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Associative Law of Binary Operation

Compute the a...

Question

Compute the adjoint of each of the following matrices:

(i) $[\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}]$

(ii) $[\begin{matrix} 1 & 2 & 5 \\ 2 & 3 & 1 \\ - 1 & 1 & 1 \end{matrix}]$

(iii) $[\begin{matrix} 2 & - 1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & - 1 \end{matrix}]$

(iv) $[\begin{matrix} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{matrix}]$

(v) $[\begin{matrix} 1 & 2 & 3 \\ 0 & 5 & 0 \\ 2 & 4 & 3 \end{matrix}]$

Verify that (adj A) A = |A| I = A (adj A) for the above matrices.

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Solution

$(i) A = [\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}] Now, C_{11} = |\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix}| = - 3, C_{12} = - |\begin{matrix} 2 & 2 \\ 2 & 1 \end{matrix}| = 2 and C_{13} = |\begin{matrix} 2 & 1 \\ 2 & 2 \end{matrix}| = 2 C_{21} = - |\begin{matrix} 2 & 2 \\ 2 & 1 \end{matrix}| = 2, C_{22} = |\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix}| = - 3 and C_{23} = - |\begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix}| = 2 C_{31} = |\begin{matrix} 2 & 2 \\ 1 & 2 \end{matrix}| = 2, C_{32} = - |\begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix}| = 2 and C_{33} = |\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix}| = - 3 ∴ adj A = {[\begin{matrix} - 3 & 2 & 2 \\ 2 & - 3 & 2 \\ 2 & 2 & - 3 \end{matrix}]}^{T} = [\begin{matrix} - 3 & 2 & 2 \\ 2 & - 3 & 2 \\ 2 & 2 & - 3 \end{matrix}] and (adj A) A = [\begin{matrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{matrix}] Now, |A| = 5 ∴ |A| I = [\begin{matrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{matrix}] and A (adj A) = [\begin{matrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{matrix}] Thus, (adj A) A = |A| I = A (adj A) (ii) B = [\begin{matrix} 1 & 2 & 5 \\ 2 & 3 & 1 \\ - 1 & 1 & 1 \end{matrix}] Now, C_{11} = |\begin{matrix} 3 & 1 \\ 1 & 1 \end{matrix}| = 2, C_{12} = - |\begin{matrix} 2 & 1 \\ - 1 & 1 \end{matrix}| = - 3 and C_{13} = |\begin{matrix} 2 & 3 \\ - 1 & 1 \end{matrix}| = 5 C_{21} = - |\begin{matrix} 2 & 5 \\ 1 & 1 \end{matrix}| = 3, C_{22} = |\begin{matrix} 1 & 5 \\ - 1 & 1 \end{matrix}| = 6 and C_{23} = - |\begin{matrix} 1 & 2 \\ - 1 & 1 \end{matrix}| = - 3 C_{31} = |\begin{matrix} 2 & 5 \\ 3 & 1 \end{matrix}| = - 13, C_{32} = - |\begin{matrix} 1 & 5 \\ 2 & 1 \end{matrix}| = 9 and C_{33} = |\begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix}| = - 1 ∴ adj B = {[\begin{matrix} 2 & - 3 & 5 \\ 3 & 6 & - 3 \\ - 13 & 9 & - 1 \end{matrix}]}^{T} = [\begin{matrix} 2 & 3 & - 13 \\ - 3 & 6 & 9 \\ 5 & - 3 & - 1 \end{matrix}] (adj B) B = [\begin{matrix} 21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21 \end{matrix}] and |B| = 21 ∴ |B| I = [\begin{matrix} 21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21 \end{matrix}] and B (adj B) = [\begin{matrix} 21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21 \end{matrix}] Thus, (adj A) A = |A| I = A (adj A) (iii) C = [\begin{matrix} 2 & - 1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & - 1 \end{matrix}] Now, C_{11} = |\begin{matrix} 2 & 5 \\ 4 & - 1 \end{matrix}| = - 22, C_{12} = - |\begin{matrix} 4 & 5 \\ 0 & - 1 \end{matrix}| = 4 and C_{13} = |\begin{matrix} 4 & 2 \\ 0 & 4 \end{matrix}| = 16 C_{21} = - |\begin{matrix} - 1 & 3 \\ 4 & - 1 \end{matrix}| = 11, C_{22} = |\begin{matrix} 2 & 3 \\ 0 & - 1 \end{matrix}| = - 2 and C_{23} = - |\begin{matrix} 2 & - 1 \\ 0 & 4 \end{matrix}| = - 8 C_{31} = |\begin{matrix} - 1 & 3 \\ 2 & 5 \end{matrix}| = - 11, C_{32} = - |\begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix}| = 2 and C_{33} = |\begin{matrix} 2 & - 1 \\ 4 & 2 \end{matrix}| = 8 ∴ adj C = {[\begin{matrix} - 22 & 4 & 16 \\ 11 & - 2 & - 8 \\ - 11 & 2 & 8 \end{matrix}]}^{T} = [\begin{matrix} - 22 & 11 & - 11 \\ 4 & - 2 & 2 \\ 16 & - 8 & 8 \end{matrix}] (adj C) C = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] and |C| = 0 ∴ |C| I = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] and C (adj C) = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] Thus, (adj A) A = |A| I = A (adj A) (iv) D = [\begin{matrix} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{matrix}] Now, C_{11} = |\begin{matrix} 1 & 0 \\ 1 & 3 \end{matrix}| = 3, C_{12} = - |\begin{matrix} 5 & 0 \\ 1 & 3 \end{matrix}| = - 15 and C_{13} = |\begin{matrix} 5 & 1 \\ 1 & 1 \end{matrix}| = 4 C_{21} = - |\begin{matrix} 0 & - 1 \\ 1 & 3 \end{matrix}| = - 1, C_{22} = |\begin{matrix} 2 & - 1 \\ 1 & 3 \end{matrix}| = 7 and C_{23} = - |\begin{matrix} 2 & 0 \\ 1 & 1 \end{matrix}| = - 2 C_{31} = |\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}| = 1, C_{32} = - |\begin{matrix} 2 & - 1 \\ 5 & 0 \end{matrix}| = - 5 and C_{33} = |\begin{matrix} 2 & 0 \\ 5 & 1 \end{matrix}| = 2 ∴ adj D = {[\begin{matrix} 3 & - 15 & 4 \\ - 1 & 7 & - 2 \\ 1 & - 5 & 2 \end{matrix}]}^{T} = [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 7 & - 5 \\ 4 & - 2 & 2 \end{matrix}] (adj D) D = [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}] and |D| = 2 ∴ |D| I = [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}] and D (adj D) = [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}] Thus, (adj A) A = |A| I = A (adj A) (v) E = [\begin{matrix} 1 & 2 & 3 \\ 0 & 5 & 0 \\ 2 & 4 & 3 \end{matrix}] Now, C_{11} = |\begin{matrix} 5 & 0 \\ 4 & 3 \end{matrix}| = 15, C_{12} = - |\begin{matrix} 0 & 0 \\ 2 & 3 \end{matrix}| = 0 and C_{13} = |\begin{matrix} 0 & 5 \\ 2 & 4 \end{matrix}| = - 10 C_{21} = - |\begin{matrix} 2 & 3 \\ 4 & 3 \end{matrix}| = 6, C_{22} = |\begin{matrix} 1 & 3 \\ 2 & 3 \end{matrix}| = - 3 and C_{23} = - |\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}| = 0 C_{31} = |\begin{matrix} 2 & 3 \\ 5 & 0 \end{matrix}| = - 15, C_{32} = - |\begin{matrix} 1 & 3 \\ 0 & 0 \end{matrix}| = 0 and C_{33} = |\begin{matrix} 1 & 2 \\ 0 & 5 \end{matrix}| = 5 ∴ adj E = {[\begin{matrix} 15 & 0 & - 10 \\ 6 & - 3 & 0 \\ - 15 & 0 & 5 \end{matrix}]}^{T} = [\begin{matrix} 15 & 6 & - 15 \\ 0 & - 3 & 0 \\ - 10 & 0 & 5 \end{matrix}] (adj E) E = [\begin{matrix} - 15 & 0 & 0 \\ 0 & - 15 & 0 \\ 0 & 0 & - 15 \end{matrix}] ∴ |E| = - 15 ∴ |E| I = [\begin{matrix} - 15 & 0 & 0 \\ 0 & - 15 & 0 \\ 0 & 0 & - 15 \end{matrix}] and E (adj E) = [\begin{matrix} - 15 & 0 & 0 \\ 0 & - 15 & 0 \\ 0 & 0 & - 15 \end{matrix}] Thus, (adj A) A = |A| I = A (adj A)$

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