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Let
$A=\left[\begin{array}{cc}x+y & y \\ 2 x & x-y\end{array}\right], B=\left[\begin{array}{c}2 \\ -1\end{array}\right]$ and $C=\left[\begin{array}{l}3 \\ 2\end{array}\right]$
If $A B=C$, then what is $A^{2}$ equal to?
Options:
$A=\left[\begin{array}{cc}x+y & y \\ 2 x & x-y\end{array}\right], B=\left[\begin{array}{c}2 \\ -1\end{array}\right]$ and $C=\left[\begin{array}{l}3 \\ 2\end{array}\right]$
If $A B=C$, then what is $A^{2}$ equal to?
Solution:
2233 Upvotes
Verified Answer
The correct answer is:
$\left[\begin{array}{cc}6 & -10 \\ 4 & 26\end{array}\right]$
$A=\left[\begin{array}{cc}x+y & y \\ 2 x & x-y\end{array}\right]$
$B=\left[\begin{array}{c}2 \\ -1\end{array}\right]$ and $C=\left[\begin{array}{l}3 \\ 2\end{array}\right]$
Here $\mathrm{AB}=\mathrm{C}$
$\therefore \quad\left[\begin{array}{cc}\mathrm{x}+\mathrm{y} & \mathrm{y} \\ 2 \mathrm{x} & \mathrm{x}-\mathrm{y}\end{array}\right]\left[\begin{array}{c}2 \\ -1\end{array}\right]=\left[\begin{array}{l}3 \\ 2\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}2(\mathrm{x}+\mathrm{y}) & -\mathrm{y} \\ 4 \mathrm{x} & -\mathrm{x}+\mathrm{y}\end{array}\right]=\left[\begin{array}{l}3 \\ 2\end{array}\right]$
$2 x+y=3$ ...(i)
$3 x+y=2$ ...(ii)
From equations (i) and (ii), we get $x=-1$ and $y=5$
$\therefore A=\left[\begin{array}{cc}4 & 5 \\ -2 & -6\end{array}\right]$
Now, $A^{2}=\left[\begin{array}{cc}4 & 5 \\ -2 & -6\end{array}\right]\left[\begin{array}{cc}4 & 5 \\ -2 & -6\end{array}\right]$
$=\left[\begin{array}{cc}16-10 & 20-30 \\ -8+12 & -10+36\end{array}\right]=\left[\begin{array}{cc}6 & -10 \\ 4 & 26\end{array}\right]$
$\therefore$ Option (a) is correct.
$B=\left[\begin{array}{c}2 \\ -1\end{array}\right]$ and $C=\left[\begin{array}{l}3 \\ 2\end{array}\right]$
Here $\mathrm{AB}=\mathrm{C}$
$\therefore \quad\left[\begin{array}{cc}\mathrm{x}+\mathrm{y} & \mathrm{y} \\ 2 \mathrm{x} & \mathrm{x}-\mathrm{y}\end{array}\right]\left[\begin{array}{c}2 \\ -1\end{array}\right]=\left[\begin{array}{l}3 \\ 2\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}2(\mathrm{x}+\mathrm{y}) & -\mathrm{y} \\ 4 \mathrm{x} & -\mathrm{x}+\mathrm{y}\end{array}\right]=\left[\begin{array}{l}3 \\ 2\end{array}\right]$
$2 x+y=3$ ...(i)
$3 x+y=2$ ...(ii)
From equations (i) and (ii), we get $x=-1$ and $y=5$
$\therefore A=\left[\begin{array}{cc}4 & 5 \\ -2 & -6\end{array}\right]$
Now, $A^{2}=\left[\begin{array}{cc}4 & 5 \\ -2 & -6\end{array}\right]\left[\begin{array}{cc}4 & 5 \\ -2 & -6\end{array}\right]$
$=\left[\begin{array}{cc}16-10 & 20-30 \\ -8+12 & -10+36\end{array}\right]=\left[\begin{array}{cc}6 & -10 \\ 4 & 26\end{array}\right]$
$\therefore$ Option (a) is correct.
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