A= [ array cc x+y & y x & x-y array ], B= [ array l 3 -2 array ] and C=[ array l 4 -2 array ] . If A B=C , then what is A^2 equal to?

Correct answer: D. [ array cc -4 & -8 8 & 12 array ] aligned & A=[ array cc x+y & y x & x-y array ], B=[ array c 3 -2 array ], c=[ array c 4 -2 array ] & A B=C & [ array c 3 x+3 y-2 y 3 x-2 x+2 y array ]=[ array c 4 -2 array ] & 3 x+y=4 ; x+2 y=-2 aligned Solving these equations, gathered 3 x+y=4 3 x+6 y=-6 (-)(-)(+) -5 y=10 gathered aligned & y=-2 & So, 3 x-2=4 3 x=6 x=2 & A=[ array cc 0 & -2 2 & 4 array ], A^2=[ array cc 0 & -2 2 & 4 array ][ array cc 0 & -2 2 & 4 array ] & =[ array cc -4 & -8 8 & 12 array ] aligned

NDA 2017

Exam: NDA
Year: 2017
Subject: Mathematics
Chapter: Matrices

NDA 2017 · Mathematics · Matrices

A= [ array cc x+y & y x & x-y array ], B= [ array l 3 -2 array ]...

A=

\left[

\begin{array}{cc}x+y & y \\ x & x-y\end{array}

\right], B=

\left[

\begin{array}{l}3 \\ -2\end{array}

\right]

and

C=\left[

\begin{array}{l}4 \\ -2\end{array}

\right]

. If

A B=C

, then what is

\mathrm{A}^2

equal to?

Options

A $\left[$ $\begin{array}{cc}4 & 8 \\ -4 & -16\end{array}$ $\right]$
B $\left[$ $\begin{array}{cc}4 & -4 \\ 8 & -16\end{array}$ $\right]$
C $\left[$ $\begin{array}{cc}-4 & -8 \\ 4 & 12\end{array}$ $\right]$
D $\left[$ $\begin{array}{cc}-4 & -8 \\ 8 & 12\end{array}$ $\right]$

Correct answer

D. $\left[$ $\begin{array}{cc}-4 & -8 \\ 8 & 12\end{array}$ $\right]$

Step-by-step solution

\begin{aligned} & A=\left[\begin{array}{cc} x+y & y \\ x & x-y \end{array}\right], B=\left[\begin{array}{c} 3 \\ -2 \end{array}\right], c=\left[\begin{array}{c} 4 \\ -2 \end{array}\right] \\ & A B=C \\ & \Rightarrow\left[\begin{array}{c} 3 x+3 y-2 y \\ 3 x-2 x+2 y \end{array}\right]=\left[\begin{array}{c} 4 \\ -2 \end{array}\right] \\ & \Rightarrow 3 x+y=4 ; x+2 y=-2 \end{aligned}

Solving these equations,

\begin{gathered} 3 x+y=4 \\ 3 x+6 y=-6 \\ (-)(-)(+) \\ \hline-5 y=10 \end{gathered}

\begin{aligned} & \Rightarrow y=-2 \\ & \text{ So, } 3 x-2=4 \Rightarrow 3 x=6 \Rightarrow x=2 \\ & \therefore A=\left[\begin{array}{cc} 0 & -2 \\ 2 & 4 \end{array}\right], \quad A^2=\left[\begin{array}{cc} 0 & -2 \\ 2 & 4 \end{array}\right]\left[\begin{array}{cc} 0 & -2 \\ 2 & 4 \end{array}\right] \\ & =\left[\begin{array}{cc} -4 & -8 \\ 8 & 12 \end{array}\right] \end{aligned}

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