Question

Which of the following relations is incorrect?

A. \[\left( A+B+......+l \right)'=A'+B'+....+l'\]
B. \[\left( AB....l \right)'=A'B'....l'\]
C. \[\left( kA \right)'=kA'\]
D. \[\left( A \right)'=A\]

Answer 1

Hint: In this question we have to find which of the given relation is incorrect. Then, we have to find the transpose of the resultant matrix. To find the answer to the given question, we must know the basic operations on matrices, such as addition, subtraction, multiplication, and transpose. In this question, we will assume the matrices,$A$ and $B$, and then check which of the given relations are correct and incorrect to find the correct solution.

Formula Used:
For scalar multiplication of a matrix, multiply the constant with every element of the matrix.
Let \[k\] be a constant and \[A\] be a \[2 \times 2\] matrix,
\[A=\left[ \begin{matrix}
 {{a}_{11}} & {{a}_{12}} \\
 \text{ }\!\!~\!\!\text{ }{{a}_{21}} & {{a}_{22}}\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]\]
So,
\[kA=\left[ \begin{matrix}
  \text{ }\!\!~\!\!\text{ }(k\times {{a}_{11}}) & (k\times {{a}_{12}}) \\
  \text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }(k\times {{a}_{21}}) & (k\times {{a}_{22}})\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]\]

Complete step by step Solution:
Let us assume two matrices of order$2
\times 2$, matrix$\ A=\ \left[
\begin{matrix}
   1 & 2 \\
   3 & 4 \\
\end{matrix} \right]$ and matrix
\[B=\ \left[ \begin{matrix}
   5 & 6 \\
   7 & 8 \\
\end{matrix} \right]\]
Checking the condition of option,
A:
  \[(A+B+\text{ }\!\!~\!\!\text{ }\ldots
.+l)'=\text{ }A'+\text{ }B'+\text{ }\!\!~\!\!\text{ }\ldots .+\text{ }l'\]
                                                                           \[(A+B)=\
\left[ \begin{matrix}
   1 & 2 \\
   3 & 4 \\
\end{matrix} \right]+\ \left[ \begin{matrix}
   5 & 6 \\
   7 & 8 \\
\end{matrix} \right]\]
In order to add two matrices, the first element of matrix$A$will be added to the first element of matrix$B$, and so on…
\[(A+B)=\ \left[ \begin{matrix}
   1+5 & 2+6 \\
   3+7 & 4+8\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]\]
                 \[=\ \left[ \begin{matrix}
   6 & 8 \\
   10 & 12\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]\]

In order to find the transpose of$(A + B)$, that is,$(A+B)'$, we have to interchange the rows with columns as:
\[(A+B)'=\left[ \begin{matrix}
   6 & 10 \\
   8 & 12\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]\ldots \text{ }\!\!~\!\!\text{ }\ldots \text{ }\!\!~\!\!\text{ }\ldots \text{ }\!\!~\!\!\text{ }\ldots eq(1)\]
Now, since matrix$\ A=\ \left[ \begin{matrix}
   1 & 2 \\
   3 & 4 \\
\end{matrix} \right]$, transpose of matrix$A$,that is,$\ A'=\ \left[ \begin{matrix}
   1 & 3 \\
   2 & 4 \\
\end{matrix} \right]$
Similarly, since matrix\[B=\ \left[ \begin{matrix}
   5 & 6 \\
   7 & 8 \\
\end{matrix} \right]\], transpose of matrix$B$,that is,\[B'=\ \left[ \begin{matrix}
   5 & 7 \\
   6 & 8 \\
\end{matrix} \right]\]
\[\begin{align}
  & (A'+B')=\ \left[ \begin{matrix}
   1 & 3 \\
   2 & 4 \\
\end{matrix} \right]+\ \left[ \begin{matrix}
   5 & 7 \\
   6 & 8 \\
\end{matrix} \right] \\
 & =\ \left[ \begin{matrix}
   1+5 & 3+7 \\
   2+6 & 4+8 \\
\end{matrix} \right]
\end{align}\]
$(A'+B')=\ \left[ \begin{matrix}
   6 & 10 \\
   8 & 12 \\
\end{matrix} \right]\ldots \text{ }\!\!~\!\!\text{ }\ldots \text{ }\!\!~\!\!\text{ }\ldots \text{ }\!\!~\!\!\text{ }\ldots eq(2)$
Here,$eq(1) = eq(2)$, which means \[\left( A+B+......+l \right)'=A'+B'+....+l'\].
So, option A is correct.

Checking the condition of option B:
\[\left( AB....l \right)'=A'B'....l'\]
\[AB=\ \ \left[ \begin{matrix}
   1 & 2 \\
   3 & 4 \\
\end{matrix} \right]\times \ \left[ \begin{matrix}
   5 & 6 \\
   7 & 8 \\
\end{matrix} \right]\]
In order to multiply two matrices, the elements of the first row of matrix$A$will be multiplied to the elements of the first column of matrix $B$, then the elements of the first row of matrix$A$will be multiplied to the elements of the second column of matrix$B$, then the elements of the second row of matrix$A$will be multiplied to the elements of the first column matrix$B$, and in the end, the elements of the second row of matrix$A$will be multiplied to the elements of the second column matrix$B$.
\[AB=\ \left[ \begin{matrix}
   (1\times 5)+(2\times 7) & (1\times 6)+(2\times 8) \\
   (3\times 5)+(4\times 7) & (3\times 6)+(4\times 8)\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]\]
        \[=\ \left[ \begin{matrix}
   5+14 & 6+16 \\
   15+28 & 18+32 \\
\end{matrix} \right]\]
      \[=\ \left[ \begin{matrix}
   19 & 22 \\
   43 & 50\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]\]
In order to find the transpose of$(AB)$, that is,$(AB)'$, we have to interchange the rows with columns as:
$(AB)'=\ \left[ \begin{matrix}
   19 & 43 \\
   22 & 50\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]\ldots \text{ }\!\!~\!\!\text{ }\ldots \text{ }\!\!~\!\!\text{ }\ldots \text{ }\!\!~\!\!\text{ }\ldots eq(3)$
Now, since matrix$\ A=\ \left[ \begin{matrix}
   1 & 2 \\
   3 & 4 \\
\end{matrix} \right]$, transpose of matrix$A$,that is,$\ A'=\ \left[ \begin{matrix}
   1 & 3 \\
   2 & 4 \\
\end{matrix} \right]$
Similarly, since matrix\[B=\ \left[ \begin{matrix}
   5 & 6 \\
   7 & 8 \\
\end{matrix} \right]\], transpose of matrix$B$,that is,\[B'=\ \left[ \begin{matrix}
   5 & 7 \\
   6 & 8 \\
\end{matrix} \right]\].
$\begin{align}
  & AB'=\ \ \left[ \begin{matrix}
   1 & 3 \\
   2 & 4 \\
\end{matrix} \right]\ \times \left[ \begin{matrix}
   5 & 7 \\
   6 & 8 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   (1\times 5)+(3\times 6) & (1\times 7)+(3\times 8) \\
   (2\times 5)+(4\times 6) & (2\times 7)+(4\times 8) \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   5+18 & 7+24 \\
   10+24 & 14+32\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]
\end{align}$
$A'B'=\ \left[ \begin{matrix}
   23 & 31 \\
   34 & 46 \\
\end{matrix} \right]\ldots \text{ }\!\!~\!\!\text{ }\ldots \text{ }\!\!~\!\!\text{ }\ldots \text{ }\!\!~\!\!\text{ }\ldots eq(4)$
Here,$eq(3) \ne eq(4)$, which means\[{{\left( AB....l \right)}^{\prime }}\ne {A}'{B}'....{l}'\].
So, option B is incorrect.

Checking the condition of option C:
$(kA)'=kA'$
Here, $k$is a non-zero scalar quantity and$A$can be represented as ${[{A_{ij}}]_{m \times n}}$of order $m \times n$.
Let $(kA)'=L.H.S$and $kA'=R.H.S$
Transpose of matrix$A$,that is,$A'$will be of order $n \times m$.
So, $L.H.S\Rightarrow \text{ }(kA)'$will still be of order $n \times m$.
\[R.H.S\Rightarrow kA'\]will also be of order $n \times m$
Since, $(kA)'$and $kA'$both are of same order $(n \times m)$ .
So. $L.H.S = R.H.S$, option C is correct.

Checking the condition of option D:
$\left( A' \right)'=A$
In order to find the transpose of$A$, that is,$A'$, we have to interchange the rows with columns as:
$\ A'=\ \left[ \begin{matrix}
   1 & 3 \\
   2 & 4 \\
\end{matrix} \right]$Now,
$\ \left( A' \right)'=\ \left[ \begin{matrix}
   1 & 2 \\
   3 & 4 \\
\end{matrix} \right]$ which is equal to$A$.
So, option D is correct.
Therefore, the correct option is (B).

Additional Information: To solve this type of question, we must have a clear knowledge about order of matrices, like if a matrix A is 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.
And, another matrix B is 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\[\left( {m \times p} \right)\].



Note: Whenever we have to perform matrix multiplication in any question, keep in mind that, if the number of columns of first matrix is equal to the number of rows of second matrix, then multiplication is possible. Otherwise, multiplication is not possible.