If A = [21201]\begin{bmatrix} 2 & 12 \ 0 & 1 \end{bmatrix}[20121] and B = [4x01]\begin{bmatrix} 4 & x \ 0 & 1 \end{bmatrix}[40x1] and A2 = B, find the value of x.
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Let's find A2,
⇒A2=[21201]×[21201]⇒[2×2+12×02×12+12×10×2+1×00×12+1×1]⇒[424+1201]⇒[43601].\Rightarrow A^2 = \begin{bmatrix} 2 & 12 \ 0 & 1 \end{bmatrix} \times \begin{bmatrix} 2 & 12 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \times 2 + 12 \times 0 & 2 \times 12 + 12 \times 1 \ 0 \times 2 + 1 \times 0 & 0 \times 12 + 1 \times 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 24 + 12 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 36 \ 0 & 1 \end{bmatrix}.⇒A2=[20121]×[20121]⇒[2×2+12×00×2+1×02×12+12×10×12+1×1]⇒[4024+121]⇒[40361].
Since, A2 = B,
⇒[43601]=[4x01]\Rightarrow \begin{bmatrix} 4 & 36 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & x \ 0 & 1 \end{bmatrix}⇒[40361]=[40x1]
∴ x = 36
Hence, x = 36.
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