Find the adjoint of the matrix
$open square brackets table row 1 cell space space space minus 1 end cell cell space space space space space space 2 end cell row 2 cell space space space space space 3 end cell cell space space space space space 5 end cell row cell negative 2 end cell cell space space space space space 0 end cell cell space space space space 1 end cell end table close square brackets.$

Solution

Let space straight A space equals space open square brackets table row 1 cell negative 1 end cell cell space space 2 end cell row 2 cell space space 3 end cell cell space space 5 end cell row cell negative 2 end cell cell space space 0 end cell cell space 1 end cell end table close square brackets therefore space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space minus 1 end cell cell space space space space 2 space end cell row 2 cell space space space space space 3 end cell cell space space 5 end cell row cell negative 2 end cell cell space space space space space 0 end cell cell space space 1 end cell end table close vertical bar

Co-factors of the elements of first row of | A | are

open vertical bar table row 3 cell space space space 5 end cell row 0 cell space space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space space space 5 space end cell row cell negative 2 end cell cell space space space 1 end cell end table close vertical bar comma space space open vertical bar table row 2 cell space space space space 3 end cell row cell negative 2 end cell cell space space space space 0 end cell end table close vertical bar

i.e. 3 – 0, – (2 + 10), 0 + 6 i.e. 3, – 12, 6 respectively.
Co-factors of the elements of second row of | A | are

negative open vertical bar table row cell negative 1 end cell cell space space 2 end cell row 0 cell space space space 1 space space end cell end table close vertical bar comma space space open vertical bar table row 1 cell space space space space 2 end cell row cell negative 2 end cell cell space space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space space minus 1 end cell row cell negative 2 end cell cell space space space space space space space 0 end cell end table close vertical bar

i.e. – (– 1 – 0), 1 + 4, – (0 – 2) i.e. 1, 5, 2 respectively.
Cofactors of the elements of third row of | A | are
$open vertical bar table row cell negative 1 end cell cell space space space space 2 end cell row 3 cell space space space 5 end cell end table close vertical bar comma space space space space space minus open vertical bar table row 1 cell space space space space 2 end cell row 2 cell space space space space 5 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space space minus 1 end cell row 2 cell space space space space space space space 3 end cell end table close vertical bar$
i.e. – 5 – 6, – (5 – 4), 3 + 2
∴ –11,–1, 5 respectively.
$therefore space space space adj. space straight A space equals space open square brackets table row 3 cell space space minus 12 end cell cell space space space 6 space end cell row 1 cell space space space 5 end cell cell space space 2 end cell row cell negative 11 end cell cell negative 1 end cell cell space space 5 end cell end table close square brackets space equals space open square brackets table row 3 cell space space space space 1 end cell cell space space space space space minus 11 end cell row cell negative 12 end cell cell space space space space 5 end cell cell space space space space minus 1 end cell row 6 cell space space space space 2 end cell cell space space space space space space space 5 end cell end table close square brackets$

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