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ICSE Class X Math 2015 Paper

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Question : 5 of 52
Marks: +1, -0
If A=[3x01]A=\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} and B=[9160−y]B=\begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}, find xx and yy when A2=BA^2=B.
Solution:  
     Given:   A=[3x01]   and   B=[9160−y]\;\;\text{ Given: }\; A=\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} \;\text{ and }\; B=\begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}
  A2=B\; A^2 = B
  [3x01][3x01]\; \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} =[9160−y]= \begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}
  [3×3+x×03×x+x×10×3+1×00×x+1×1]=[9160−y]\; \begin{bmatrix} 3 \times 3 + x \times 0 & 3 \times x + x \times 1 \\ 0 \times 3 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix} = \begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}
  [94x01]=[9160−y]\; \begin{bmatrix} 9 & 4x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}
By equality of matrices,
4x=16     and     1=−y4x = 16 \;\;\text{ and }\;\; 1 = -y
x=4     and     y=−1x = 4 \;\;\text{ and }\;\; y = -1
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