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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Matrices

NDA 2015 · Mathematics · Matrices

Let A= [ array cc x+y & y 2 x & x-y array ], B= [ array c 2 -1...

Let A=\left[x+yy2xx−y\begin{array}{cc} x+y & y \\ 2 x & x-y \end{array}\right], B=\left[2−1\begin{array}{c} 2 \\ -1 \end{array}\right]\text{ and } C=\left[32\begin{array}{l} 3 \\ 2 \end{array}\right] If A B=C,thenwhatisA2, then what is A^2 equal to?

Options

  1. A\left[6−10426\begin{array}{cc}6 & -10 \\ 4 & 26\end{array}\right]
  2. B\left[−105424\begin{array}{cc}-10 & 5 \\ 4 & 24\end{array}\right]
  3. C\left[−5−6−4−20\begin{array}{cc}-5 & -6 \\ -4 & -20\end{array}\right]
  4. D\left[−5−7−520\begin{array}{ll}-5 & -7 \\ -5 & 20\end{array}\right]

Correct answer

A. \left[6−10426\begin{array}{cc}6 & -10 \\ 4 & 26\end{array}\right]

Step-by-step solution

A=[x+yy2xx−y]B=[2−1] and C=[32]\begin{aligned} & \mathrm{A}=\left[\begin{array}{cc} \mathrm{x}+\mathrm{y} & \mathrm{y} \\ 2 \mathrm{x} & \mathrm{x}-\mathrm{y} \end{array}\right] \\ & \mathrm{B}=\left[\begin{array}{c} 2 \\ -1 \end{array}\right] \text{ and } \mathrm{C}=\left[\begin{array}{l} 3 \\ 2 \end{array}\right] \end{aligned} Here AB=C\mathrm{AB}=\mathrm{C}∴[x+yy2xx−y][2−1]=[32]⇒[2(x+y)−y4x−x+y]=[32]2x+y=3...(i)3x+y=2...(ii)\begin{aligned} \therefore & {\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] } \\ \Rightarrow & {\left[\begin{array}{cc} 2(x+y) & -y \\ 4 x & -x+y \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \end{array}\right] } \\ & 2 x+y=3 ...(i)\\ & 3 x+y=2...(ii) \end{aligned}Fromequations(i)and(ii),wegetFrom equations (i) and (ii), we getx=-1andandy=5∴A=[45−2−6] Now, A2=[45−2−6][45−2−6]=[16−1020−30−8+12−10+36]=[6−10426]\begin{aligned} \therefore & A=\left[\begin{array}{cc} 4 & 5 \\ -2 & -6 \end{array}\right] \\ & \text{ 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] \end{aligned}\therefore \quad$ Option (A) is correct.
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