Question
Let $\text{A}=\begin{bmatrix}2&-3\\-7&5\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&0\\2&-4\end{bmatrix},$ verify that
$(\text{A}+\text{B})^\text{T}=\text{A}^\text{T}+\text{B}^\text{T}$

Answer

Given: $\text{A}=\begin{bmatrix}2&-3\\-7&5\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}2&-7\\-3&5\end{bmatrix}$
$\text{B}=\begin{bmatrix}1&0\\2&-4\end{bmatrix}$
$\text{B}^\text{T}=\begin{bmatrix}1&2\\0&-4\end{bmatrix}$
Given,
$(\text{A}+\text{B})^\text{T}=\text{A}^\text{T}+\text{B}^\text{T}$
$\begin{pmatrix} \begin{bmatrix} 2&-3\\-7&5\end{bmatrix}+\begin{bmatrix} 1&0\\2&-4\end{bmatrix}\end{pmatrix}^\text{T}$ $=\begin{bmatrix}2&-3\\-7&5\end{bmatrix}^\text{T}+\begin{bmatrix}1&0\\2&-4\end{bmatrix}^\text{T}$
$\Rightarrow\begin{bmatrix}2+1&-3+0\\-7+2&5-4\end{bmatrix}^\text{T}$ $=\begin{bmatrix}2&-7\\-3&5\end{bmatrix}+\begin{bmatrix}1&2\\0&-4\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3&-3\\-5&1\end{bmatrix}^\text{T}=\begin{bmatrix}2+1&-7+2\\-3+0&5-4\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3&-5\\-3&1\end{bmatrix}=\begin{bmatrix}3&-5\\-3&1\end{bmatrix}$
$\Rightarrow\text{LHS}=\text{RHS}$
So,
$(\text{A}+\text{B})^\text{T}=\text{A}^\text{T}+\text{B}^\text{T}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Find the area of the parallelogram determinrd by the vectors:
$3\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
If $\text{y}=\log\sqrt{\tan\text{x}},$ write $\frac{\text{dy}}{\text{dx}}.$
The following relation are defined on the set of real numbers.
aRb if 1 + ab > 0
Find whether these relation are reflexive, symmetric or transitive.
Find the angle between the lines
$
\begin{array}{ll}
& 2 x=3 y=-z \\
\text { and } & 6 x=-y=-4 z .
\end{array}
$
Find the integral of the function ${\frac{{\cos 2x - \cos 2\alpha }}{{\cos x - \cos \alpha }}}$
Find the perpendicular distence of the point (3, -1, 11) from the line $\frac{\text{x}}{2}=\frac{\text{y}-2}{-3}=\frac{\text{z}-3}{4}.$
An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
If  $\vec{\text{a}}=2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}},$and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}$ are such that $\vec{\text{a}}+\lambda\vec{\text{b}}$ is perpendicular to $\vec{\text{c}},$ then find the value of $\lambda.$
An unbiased die is thrown twice. A success is getting a number greater than 4. Find the probability distribution of the number of successes.
Find the values of p so that the lines $\frac{1-\text{x}}{3}=\frac{7\text{y}-14}{2\text{p}}=\frac{\text{z}-3}{2}\ \text{and}\ \frac{7-7\text{x}}{3\text{p}}=\frac{\text{y}-5}{1}=\frac{6-\text{z}}{5}$ are at right angles.