Compute the adjoint of the following matrice :Verify that (adjA)A = |A|I = A(adjA) for the above matrice.

Engineering

Mathematics

Introduction to Determinants

Properties of Determinant

Transpose and Adjoint of a Matrix

Question

Compute the adjoint of the following matrice :
$[\begin{array}{ccc} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{array}]$
Verify that (adjA)A = |A|I = A(adjA) for the above matrice.

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$A = [\begin{array}{ccc} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{array}]$

We know that, A_ij = (–1)ⁱ⁺^jM_ij

$A_{11} = (- 1)^{2} |\begin{array}{ll} 1 0 \\ 1 3 \end{array}| = 3; A_{12} = (- 1)^{3} |\begin{array}{ll} 5 0 \\ 1 3 \end{array}| = - 15; A_{13} = (- 1)^{4} |\begin{array}{ll} 5 1 \\ 1 1 \end{array}| = 4$

$A_{21} = (- 1)^{3} |\begin{array}{cc} 0 & - 1 \\ 1 & 3 \end{array}| = - 1; A_{22} = (- 1)^{4} |\begin{array}{cc} 2 & - 1 \\ 1 & 3 \end{array}| = 7; A_{23} = (- 1)^{5} |\begin{array}{ll} 2 0 \\ 1 1 \end{array}| = - 2$

$A_{31} = (- 1)^{4} |\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}| = 1; A_{32} = (- 1)^{5} |\begin{array}{cc} 2 & - 1 \\ 5 & 0 \end{array}| = - 5; A_{33} = (- 1)^{6} |\begin{array}{cc} 2 & 0 \\ 5 & 1 \end{array}| = 2$

We have,

$(adj A) = [\begin{array}{lll} A_{11} A_{12} A_{13} \\ A_{21} A_{22} A_{23} \\ A_{21} A_{32} A_{33} \end{array}] = [\begin{array}{ccc} 3 & - 15 & 4 \\ - 1 & 7 & - 2 \\ 1 & - 5 & 2 \end{array}] = [\begin{array}{ccc} 3 & - 1 & 1 \\ - 15 & 7 & - 5 \\ 4 & - 2 & 2 \end{array}]$

$∴ (adj A) = [\begin{array}{ccc} 3 & - 1 & 1 \\ - 15 & 7 & - 5 \\ 4 & - 2 & 2 \end{array}]$

Now, we have to verify that : (adjA) = |A|I = A(adjA)

$(adj A) A = [\begin{array}{ccc} 3 & - 1 & 1 \\ - 15 & 7 & - 5 \\ 4 & - 2 & 2 \end{array}] [\begin{array}{ccc} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{array}] = [\begin{array}{ccc} 6 - 5 + 1 & - 1 + 1 & - 3 + 3 \\ - 30 + 35 + 5 & 7 - 5 & 15 - 15 \\ 8 - 10 + 2 & - 2 + 2 & - 4 + 6 \end{array}]$

$(adj A) A = [\begin{array}{lll} 2 0 0 \\ 0 2 0 \\ 0 0 2 \end{array}]$ ....(1)

$| A | = |\begin{array}{ccc} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{array}|$

= 2(3 – 0) – 0(15 – 0) + (1) 5 – 1

= 6 – 4

= +2

∴ |A|I = 2I

$∴ | A | I = [\begin{array}{lll} 2 0 0 \\ 0 2 0 \\ 0 0 2 \end{array}]$ ....(2)

$A (adj A) = [\begin{array}{ccc} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{array}] [\begin{array}{ccc} 3 & - 1 & 1 \\ - 15 & 7 & - 5 \\ 4 & - 2 & 2 \end{array}] = [\begin{array}{ccc} 6 - 4 & - 2 + 2 & 2 - 2 \\ 15 - 15 & - 5 + 7 & 5 - 5 \\ 3 - 15 + 12 & - 1 + 7 - 6 & 1 - 5 + 6 \end{array}]$

$∴ A (adj A) = [\begin{array}{lll} 2 0 0 \\ 0 2 0 \\ 0 0 2 \end{array}]$ ....(3)

From (1), (2) and (3) we have,

(adjA)A = |A|I = A(adjA)