If then Adj (A) =

Engineering

Mathematics

Inverse of a Matrix

Properties of Determinant

Transpose and Adjoint of a Matrix

Question

If $A = [\begin{array}{ccc} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}]$ then Adj (A) =

$[\begin{array}{ccc} - 1 & + 1 & - 1 \\ 8 & - 6 & 2 \\ - 5 & 3 & - 1 \end{array}]$

$[\begin{array}{ccc} - 1 & + 8 & - 5 \\ 1 & - 6 & 3 \\ - 1 & 2 & - 1 \end{array}]$

$[\begin{array}{ccc} - 1 & - 8 & 5 \\ - 1 & 6 & - 3 \\ 1 & - 2 & 1 \end{array}]$

$[\begin{array}{ccc} 1 & - 1 & 1 \\ 8 & - 6 & 2 \\ - 5 & 3 & - 1 \end{array}]$

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Cofactor of the matrix is given by

[\begin{array}{ccc} M_{11} & - M_{12} & M_{13} \\ - M_{21} & M_{22} & - M_{23} \\ M_{31} & - M_{32} & M_{33} \end{array}]

where

M_{11} = [\begin{array}{cc} 2 & 3 \\ 1 & 1 \end{array}] = 2 - 3 = - 1

M_{12} = [\begin{array}{cc} 1 & 2 \\ 3 & 1 \end{array}] = 1 - 9 = - 8

M_{13} = [\begin{array}{cc} 1 & 2 \\ 3 & 1 \end{array}] = 1 - 6 = - 5

M_{21} = [\begin{array}{cc} 1 & 2 \\ 1 & 1 \end{array}] = 1 - 2 = - 1

M_{22} = [\begin{array}{cc} 0 & 2 \\ 3 & 1 \end{array}] = 0 - 6 = - 6

M_{23} = [\begin{array}{cc} 0 & 1 \\ 3 & 1 \end{array}] = 0 - 3 = - 3

M_{31} = [\begin{array}{cc} 1 & 2 \\ 2 & 3 \end{array}] = 3 - 4 = - 1

M_{32} = [\begin{array}{cc} 0 & 2 \\ 1 & 3 \end{array}] = 0 - 2 = - 2

M_{33} = [\begin{array}{cc} 0 & 2 \\ 1 & 3 \end{array}] = 0 - 2 = - 2

substituting all the values in the Cofactor of the matrix formula we get,

= [\begin{array}{ccc} - 1 & 8 & - 5 \\ 1 & - 6 & 3 \\ - 1 & 2 & - 1 \end{array}]

Hence the adjoint of the matrix is transpose of the cofactor of the given matrix

∴ adjoint of the given matrix is

[\begin{array}{ccc} - 1 & 1 & - 1 \\ 8 & - 6 & 2 \\ - 5 & 3 & - 1 \end{array}]

Hence the answer is option B.