Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 116 Question 116 of 472 Medium NUMERICAL
Let I I I be the identity matrix of order 3 × 3 3\times 3 3 × 3 and for the matrix A = [ [ ∣ λ 2 3 4 5 6 7 − 1 2 ] ] ∣ , ∣ A ∣ = − 1 A=[[|\begin{gathered} \lambda \quad 2 \quad 3 \\ 4 \quad 5 \quad 6 \\ 7 \quad -1 \quad 2 \end{gathered}]]|,|A|=-1 A = [[ ∣ λ 2 3 4 5 6 7 − 1 2 ]] ∣ , ∣ A ∣ = − 1 . Let B B B be the inverse of the matrix a d j ( A a d j ( A 2 ) ) adj(Aadj({A}^{2})) a d j ( A a d j ( A 2 )) . Then ∣ ( λ B + I ) ∣ |(\lambda B+I)| ∣ ( λ B + I ) ∣ is equal to\_\_\_\_\_\_
Answer: 38OR-38
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B = [ a d j ( A a d j ( A 2 ) ) ] − 1 A d j ( A 2 ) = ( a d j A ) 2 ⇒ A a d j ( A 2 ) = A a d j ( A ) ⋅ ( a d j A ) = A ( ∣ A ∣ A − 1 ) 2 = ∣ A ∣ 2 ( A − 1 ) = A − 1 ⇒ B = ( a d j ( A − 1 ) ) − 1 = ( ∣ ( A − 1 ) ∣ A ) − 1 = A − 1 − 1 = − A − 1 ⇒ B = − A − 1 \begin{gathered} B={[adj(Aadj({A}^{2}))]}^{-1} \\ Adj({A}^{2})=(adjA{)}^{2}\Rightarrow Aadj({A}^{2})=Aadj(A)\cdot (adjA) \\ =A{(|A|{A}^{-1})}^{2}=|A{|}^{2}({A}^{-1})={A}^{-1} \\ \Rightarrow B={(adj({A}^{-1}))}^{-1}={(|({A}^{-1})|A)}^{-1}=\frac{{A}^{-1}}{-1}=-{A}^{-1} \\ \Rightarrow B=-{A}^{-1} \end{gathered} B = [ a d j ( A a d j ( A 2 ))] − 1 A d j ( A 2 ) = ( a d j A ) 2 ⇒ A a d j ( A 2 ) = A a d j ( A ) ⋅ ( a d j A ) = A ( ∣ A ∣ A − 1 ) 2 = ∣ A ∣ 2 ( A − 1 ) = A − 1 ⇒ B = ( a d j ( A − 1 )) − 1 = ( ∣ ( A − 1 ) ∣ A ) − 1 = − 1 A − 1 = − A − 1 ⇒ B = − A − 1
∣ A ∣ = − 1 = ∣ λ 2 3 4 5 6 7 − 1 2 ∣ = − 1 ⇒ λ = 3 ∣ 3 B + I ∣ = ∣ I − 3 A − 1 ∣ = ∣ A ∣ ∣ I − 3 A − 1 ∣ ∣ A ∣ = ∣ A − 3 I ∣ ∣ A ∣ = ∣ A − 3 I ∣ − 1 = ∣ 0 2 3 4 2 6 7 − 1 − 1 ∣ − 1 = 38 ⇒ ∣ 3 B + I ∣ = 38 \begin{gathered} |A|=-1=|\begin{array}{|c|}\lambda \quad 2 \quad 3 \\ 4 \quad 5 \quad 6 \\ 7 \quad -1 \quad 2\end{array}|=-1\Rightarrow \lambda =3 \\ |3B+I|=|I-3{A}^{-1}|=\frac{|A||I-3{A}^{-1}|}{|A|}=\frac{|A-3I|}{|A|} \\ =\frac{|A-3I|}{-1}=\frac{|\begin{array}{|c|}0 \quad 2 \quad 3 \\ 4 \quad 2 \quad 6 \\ 7 \quad -1 \quad -1\end{array}|}{-1}=38 \\ \Rightarrow |3B+I|=38 \end{gathered} ∣ A ∣ = − 1 = ∣ λ 2 3 4 5 6 7 − 1 2 ∣ = − 1 ⇒ λ = 3 ∣3 B + I ∣ = ∣ I − 3 A − 1 ∣ = ∣ A ∣ ∣ A ∣∣ I − 3 A − 1 ∣ = ∣ A ∣ ∣ A − 3 I ∣ = − 1 ∣ A − 3 I ∣ = − 1 ∣ 0 2 3 4 2 6 7 − 1 − 1 ∣ = 38 ⇒ ∣3 B + I ∣ = 38 Answer the question to unlock