times 8 + 1 \times 3 \\ &= -3 + 40 + 3 = 40 \end{aligned} \]

Step 4: Row 2 of A times column 2 of B:

\[ \begin{aligned} & 3 \times 4 + 5 \times 2 + 1 \times 7 \\ &= 12 + 10 + 7 = 29 \end{aligned} \]

Step 5: Placing the entries:

\[ AB = \begin{bmatrix} 30 & 16 \\ 40 & 29 \end{bmatrix} \]

For BA: the number of columns of B is 2 and the number of rows of A is 2. They are equal, so BA is also defined. Its order will be \( 3 \times 3 \), so there are nine entries to find.

\[ BA = \begin{bmatrix} -1 & 4 \\ 8 & 2 \\ 3 & 7 \end{bmatrix} \begin{bmatrix} 2 & 4 & 0 \\ 3 & 5 & 1 \end{bmatrix} \]

Step 1: Row 1 of B is \( -1 \), 4. Multiplying it by each of the three columns of A gives the first row of BA:

\[ \begin{aligned} (-1) \times 2 + 4 \times 3 &= 10 \\ (-1) \times 4 + 4 \times 5 &= 16 \\ (-1) \times 0 + 4 \times 1 &= 4 \end{aligned} \]

Step 2: Row 2 of B is 8, 2. Multiplying it by the three columns of A gives the second row:

\[ \begin{aligned} 8 \times 2 + 2 \times 3 &= 22 \\ 8 \times 4 + 2 \times 5 &= 42 \\ 8 \times 0 + 2 \times 1 &= 2 \end{aligned} \]

Step 3: Row 3 of B is 3, 7. Multiplying it by the three columns of A gives the third row:

\[ \begin{aligned} 3 \times 2 + 7 \times 3 &= 27 \\ 3 \times 4 + 7 \times 5 &= 47 \\ 3 \times 0 + 7 \times 1 &= 7 \end{aligned} \]

Step 4: Writing the three rows together:

\[ BA = \begin{bmatrix} 10 & 16 & 4 \\ 22 & 42 & 2 \\ 27 & 47 & 7 \end{bmatrix} \]

AB is a \( 2 \times 2 \) matrix while BA is a \( 3 \times 3 \) matrix. Both products exist, yet they are not even the same size, which is the clearest possible sign that AB and BA are different.

If \( A = \begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix} \), prove that \( A^2 - 6A + 9I = O \), where I and O are the identity matrix and the zero matrix of order \( 2 \times 2 \).

Here \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) and \( O = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).

Step 1: Finding \( A^2 \) by multiplying A by A. The top left entry is \( 4 \times 4 + 1 \times (-1) = 16 - 1 \), and the others are found the same way:

\[ \begin{aligned} A^2 &= \begin{bmatrix} 16 - 1 & 4 + 2 \\ -4 - 2 & -1 + 4 \end{bmatrix} \\ &= \begin{bmatrix} 15 & 6 \\ -6 & 3 \end{bmatrix} \end{aligned} \]

Step 2: Multiplying A by the number 6, which multiplies every entry by 6:

\[ 6A = \begin{bmatrix} 24 & 6 \\ -6 & 12 \end{bmatrix} \]

Step 3: Multiplying I by the number 9:

\[ 9I = \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} \]

Step 4: Putting these into the left hand side and combining the entries in each position:

\[ \begin{aligned} & A^2 - 6A + 9I \\ &= \begin{bmatrix} 15 - 24 + 9 & 6 - 6 + 0 \\ -6 + 6 + 0 & 3 - 12 + 9 \end{bmatrix} \\ &= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \end{aligned} \]

The left hand side equals O, which is the right hand side. Hence \( A^2 - 6A + 9I = O \) is proved.

If \( A = \begin{bmatrix} 4 & 0 \\ 0 & 5 \end{bmatrix} \), \( B = \begin{bmatrix} p & q \\ 0 & r \end{bmatrix} \) and \( AB = A + B \), find the values of p, q and r.

The plan is to work out each side as a single matrix. Two equal matrices have equal entries in the same positions, so matching entries give simple equations.

Step 1: Finding AB, row by column. The top left entry is \( 4 \times p + 0 \times 0 = 4p + 0 \):

\[ \begin{aligned} AB &= \begin{bmatrix} 4p + 0 & 4q + 0 \\ 0 + 0 & 0 + 5r \end{bmatrix} \\ &= \begin{bmatrix} 4p & 4q \\ 0 & 5r \end{bmatrix} \end{aligned} \]

Step 2: Finding A + B by adding the entries in the same positions:

\[ A + B = \begin{bmatrix} 4 + p & q \\ 0 & 5 + r \end{bmatrix} \]

Step 3: Since AB = A + B, the two matrices are equal:

\[ \begin{bmatrix} 4p & 4q \\ 0 & 5r \end{bmatrix} = \begin{bmatrix} 4 + p & q \\ 0 & 5 + r \end{bmatrix} \]

Step 4: Equating the entries in row 1, column 1 and solving for p:

\[ \begin{aligned} 4p &= 4 + p \\ 3p &= 4 \\ p &= \frac{4}{3} \end{aligned} \]

Step 5: Equating the entries in row 1, column 2 and solving for q:

\[ \begin{aligned} 4q &= q \\ 3q &= 0 \\ q &= 0 \end{aligned} \]

Step 6: Equating the entries in row 2, column 2 and solving for r:

\[ \begin{aligned} 5r &= 5 + r \\ 4r &= 5 \\ r &= \frac{5}{4} \end{aligned} \]

Hence \( p = \frac{4}{3} \), \( q = 0 \) and \( r = \frac{5}{4} \).

If \( A = \begin{bmatrix} 1 & 2 \\ 0 & 9 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & 4 \\ 2 & 1 \end{bmatrix} \), prove that \( (AB)^T = B^T.A^T \).

Step 1: Finding AB. The top left entry is \( 1 \times 3 + 2 \times 2 = 3 + 4 \):

\[ \begin{aligned} AB &= \begin{bmatrix} 3 + 4 & 4 + 2 \\ 0 + 18 & 0 + 9 \end{bmatrix} \\ &= \begin{bmatrix} 7 & 6 \\ 18 & 9 \end{bmatrix} \end{aligned} \]

Step 2: Taking the transpose of AB, which turns its rows into columns. Call this result (i):

\[ (AB)^T = \begin{bmatrix} 7 & 18 \\ 6 & 9 \end{bmatrix} \]

Step 3: Finding the transpose of A:

\[ A^T = \begin{bmatrix} 1 & 0 \\ 2 & 9 \end{bmatrix} \]

Step 4: Finding the transpose of B:

\[ B^T = \begin{bmatrix} 3 & 2 \\ 4 & 1 \end{bmatrix} \]

Step 5: Multiplying \( B^T \) by \( A^T \), in that order. The top left entry is \( 3 \times 1 + 2 \times 2 = 3 + 4 \). Call this result (ii):

\[ \begin{aligned} B^T.A^T &= \begin{bmatrix} 3 + 4 & 0 + 18 \\ 4 + 2 & 0 + 9 \end{bmatrix} \\ &= \begin{bmatrix} 7 & 18 \\ 6 & 9 \end{bmatrix} \end{aligned} \]

Results (i) and (ii) are the same matrix. Hence \( (AB)^T = B^T.A^T \) is proved.