Question
If $\text{X}-\text{Y}=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}$ and $\text{X}+\text{Y}=\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix},$ find X and Y.
✓
Answer
Here,
$\text{X}-\text{Y}+\text{X}+\text{Y}=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}+\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}1+3&1+5&1+1\\1-1&1+1&0+4\\1+11&0+8&0+0\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}4&6&2\\0&2&4\\12&8&0\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{2}\begin{bmatrix}4&6&2\\0&2&4\\12&8&0\end{bmatrix}$
$\Rightarrow\text{X}=\begin{bmatrix}2&3&1\\0&1&2\\6&4&0\end{bmatrix}$
Now,
$(\text{X}-\text{Y})-(\text{X}+\text{Y})=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}-\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix}$
$\Rightarrow\text{X}-\text{Y}-\text{X}-\text{Y}=\begin{bmatrix}1-3&1-5&1-1\\1+1&1-1&0-4\\1-11&0-8&0-0\end{bmatrix}$
$\Rightarrow-2\text{Y}=\begin{bmatrix}-2&-4&0\\2&0&-4\\-10&-8&0\end{bmatrix}$
$\Rightarrow\text{Y}=-\frac{1}{2}\begin{bmatrix}-2&-4&0\\2&0&-4\\-10&-8&0\end{bmatrix}$
$\Rightarrow\text{Y}=\begin{bmatrix}1&2&0\\-1&0&2\\5&4&0\end{bmatrix}$
$\therefore\ \text{X}=\begin{bmatrix}2&3&1\\0&1&2\\6&4&0\end{bmatrix}$ and $\text{Y}=\begin{bmatrix}1&2&0\\-1&0&2\\5&4&0\end{bmatrix}$
$\text{X}-\text{Y}+\text{X}+\text{Y}=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}+\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}1+3&1+5&1+1\\1-1&1+1&0+4\\1+11&0+8&0+0\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}4&6&2\\0&2&4\\12&8&0\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{2}\begin{bmatrix}4&6&2\\0&2&4\\12&8&0\end{bmatrix}$
$\Rightarrow\text{X}=\begin{bmatrix}2&3&1\\0&1&2\\6&4&0\end{bmatrix}$
Now,
$(\text{X}-\text{Y})-(\text{X}+\text{Y})=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}-\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix}$
$\Rightarrow\text{X}-\text{Y}-\text{X}-\text{Y}=\begin{bmatrix}1-3&1-5&1-1\\1+1&1-1&0-4\\1-11&0-8&0-0\end{bmatrix}$
$\Rightarrow-2\text{Y}=\begin{bmatrix}-2&-4&0\\2&0&-4\\-10&-8&0\end{bmatrix}$
$\Rightarrow\text{Y}=-\frac{1}{2}\begin{bmatrix}-2&-4&0\\2&0&-4\\-10&-8&0\end{bmatrix}$
$\Rightarrow\text{Y}=\begin{bmatrix}1&2&0\\-1&0&2\\5&4&0\end{bmatrix}$
$\therefore\ \text{X}=\begin{bmatrix}2&3&1\\0&1&2\\6&4&0\end{bmatrix}$ and $\text{Y}=\begin{bmatrix}1&2&0\\-1&0&2\\5&4&0\end{bmatrix}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Let $f: R \rightarrow R^{+}$be defined by $f(x)=a x, a>0$ and $a \neq 1$. Write $f^{-1}(x)$.
→Find $\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big), $ if $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$.
→Form the differential equation of family of lines having intercepts a and b on the coordinate axes respectively.
Find the mean number of heads in three tosses of a fair coin.
Discuss the applicability of the Rolle's theorem for the following function on the indicated interval
$\text{f}(\text{x})=2\text{x}^2-5\text{x}+3\text{ on }[1,3]$
→$\text{f}(\text{x})=2\text{x}^2-5\text{x}+3\text{ on }[1,3]$
If $\text{A}=\begin{bmatrix}1&-3&2\\2&0&2\end{bmatrix}$ and $\text{B}=\begin{bmatrix}2&-1&-1\\1&0&-1\end{bmatrix},$ find the matrix C such that A + B + C is zeor matrix.
→Write the equation of the plane passing through (2, −1, 1) and parallel to the plane 3x + 2y − z = 7.
Let $'o'$ be a binary operation on the set $Q_0$ of all non-zero rational numbers defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{2}$ for all $\text{a},\text{b}\in\text{Q}_0.$
Show that $'o'$ is both commutative and associate.
→Show that $'o'$ is both commutative and associate.
Simplify the following so that the new circuit circuit.
Evaluate :
→$\int_0^{\infty} x \cdot e^{-x} \cdot d x$