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If $A=\left[\begin{array}{rrr}3 & 2 & -1 \\ -5 & 0 & -6\end{array}\right]$ and $B=\left[\begin{array}{ccc}-4 & -5 & -2 \\ 3 & 1 & 8\end{array}\right]$, verify that $(A+B)^{\prime}=\left(A^{\prime}+B^{\prime}\right)$.
If $A=\left[\begin{array}{rrr}3 & 2 & -1 \\ -5 & 0 & -6\end{array}\right]$ and $B=\left[\begin{array}{ccc}-4 & -5 & -2 \\ 3 & 1 & 8\end{array}\right]$, verify that $(A+B)^{\prime}=\left(A^{\prime}+B^{\prime}\right)$.
CBSE | Class 12 | Exercise 5D | Maths | Matrices | RS Aggarwal
If $A=\left[\begin{array}{rrr}3 & 2 & -1 \\ -5 & 0 & -6\end{array}\right]$ and $B=\left[\begin{array}{ccc}-4 & -5 & -2 \\ 3 & 1 & 8\end{array}\right]$, verify that $(A+B)^{\prime}=\left(A^{\prime}+B^{\prime}\right)$.

Solution:

We have $A=\left(\begin{array}{ccc}3 & 2 & -1 \\ -5 & 0 & -6\end{array}\right)$ and $B=\left(\begin{array}{ccc}-4 & -5 & -2 \\ 3 & 1 & 8\end{array}\right)$.
The transpose of the matrix is an operation of making interchange of elements by the rule on positioned element $a_{i j}$ shifted to new position $a_{j i}$.

(i) The transpose of $A$ and $B$ are given by $A^{T}$ and $B^{T}$ that can be written as
$\begin{array}{l}
A^{T}=\left(\begin{array}{ccc}
3 & 2 & -1 \\
-5 & 0 & -6
\end{array}\right)^{T}=\left(\begin{array}{cc}
3 & -5 \\
2 & 0 \\
-1 & -6
\end{array}\right) \\
B^{T}=\left(\begin{array}{ccc}
-4 & -5 & -2 \\
3 & 1 & 8
\end{array}\right)^{T}=\left(\begin{array}{cc}
-4 & 3 \\
-5 & 1 \\
-2 & 8
\end{array}\right)
\end{array}$

(ii) Next the addition of $A$ and $B$ can be processed as
$A+B=\left(\begin{array}{ccc}3 & 2 & -1 \\ -5 & 0 & -6\end{array}\right)+\left(\begin{array}{ccc}-4 & -5 & -2 \\ 3 & 1 & 8\end{array}\right)=\left(\begin{array}{ccc}-1 & -3 & -3 \\ -2 & 1 & 2\end{array}\right)$

(iii) Thus $A^{T}+B^{T}=\left(\begin{array}{cc}3 & -5 \\ 2 & 0 \\ -1 & -6\end{array}\right)+\left(\begin{array}{cc}-4 & 3 \\ -5 & 1 \\ -2 & 8\end{array}\right)=\left(\begin{array}{cc}-1 & -2 \\ -3 & 1 \\ -3 & 2\end{array}\right)$

(iv) The transpose of $A+B$ is given by
$(A+B)^{T}=\left(\begin{array}{ccc}
-1 & -3 & -3 \\
-2 & 1 & 2
\end{array}\right)^{T}=\left(\begin{array}{cc}
-1 & -2 \\
-3 & 1 \\
-3 & 2
\end{array}\right)$
From eq.(iii) and eq.(iv) it can said that $(A+B)^{T}=A^{T}+B^{T}$.

i

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