Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 119 Question 119 of 472 Medium NUMERICAL
Let A be a 3 × 3 3\times 3 3 × 3 matrix such that X T A X = O {X}^{T}AX=O X T A X = O for all nonzero 3 × 1 3\times 1 3 × 1 matrices X = [ [ ∣ x y z ] ] ∣ X=[[|\begin{gathered} x \\ y \\ z \end{gathered}]]| X = [[ ∣ x y z ]] ∣ . If A [ [ ∣ 1 1 1 ] ] ∣ = [ [ ∣ 1 4 − 5 ] ] ∣ , A [ [ ∣ 1 2 1 ] ] ∣ = [ [ ∣ 0 4 − 8 ] ] ∣ A[[|\begin{gathered} 1 \\ 1 \\ 1 \end{gathered}]]|=[[|\begin{gathered} 1 \\ 4 \\ -5 \end{gathered}]]|,A[[|\begin{gathered} 1 \\ 2 \\ 1 \end{gathered}]]|=[[|\begin{gathered} 0 \\ 4 \\ -8 \end{gathered}]]| A [[ ∣ 1 1 1 ]] ∣ = [[ ∣ 1 4 − 5 ]] ∣ , A [[ ∣ 1 2 1 ]] ∣ = [[ ∣ 0 4 − 8 ]] ∣ , and det ( a d j ( 2 ( A + I ) ) ) = 2 α 3 β 5 γ , α , β , γ ∈ N \det (adj(2(A+I)))={2}^{\alpha }{3}^{\beta }{5}^{\gamma },\alpha ,\beta ,\gamma \in N det ( a d j ( 2 ( A + I ))) = 2 α 3 β 5 γ , α , β , γ ∈ N , then α 2 + β 2 + γ 2 {\alpha }^{2}+{\beta }^{2}+{\gamma }^{2} α 2 + β 2 + γ 2 is
Answer: 44
Solutions · 1 Post your solution Official solution
X T A X = 0 ( x y z ) ( ( ∣ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ) ) ∣ ( ( ∣ x y z ) ) ∣ = 0 ( x y z ) ( ( ∣ a 1 x + a 2 y + a 3 z b 1 x + b 2 y + b 3 z c 1 x + c 2 y + c 3 z ) ) ∣ = 0 \begin{gathered} {X}^{T}AX=0 \\ (xyz)((|\begin{array}{|c|}{a}_{1} \quad {a}_{2} \quad {a}_{3} \\ {b}_{1} \quad {b}_{2} \quad {b}_{3} \\ {c}_{1} \quad {c}_{2} \quad {c}_{3}\end{array}))|((|\begin{array}{|c|}x \\ y \\ z\end{array}))|=0 \\ (xyz)((|\begin{array}{|c|}{a}_{1}x+{a}_{2}y+{a}_{3}z \\ {b}_{1}x+{b}_{2}y+{b}_{3}z \\ {c}_{1}x+{c}_{2}y+{c}_{3}z\end{array}))|=0 \end{gathered} X T A X = 0 ( x y z ) (( ∣ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 )) ∣ (( ∣ x y z )) ∣ = 0 ( x y z ) (( ∣ a 1 x + a 2 y + a 3 z b 1 x + b 2 y + b 3 z c 1 x + c 2 y + c 3 z )) ∣ = 0
x ( a 1 x + a 2 y + a 3 z ) + y ( b 1 x + b 2 y + b 3 z ) + z ( c 1 x + c 2 y + c 3 z ) = 0 a 1 = 0 , b 2 = 0 c 3 = 0 a 2 + b 1 = 0 , a 3 + c 1 = 0 , b 3 = c 2 = 0 \begin{gathered} x({a}_{1}x+{a}_{2}y+{a}_{3}z)+y({b}_{1}x+{b}_{2}y+{b}_{3}z) \\ +z({c}_{1}x+{c}_{2}y+{c}_{3}z)=0 \\ {a}_{1}=0,{b}_{2}=0{c}_{3}=0 \\ {a}_{2}+{b}_{1}=0,{a}_{3}+{c}_{1}=0,{b}_{3}={c}_{2}=0 \end{gathered} x ( a 1 x + a 2 y + a 3 z ) + y ( b 1 x + b 2 y + b 3 z ) + z ( c 1 x + c 2 y + c 3 z ) = 0 a 1 = 0 , b 2 = 0 c 3 = 0 a 2 + b 1 = 0 , a 3 + c 1 = 0 , b 3 = c 2 = 0
A = A = ( ( ∣ 0 x y − x 0 z − y − z 0 ) ) ∣ ; A = ( ( ∣ 1 1 1 ) ) ∣ = ( ( ∣ 1 4 − 5 ) ) ∣ ⇒ A = ( ( ∣ 0 x y − x 0 z − y − z 0 ) ) ∣ ( ( ∣ 1 1 1 ) ) ∣ = ( ( ∣ 1 4 − 5 ) ) ∣ \begin{gathered} A= \\ A=((|\begin{array}{|c|}0 \quad x \quad y \\ -x \quad 0 \quad z \\ -y \quad -z \quad 0\end{array}))|;\, A=((|\begin{array}{|c|}1 \\ 1 \\ 1\end{array}))|=((|\begin{array}{|c|}1 \\ 4 \\ -5\end{array}))| \\ \Rightarrow A=((|\begin{array}{|c|}0 \quad x \quad y \\ -x \quad 0 \quad z \\ -y \quad -z \quad 0\end{array}))|((|\begin{array}{|c|}1 \\ 1 \\ 1\end{array}))|=((|\begin{array}{|c|}1 \\ 4 \\ -5\end{array}))| \end{gathered} A = A = (( ∣ 0 x y − x 0 z − y − z 0 )) ∣ ; A = (( ∣ 1 1 1 )) ∣ = (( ∣ 1 4 − 5 )) ∣ ⇒ A = (( ∣ 0 x y − x 0 z − y − z 0 )) ∣ (( ∣ 1 1 1 )) ∣ = (( ∣ 1 4 − 5 )) ∣
x + y = 1 − x + z = 4 y + z = 5 ( ( ∣ 0 x y − x 0 z − y − z 0 ) ) ∣ ( ( ∣ 1 2 1 ) ) ∣ = ( ( ∣ 1 4 − 8 ) ) ∣ \begin{gathered} x+y=1 \\ -x+z=4 \\ y+z=5 \\ ((|\begin{array}{|c|}0 \quad x \quad y \\ -x \quad 0 \quad z \\ -y \quad -z \quad 0\end{array}))|((|\begin{array}{|c|}1 \\ 2 \\ 1\end{array}))|=((|\begin{array}{|c|}1 \\ 4 \\ -8\end{array}))| \end{gathered} x + y = 1 − x + z = 4 y + z = 5 (( ∣ 0 x y − x 0 z − y − z 0 )) ∣ (( ∣ 1 2 1 )) ∣ = (( ∣ 1 4 − 8 )) ∣
2 x + y = 0 x = − 1 − x + z = 4 y = 2 − y − 2 z = − 8 z = 3 \begin{gathered} 2x+y=0 x=-1 \\ -x+z=4 y=2 \\ -y-2z=-8 z=3 \end{gathered} 2 x + y = 0 x = − 1 − x + z = 4 y = 2 − y − 2 z = − 8 z = 3
A = ( ( ∣ 0 − 1 2 1 0 3 − 2 − 3 0 ) ) ∣ 2 ( A + I ) = ( ( ∣ 2 − 2 4 2 2 6 − 2 − 6 2 ) ) ∣ 2 ( A + I ) = 120 ⇒ det ∣ a d i ( 2 ( A + I ) ) ∣ = 120 2 = 2 6 ⋅ 3 2 ⋅ 5 2 α = 6 , β = 2 , γ = 2 \begin{gathered} A=((|\begin{array}{|c|}0 \quad -1 \quad 2 \\ 1 \quad 0 \quad 3 \\ -2 \quad -3 \quad 0\end{array}))| \\ 2(A+I)=((|\begin{array}{|c|}2 \quad -2 \quad 4 \\ 2 \quad 2 \quad 6 \\ -2 \quad -6 \quad 2\end{array}))| \\ 2(A+I)=120\Rightarrow \det |adi(2(A+I))| \\ ={120}^{2}={2}^{6}\cdot {3}^{2}\cdot {5}^{2} \\ \alpha =6,\beta =2,\gamma =2 \end{gathered} A = (( ∣ 0 − 1 2 1 0 3 − 2 − 3 0 )) ∣ 2 ( A + I ) = (( ∣ 2 − 2 4 2 2 6 − 2 − 6 2 )) ∣ 2 ( A + I ) = 120 ⇒ det ∣ a d i ( 2 ( A + I )) ∣ = 120 2 = 2 6 ⋅ 3 2 ⋅ 5 2 α = 6 , β = 2 , γ = 2 Answer the question to unlock