If A=\left[\begin{array}{ccc}-1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 ...

CBSE Class 12 Math 2025 All Sets Solved Paper

Question : 5 of 20

Marks: +1, -0

A={[\table -1 , 0 , 0 ; 0 , 1 , 0 ; 0 , 0 , -1]},

then

A^{-1}

Solution:

A={[\table -1 , 0 , 0 ; 0 , 1 , 0 ; 0 , 0 , 1]}

| A | =-1 ≠ 0 ⇒ A^{-1}

exists

A^{-1}=\;{{\adj}/{A}}{ | A | }

{\adj} A={[\table C_{11} , C_{12} , C_{13} ; C_{21} , C_{22} , C_{23} ; C_{31} , C_{32} , C_{33}]}

where

C_{i j}=(-1)^{i+j} M_{i j}

where

M_{i j}

is the minor

C_{11}=(-1)^{1+1}{|\table 1 , 0 ; 0 , 1|}=1

C_{12}=(-1)^{1+2}{|\table 0 , 0 ; 0 , 1|}=0

C_{13}=(-1)^{1+3}{|\table 0 , 1 ; 0 , 0|}=0

C_{21}=(-1)^{2+1}{|\table 0 , 0 ; 0 , 1|}=0

C_{22}=(-1)^{2+2}{|\table -1 , 0 ; 0 , 1|}=-1

C_{23}=(-1)^{2+3}{|\table -1 , 0 ; 0 , 0|}=0

C_{31}=(-1)^{3+1}{|\table 0 , 0 ; 1 , 0|}=0

C_{32}=(-1)^{3+2}{|\table -1 , 0 ; 0 , 0|}=0

C_{33}=(-1)^{3+3}{|\table -1 , 0 ; 0 , 1|}=-1

{\adj} A={[\table 1 , 0 , 0 ; 0 , -1 , 0 ; 0 , 0 , -1]}

A^{-1}=\;{{\adj}/{A}}{ | A | }

A^{-1}=-1{[\table 1 , 0 , 0 ; 0 , -1 , 0 ; 0 , 0 , -1]}

A^{-1}={[\table -1 , 0 , 0 ; 0 , 1 , 0 ; 0 , 0 , 1]}

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