If and then det (A6) =

Engineering

Mathematics

Introduction to Determinants

Question

If $θ = \frac{π}{12}$ and $A = [\begin{array}{cc} \cos θ & \sin θ \\ \sin θ & \cos θ \end{array}]$ then det (A⁶) =

$\frac{27}{64}$

$\frac{3}{2}$

– 1

$\frac{9}{16}$

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$A = [\begin{array}{cc} \cos θ & \sin θ \\ \sin θ & \cos θ \end{array}]$

A^{2} = A \times A = [\begin{array}{cc} \cos θ & \sin θ \\ \sin θ & \cos θ \end{array}] \times [\begin{array}{cc} \cos θ & \sin θ \\ \sin θ & \cos θ \end{array}]

= [\begin{array}{cc} \cos^{2} θ \sin^{2} θ & \cos θsin θ + \sin θcos θ \\ \sin θcos θ + \sin θcos θ & \cos^{2} θ + \sin^{2} θ \end{array}]

= [\begin{array}{cc} 1 & \sin^{2} θ \\ \sin^{2} θ & 1 \end{array}]

θ = \frac{π}{12}

∴ A^{2} = [\begin{array}{cc} 1 & \sin (\frac{2 π}{12}) \\ \sin (\frac{2 π}{12}) & 1 \end{array}] = [\begin{array}{cc} 1 & \sin \frac{π}{6} \\ \sin \frac{π}{6} & 1 \end{array}] = [\begin{array}{cc} 1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{array}]

∴ A^{4} = A^{2} \times A^{2} = [\begin{array}{cc} 1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{array}] [\begin{array}{cc} 1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{array}] = [\begin{array}{cc} 1 + \frac{1}{4} & \frac{1}{2} + \frac{1}{2} \\ \frac{1}{2} + \frac{1}{2} & \frac{1}{4} + 1 \end{array}] = [\begin{array}{cc} \frac{5}{4} & + 1 \\ 1 & \frac{5}{4} \end{array}]

Hence

A^{6} = A^{4} \times A^{2} = [\begin{array}{cc} \frac{5}{4} & 1 \\ 1 & \frac{5}{4} \end{array}] [\begin{array}{cc} 1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{array}]

= [\begin{array}{cc} \frac{5}{4} + \frac{1}{2} & \frac{5}{8} + 1 \\ 1 + \frac{5}{8} & \frac{1}{2} + \frac{5}{4} \end{array}] = [\begin{array}{cc} \frac{7}{4} & \frac{13}{8} \\ \frac{13}{8} & \frac{7}{4} \end{array}]

= \frac{7}{4} \times \frac{7}{4} - \frac{13}{8} \times \frac{13}{8} = \frac{27}{64}

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