Question

If $A = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$, then
A. ${A^2} = A$
B. ${B^2} = B$
C. $AB \ne BA$
D. $AB = BA$

Answer 1

Hint: In this question, we have to check that the two given matrices are satisfying which condition is given in the options. To solve these types of questions, we must know how to do the multiplication of the matrix.

Complete step by step Solution:
Given: $\;A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$ and $B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$
For checking whether the two given matrices are satisfying the condition of option A or not, that is, whether ${A^2} = A$ or not:
Step 1: To find ${A^2}$, we have to multiply $A$ with $A$. In order to do that, multiply and add each element of the first row of matrix $A$ with the respective elements of the first column of matrix $A$, as shown below:
$A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$ $A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$
Result for this will be, $R1 = \left[ {(1 \times 1) + (2 \times ( - 3))} \right]$
Step 2: Multiply and add each element of the first row of matrix $A$ with the respective elements of the second column of matrix $A$, as shown below:
$A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$ $A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$
Result for this will be, $R2 = \left[ {(1 \times 2) + (2 \times 0)} \right]$
Step 3: Multiply and add each element of the second row of matrix $A$ with the respective elements of the first column of matrix $A$, as shown below:
$A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$ $A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$
Result for this will be, $R3 = \left[ {(( - 3) \times 1) + (0 \times ( - 3))} \right]$
Step 4: Multiply and add each element of the second row of matrix $A$ with the respective elements of the second column of matrix $A$, as shown below:
$A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$ $A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$
Result for this will be, $R4 = \left[ {(( - 3) \times 2) + (0 \times 0)} \right]$
Step 5: Matrix multiplication of matrix $A$ and $A$, that is,${A^2}$will be,
${A^2} = \left[ {\begin{array}{*{20}{c}}
  {R1}&{R2} \\
  {R3}&{R4}
\end{array}} \right]$
Now, substituting the values of $R1,R2,R3$ and $R4,$ we get:
\[{A^2} = \left[ {\begin{array}{*{20}{c}}
  {(1 \times 1) + (2 \times ( - 3))}&{(1 \times 2) + (2 \times 0)} \\
  {(( - 3) \times 1) + (0 \times ( - 3))}&{(( - 3) \times 2) + (0 \times 0)}
\end{array}} \right]\]
After solving the above expressions, we get:
${A^2} = \left[ {\begin{array}{*{20}{c}}
  {1 + \left( { - 6} \right)}&{\;\;\;\;\;\;2 + 0} \\
  {( - 3) + 0}&{( - 6) + 0}
\end{array}} \right]$
${A^2} = \left[ {\begin{array}{*{20}{c}}
  { - 5}&2 \\
  { - 3}&{ - 6}
\end{array}} \right] \ne A$ $(\because A = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right])$
So, the condition of option A is not satisfied.

For checking whether the two given matrices are satisfying the condition of option B or not, that is, whether ${B^2} = B$ or not:
Step 1: To find \[{B^2}\], we have to multiply $B$ with $B$. In order to do that, multiply and add each element of the first row of matrix $B$ with the respective elements of the first column of matrix $B$, as shown below:
$B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$ $B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$
Result for this will be, $R1 = \left[ {(( - 1) \times ( - 1)) + (0 \times 2)} \right]$
Step 2: Multiply and add each element of the first row of matrix $B$ with the respective elements of the second column of matrix $B$, as shown below:
$B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$ $B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$
Result for this will be, $R2 = \left[ {(( - 1) \times 0) + (0 \times 3)} \right]$
Step 3: Multiply and add each element of the second row of matrix $B$ with the respective elements of the first column of matrix $B$, as shown below:
$B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$ $B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$
Result for this will be, $R3 = \left[ {(2 \times ( - 1) + (3 \times 2)} \right]$
Step 4: Multiply and add each element of the second row of matrix $B$ with the respective elements of the second column of matrix $B$, as shown below:
$B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$ $B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$
Result for this will be, $R4 = \left[ {(2 \times 0) + (3 \times 3)} \right]$
Step 5: Matrix multiplication of matrix $B$ and $B$, that is,\[{B^2}\]will be,
${B^2} = \left[ {\begin{array}{*{20}{c}}
  {R1}&{R2} \\
  {R3}&{R4}
\end{array}} \right]$
Now, substituting the values of $R1,R2,R3$ and $R4,$ we get:
\[{B^2} = \left[ {\begin{array}{*{20}{c}}
  {(( - 1) \times ( - 1)) + (0 \times 2)}&{(( - 1) \times 0) + (0 \times 3)} \\
  {(2 \times ( - 1)) + (3 \times 2)}&{(2 \times 0) + (3 \times 3)}
\end{array}} \right]\]
After solving the above expressions, we get:
${B^2} = \left[ {\begin{array}{*{20}{c}}
  {1 + 0}&{0 + 0} \\
  {( - 2) + 6}&{0 + 9}
\end{array}} \right]$
${B^2} = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  4&9
\end{array}} \right] \ne B$ $(\because B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right])$
So, the condition of option B is not satisfied.

For checking whether the two given matrices are satisfying the condition of options C and D or not, that is, whether $AB \ne BA$ and $AB = BA$:
For finding$AB$,
Step 1: Multiply and add each element of the first row of matrix $A$with the respective elements of the first column of matrix $B$, as shown below:
$A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$ $B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$
Result for this will be, $R1 = \left[ {(1 \times ( - 1)) + (2 \times 2)} \right]$
Step 2: Multiply and add each element of the first row of matrix $A$ with the respective elements of the second column of matrix $B$, as shown below:
$A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$ $B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$
Result for this will be, $R2 = \left[ {(1 \times 0) + (2 \times 3)} \right]$
Step 3: Multiply and add each element of the second row of matrix $A$ with the respective elements of the first column of matrix $B$, as shown below:
$A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$ $B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$
Result for this will be, $R3 = \left[ {(( - 3) \times ( - 1)) + (0 \times 2)} \right]$
Step 4: Multiply and add each element of the second row of matrix $A$ with the respective elements of the second column of matrix $B$, as shown below:
$A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$$B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$
Result for this will be, $R4 = \left[ {(( - 3) \times 0) + (0 \times 3)} \right]$
Step 5: Matrix multiplication of matrix $A$ and $B$ will be,
$AB = \left[ {\begin{array}{*{20}{c}}
  {R1}&{R2} \\
  {R3}&{R4}
\end{array}} \right]$
Now, substituting the values of $R1,R2,R3$ and $R4,$ we get:
$AB = \left[ {\begin{array}{*{20}{c}}
  {(1 \times ( - 1)) + (2 \times 2)}&{(1 \times 0) + (2 \times 3)} \\
  {(( - 3) \times ( - 1)) + (0 \times 2)}&{(( - 3) \times 0) + (0 \times 3)}
\end{array}} \right]$
After solving the above expressions, we get:
$AB = \left[ {\begin{array}{*{20}{c}}
  {( - 1) + 4}&{0 + 6} \\
  {3 + 0}&{0 + 0}
\end{array}} \right]$
$AB = \left[ {\begin{array}{*{20}{c}}
  3&6 \\
  3&0
\end{array}} \right]$

Now, for finding $BA$,
Step 1: Multiply and add each element of the first row of matrix $B$ with the respective elements of the first column of matrix $A$, as shown below:
$B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$ $A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$
Result for this will be, $R1 = \left[ {(( - 1) \times 1) + (0 \times ( - 3))} \right]$
Step 2: Multiply and add each element of the first row of matrix $B$ with the respective elements of the second column of matrix $A$, as shown below:
$B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$ $A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$
Result for this will be, $R2 = \left[ {(( - 1) \times 2) + (0 \times 0)} \right]$
Step 3: Multiply and add each element of the second row of matrix $B$ with the respective elements of the first column of matrix $A$, as shown below:
$B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$ $A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$
Result for this will be, $R3 = \left[ {(2 \times 1) + (3 \times ( - 3))} \right]$
Step 4: Multiply and add each element of the second row of matrix $B$ with the respective elements of the second column of matrix $A$, as shown below:
$B = \;\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  2&3
\end{array}} \right]$ $A = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 3}&0
\end{array}} \right]$
Result for this will be, $R4 = \left[ {(2 \times 2) + (3 \times 0)} \right]$
Step 5: Matrix multiplication of matrix $B$ and $A$ will be,
$BA = \left[ {\begin{array}{*{20}{c}}
  {R1}&{R2} \\
  {R3}&{R4}
\end{array}} \right]$
Now, substituting the values of $R1,R2,R3$ and $R4,$ we get:
$BA = \left[ {\begin{array}{*{20}{c}}
  {(( - 1) \times 1) + (0 \times ( - 3))}&{(( - 1) \times 2) + (0 \times 0)} \\
  {(2 \times 1) + (3 \times ( - 3))}&{(2 \times 2) + (3 \times 0)}
\end{array}} \right]$
After solving the above expressions, we get:
$BA = \left[ {\begin{array}{*{20}{c}}
  {( - 1) + 0}&{( - 2) + 0} \\
  {2 + ( - 9)}&{4 + 0}
\end{array}} \right]$
$BA = \left[ {\begin{array}{*{20}{c}}
  { - 1}&{ - 2} \\
  { - 7}&4
\end{array}} \right]$
Now, on comparing $AB$ and $BA$, we can conclude that,
$AB \ne BA$

Therefore, the correct option is (C).

Additional Information: In this type of question, where we have to multiply two given matrices, whose order is different, like a matrix A of order \[\left( {m \times n} \right)\], where ‘m’ is the number of rows of matrix A and ‘n’ is the number of columns of matrix A.
Another matrix B of order \[\left( {n \times p} \right)\], where ‘n’ is the number of rows of matrix B and ‘p’ is the number of columns of matrix B.
Then, the order of the resultant matrix should be\[\left( {m \times p} \right)\].
Also, if we know the properties of matrix multiplication, we don’t need to multiply and check with the options. We can directly choose the correct answer.

Note: In this type of question, where we have to multiply two given matrices, if the number of columns of the first matrix is equal to the number of rows of the second matrix, then multiplication is possible. Otherwise, multiplication is not possible.