If , then the general solution of sinθ = |A2

Engineering

Mathematics

Introduction to Determinants

Question

If $A = [\begin{array}{cc} 2 & - 1 \\ - 1 & 2 \end{array}]$ , then the general solution of sinθ = |A² – 4A + 3I| is

nπ

$2 n + 1 \frac{π}{2}$

$nπ + (- 1)^{n} \frac{π}{2}$

2n π. n ε z

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$A = [\begin{array}{cc} 2 & - 1 \\ - 1 & 2 \end{array}]$
$A^{2} = [\begin{array}{cc} 2 & - 1 \\ - 1 & 2 \end{array}] [\begin{array}{cc} 2 & - 1 \\ - 1 & 2 \end{array}] = [\begin{array}{cc} 5 & - 4 \\ - 4 & 5 \end{array}]$

$A^{2} - 4 A + 3 I = [\begin{array}{cc} 5 & - 4 \\ - 4 & 5 \end{array}] + [\begin{array}{cc} - 8 & - 4 \\ - 4 & - 8 \end{array}] + [\begin{array}{cc} 3 & 0 \\ 0 & 3 \end{array}] = [\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}]$
⇒ |A² – 4A + 3I| = 0
So, sinθ = 0

Hence, the general solution is θ = nπ

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