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प्रश्न
Find the adjoint of the following:`1/3[(2, 2, 1),(-2, 1, 2),(1, -2, 2)]`
उत्तर
A = `1/3[(2, 2, 1),(-2, 1, 2),(1, -2, 2)]_(3 xx 3)`
adj (kA) = `"k"^("n" - 1)` adj A
adj A = `(1/3)^2 [(+|(1, 2),(-2, 2)|, -|(-2, 2),(1, 2)|, +|(-2, 1),(1, -2)|),(-|(2, 1),(-2, 2)|, +|(2, 1),(1, 2)|, -|(2, 2),(1, -2)|),(+|(2, 1),(1, 2)|, -|(2, 1),(-2, 2)|, +|(2, 2),(-2, 1)|)]^"T"`
= `1/9[(+(2 + 4), -(-4 - 2), +(4 - 1)),(-4(4 + 2), +(4 - 1), -(-4 - 2)),(+(4 - 1), -(4 + 2), +(2 + 4))]^"T"`
= `1/9[(6, 6, 3),(-6, 3, 6),(, -6, 6)]^"T"`
= `1/9[(6, -6, 3),(6, 3, -6),(3, 6, 6)]`
= `1/3[(2, -2, 1),(2, 1, 2),(1, 2, 2)]`
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