then Adj(A) =

Engineering

Mathematics

Inverse of a Matrix

Transpose and Adjoint of a Matrix

Question

$A = [\begin{array}{ccc} - 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{array}]$ then Adj(A) =

A^T

A^–1

– A^T

3A^T

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Given : $A = [\begin{array}{ccc} - 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{array}]$

To find : adj(A) = ?

$A^{⊤} = [\begin{array}{ccc} - 1 & 2 & 2 \\ - 2 & 1 & - 2 \\ - 2 & - 2 & 1 \end{array}]$

$M_{11} = |\begin{array}{cc} 1 & - 2 \\ - 2 & 1 \end{array}| = 1 - 4 = - 3$

$M_{12} = |\begin{array}{cc} + 2 & - 2 \\ + 2 & 1 \end{array}| = + 2 + 4 = + 6$

$M_{13} = |\begin{array}{rr} 2 1 \\ 2 - 2 \end{array}| = - 4 - 2 = - 6$

$M_{21} = |\begin{array}{cc} - 2 & - 2 \\ - 2 & 1 \end{array}| = - 2 - 4 = - 6$

$M_{22} = |\begin{array}{cc} - 1 & - 2 \\ 2 & 1 \end{array}| = - 1 + 4 = 3$

$M_{23} = |\begin{array}{rr} - 1 - 2 \\ 2 - 2 \end{array}| = 2 + 4 = 6$

$M_{31} = |\begin{array}{cc} - 2 & - 2 \\ 1 & - 2 \end{array}| = 4 + 2 = 6$

$M_{32} = |\begin{array}{cc} - 1 & - 2 \\ 2 & - 2 \end{array}| = 2 + 4 = 6$

$M_{33} = |\begin{array}{cc} - 1 & - 2 \\ 2 & 1 \end{array}| = - 1 + 4 = 3$

$C = [\begin{array}{ccc} - 3 & 6 & - 6 \\ - 6 & 3 & 6 \\ 6 & 6 & 3 \end{array}]$

$Adj = [\begin{array}{ccc} - 3 & - 6 & 6 \\ 6 & 3 & 6 \\ - 6 & 6 & 3 \end{array}]$

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