19.7 - Non-commutativity of Multiplication of Matrices

In the previous section, we saw the basics about multiplication of matrices. In this section, we will see that, multiplication of matrices is not commutative.

This can be explained in 12 steps:
1. If AB = BA for all A and B, we can say that, multiplication is commutative.
2. If we can show atleast one example where AB is not equal to BA, we will be able to say that, multiplication of matrices is not commutative.
3. Let A = $\left[\begin{array}{r}           
3    &{1}    &{-7}    \\
-5    &{10}    &{4}    \\
\end{array}\right]           
$ and B = $\left[\begin{array}{r}       
12    &{4}    \\
2    &{3}    \\
6    &{8}    \\
\end{array}\right]       
$.
• A is of the order 2×3
• B is of the order 3×2
• So both AB and BA are defined.
4. AB can be calculated as shown below:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{AB}    & {~=~}    &{\left[\begin{array}{r}


3
&{1}
&{-7}
\\ -5
&{10}
&{4}
\\ \end{array}\right]


\left[\begin{array}{r}

12
&{4}
\\ 2
&{3}
\\ 6
&{8}
\\ \end{array}\right]

}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\begin{array}{r}

3(12)+1(2)+(-7)(6)
&{3(4)+1(3)+(-7)(8)}
\\ (-5)(12)+10(2)+4(6)
&{(-5)(4)+10(3)+4(8)}
\\ \end{array}\right]

}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\left[\begin{array}{r}

-4
&{-41}
\\ -16
&{42}
\\ \end{array}\right]

}    \\
\end{array}$                           

5. Also, BA can be calculated as shown below:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{BA}    & {~=~}    &{\left[\begin{array}{r} 12&{4}\\ 2&{3}\\ 6&{8}\\ \end{array}\right]\left[\begin{array}{r} 3&{1}&{-7}\\ -5&{10}&{4}\\ \end{array}\right]}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\begin{array}{r}


12(3)+4(-5)
&{12(1)+4(10)}
&{12(-7)+4(4)}
\\ 2(3)+3(-5)
&{2(1)+3(10)}
&{2(-7)+3(4)}
\\ 6(3)+8(-5)
&{6(1)+8(10)}
&{6(-7)+8(4)}
\\ \end{array}\right]


}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\left[\begin{array}{r}


16
&{52}
&{-68}
\\ -9
&{32}
&{-2}
\\ -22
&{86}
&{-10}
\\ \end{array}\right]


}    \\
\end{array}$                           

6. We see that AB ≠ BA.
• Here, AB is of the order 2×2 and BA is of the order 3×3. So AB will never be equal to BA.
7. There may be some cases where AB and BA are of the same order. Let us see such an example:
8. Let A = $\left[\begin{array}{r}           
1    &{0}      \\
0    &{-1}   \\
\end{array}\right]           
$ and B =  $\left[\begin{array}{r}       
0    &{1}    \\
1    &{0}    \\
\end{array}\right]       
$.
9. Then AB can be calculated as shown below:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{AB}    & {~=~}    &{\left[\begin{array}{r}           
1    &{0}      \\
0    &{-1}   \\
\end{array}\right]
\left[\begin{array}{r}       
0    &{1}    \\
1    &{0}    \\
\end{array}\right] }    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\begin{array}{r}

1(0)+0(1)
&{1(1)+0(0)}
\\ (0)(0)+(-1)(1)
&{0(1)+(-1)(0)}
\\ \end{array}\right]

}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\left[\begin{array}{r}

0
&{1}
\\ -1
&{0}
\\ \end{array}\right]

}    \\
\end{array}$   
10. Also, BA can be calculated as shown below:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{BA}    & {~=~}    &{
\left[\begin{array}{r}       
0    &{1}    \\
1    &{0}    \\
\end{array}\right] \left[\begin{array}{r}           
1    &{0}      \\
0    &{-1}   \\
\end{array}\right]}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\begin{array}{r}

0(1)+1(0)
&{0(0)+1(-1)}
\\ 1(1)+0(0)
&{1(0)+0(-1)}
\\ \end{array}\right]

}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\left[\begin{array}{r}

0
&{-1}
\\ 1
&{0}
\\ \end{array}\right]

}    \\
\end{array}$
11. Here, AB is of the order 2×2 and BA is also of the order 2×2. We see that AB ≠ BA
12. We can write:
Even if AB and BA are of the same order, we can show at least one example where AB ≠ BA. So multiplication of matrices is not commutative.

---

If A and B are both diagonal matrices, and of the same order, then AB will be equal to BA. Two examples are shown below:

Fig.19.23

• The reader may write the calculation steps of the above examples in his/her notebooks.

---

Zero matrix as a product of two non zero matrices

• If a and b are two real numbers and ab=0, then a or b has to be zero.
• But this is not applicable for matrices. An example is shown below:

$\left[\begin{array}{r}       
0    &{-1}    \\
0    &{2}    \\
\end{array}\right]\left[\begin{array}{r}       
3    &{5}    \\
0    &{0}    \\
\end{array}\right] = \left[\begin{array}{r}       
0    &{0}    \\
0    &{0}    \\
\end{array}\right]   
$                           

• So we can write:
If the product of two matrices is a zero matrix, it is not necessary that, one of the matrices is a zero matrix.

---

In the next section, we will see properties of multiplication of matrices. 

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