Question
Find $x$ and $y$ if $\left[\begin{array}{cc}x+y & y \\ 2 x & x-y\end{array}\right]\left[\begin{array}{c}2 \\ -1\end{array}\right]=\left[\begin{array}{l}3 \\ 2\end{array}\right]$
✓
Answer
Given
$\begin{aligned} & {\left[\begin{array}{cc} x+y & y \\ 2 x & x-y \end{array}\right]\left[\begin{array}{c} 2 \\\ -1 \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \end{array}\right]} \end{aligned} $
$ \Rightarrow\left[\begin{array}{cc} 2 x+2 y & -y \\ 4 x & -x+y \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \end{array}\right] $
$ \Rightarrow\left[\begin{array}{ll} 2 x & +y \\ 3 x & +y \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \end{array}\right] $
Comparing the corresponding elements
$2 x+y=3.....(1)$
$ 3 x+y=2.....(2)$
Subtracting, we get
$- x =1$
$ \Rightarrow x =-1$
Substituting the value of $x$ in $(i)$
$ 2(-1)+y=3$
$ \Rightarrow-2+y=3$
$ \Rightarrow y=3+2=5$
Hence $x=-1, y=5.$
$\begin{aligned} & {\left[\begin{array}{cc} x+y & y \\ 2 x & x-y \end{array}\right]\left[\begin{array}{c} 2 \\\ -1 \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \end{array}\right]} \end{aligned} $
$ \Rightarrow\left[\begin{array}{cc} 2 x+2 y & -y \\ 4 x & -x+y \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \end{array}\right] $
$ \Rightarrow\left[\begin{array}{ll} 2 x & +y \\ 3 x & +y \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \end{array}\right] $
Comparing the corresponding elements
$2 x+y=3.....(1)$
$ 3 x+y=2.....(2)$
Subtracting, we get
$- x =1$
$ \Rightarrow x =-1$
Substituting the value of $x$ in $(i)$
$ 2(-1)+y=3$
$ \Rightarrow-2+y=3$
$ \Rightarrow y=3+2=5$
Hence $x=-1, y=5.$
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
Which of the following cannot be the probability of an event? 0.35
Represent the following inequalities on real number lines
$8 \geq x>-3$
→$8 \geq x>-3$
P and Q are two points whose coordinate are $\left(\frac{a}{t^2}, \frac{-2 a}{t}\right)$ and S is the point $(a, 0).$ Prove that $\frac{1}{ SP }+\frac{1}{ SQ }$ is constant for all values of it.
→Solve $2x^2 – 9x + 10 =0$; when $x \in Q$
→Find $AD$
→If the replacement set is the set of real numbers solve
$\frac{x+3}{8}<\frac{x-3}{5}$
→$\frac{x+3}{8}<\frac{x-3}{5}$
Given below are part of a geometrical figure. Complete the figure so that both axes are lines of symmetry.
Prove.
$\sin ^4 A-\cos ^4 A=2 \sin ^2 A-1$
→$\sin ^4 A-\cos ^4 A=2 \sin ^2 A-1$
If $\frac{1}{12}, x$ and $\frac{1}{75}$ are in continued proportion, find $x$.
→If $4 \cos^2 A – 3 = 0$ and $\leq A \leq 90^\circ ,$ then prove that: $\cos 3 A = 4 \cos^3 A – 3 \cos A$
→