Engineering

Mathematics

Algebra of Matrices

Inverse of a Matrix

Transpose and Adjoint of a Matrix

Question

$Adj [\begin{array}{ccc} 1 & 0 & 2 \\ - 1 & 1 & - 2 \\ 0 & 2 & 1 \end{array}] = [\begin{array}{ccc} 5 & a & - 2 \\ 1 & 1 & 0 \\ - 2 & - 2 & b \end{array}] \Rightarrow [a b] =$

[– 4 – 1]

[– 4 1]

[4 – 1]

[4 1]

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Co-factor 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} 1 & - 2 \\ 2 & 1 \end{array}] = 1 + 4 = 5

M_{12} = [\begin{array}{cc} - 1 & - 2 \\ 0 & 1 \end{array}] = - 1 - 0 = - 1

M_{13} = [\begin{array}{cc} - 1 & 1 \\ 0 & 2 \end{array}] = - 2 - 0 = - 2

M_{21} = [\begin{array}{cc} 0 & 2 \\ 2 & 1 \end{array}] = 0 - 4 = - 4

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

M_{23} = [\begin{array}{cc} 1 & 0 \\ 0 & 2 \end{array}] = 2 - 0 = 2

M_{31} = [\begin{array}{cc} 0 & 2 \\ 1 & - 2 \end{array}] = 0 - 2 = - 2

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

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

Substituting all the values in the Co-factor of the matrix formula we get,

= [\begin{array}{ccc} 5 & 1 & - 2 \\ 4 & 1 & - 2 \\ - 2 & 0 & 1 \end{array}]

The adjoint of the matrix is transpose of cofactor matrix.

∴ Adjoint of the matrix is

[\begin{array}{ccc} 5 & 4 & - 2 \\ 1 & 1 & 0 \\ - 2 & - 2 & 1 \end{array}]

Here a = 4 and b = 1

Hence, the answer is option C.