Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 133 Question 133 of 472 Medium NUMERICAL
Let A = [ [ ∣ 2 0 1 1 1 0 1 0 1 ] ] ∣ , B = [ B 1 , B 2 , B 3 ] A=[[|\begin{gathered} 2 \quad 0 \quad 1 \\ 1 \quad 1 \quad 0 \\ 1 \quad 0 \quad 1 \end{gathered}]]|,B=[{B}_{1},{B}_{2},{B}_{3}] A = [[ ∣ 2 0 1 1 1 0 1 0 1 ]] ∣ , B = [ B 1 , B 2 , B 3 ] , where B 1 , B 2 , B 3 {B}_{1},{B}_{2},{B}_{3} B 1 , B 2 , B 3 are column matrics, and
A B 1 = [ [ ∣ 1 0 0 ] ] ∣ , A B 2 = [ [ ∣ 2 3 0 ] ] ∣ , A B 3 = [ [ ∣ 3 2 1 ] ] ∣ {AB}_{1}=[[|\begin{gathered} 1 \\ 0 \\ 0 \end{gathered}]]|,{AB}_{2}=[[|\begin{gathered} 2 \\ 3 \\ 0 \end{gathered}]]|,\, {AB}_{3}=[[|\begin{gathered} 3 \\ 2 \\ 1 \end{gathered}]]| A B 1 = [[ ∣ 1 0 0 ]] ∣ , A B 2 = [[ ∣ 2 3 0 ]] ∣ , A B 3 = [[ ∣ 3 2 1 ]] ∣
If α = ∣ B ∣ \alpha =|B| α = ∣ B ∣ and β \beta β is the sum of all the diagonal elements of B B B , then α 3 + β 3 {\alpha }^{3}+{\beta }^{3} α 3 + β 3 is equal to \_\_\_\_\_\_\_\_\_\_\_\_.
Answer: 28
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A = [ [ ∣ 2 0 1 1 1 0 1 0 1 ] ] ∣ B = [ B 1 , B 2 , B 3 ] A=[[|\begin{gathered} 2 \quad 0 \quad 1 \\ 1 \quad 1 \quad 0 \\ 1 \quad 0 \quad 1 \end{gathered}]]|\, B=[{B}_{1},{B}_{2},{B}_{3}] A = [[ ∣ 2 0 1 1 1 0 1 0 1 ]] ∣ B = [ B 1 , B 2 , B 3 ]
B 1 = [ [ ∣ x 1 y 1 z 1 ] ] ∣ , B 2 = [ [ ∣ x 2 y 2 z 2 ] ] ∣ , B 3 = [ [ ∣ x 3 y 3 z 3 ] ] ∣ {B}_{1}=[[|\begin{gathered} {x}_{1} \\ {y}_{1} \\ {z}_{1} \end{gathered}]]|,\, {B}_{2}=[[|\begin{gathered} {x}_{2} \\ {y}_{2} \\ {z}_{2} \end{gathered}]]|,\, {B}_{3}=[[|\begin{gathered} {x}_{3} \\ {y}_{3} \\ {z}_{3} \end{gathered}]]| B 1 = [[ ∣ x 1 y 1 z 1 ]] ∣ , B 2 = [[ ∣ x 2 y 2 z 2 ]] ∣ , B 3 = [[ ∣ x 3 y 3 z 3 ]] ∣
A B 1 = [ [ ∣ 2 0 1 1 1 0 1 0 1 ] ] ∣ [ [ ∣ x 1 y 1 z 1 ] ] ∣ = [ [ ∣ 1 0 0 ] ] ∣ {AB}_{1}=[[|\begin{gathered} 2 \quad 0 \quad 1 \\ 1 \quad 1 \quad 0 \\ 1 \quad 0 \quad 1 \end{gathered}]]|[[|\begin{gathered} {x}_{1} \\ {y}_{1} \\ {z}_{1} \end{gathered}]]|=[[|\begin{gathered} 1 \\ 0 \\ 0 \end{gathered}]]| A B 1 = [[ ∣ 2 0 1 1 1 0 1 0 1 ]] ∣ [[ ∣ x 1 y 1 z 1 ]] ∣ = [[ ∣ 1 0 0 ]] ∣
x 1 = 1 , y 1 = − 1 , z 1 = − 1 A B 2 = [ [ ∣ 2 0 1 1 1 0 1 0 1 ] ] ∣ [ [ ∣ x 2 y 2 z 2 ] ] ∣ = [ [ ∣ 2 3 0 ] ] ∣ x 2 = 2 , y 2 = 1 , z 2 = − 2 A B 3 = [ [ ∣ 2 0 1 1 1 0 1 0 1 ] ] ∣ [ [ ∣ x 3 y 3 z 3 ] ] ∣ = [ [ ∣ 3 2 1 ] ] ∣ x 3 = 2 , y 3 = 0 , z 3 = − 1 B = [ [ ∣ 1 2 2 − 1 1 0 − 1 − 2 − 1 ] ] ∣ α = ∣ B ∣ = 3 β = 1 α 3 + β 3 = 27 + 1 = 28 \begin{gathered} {x}_{1}=1,{y}_{1}=-1,{z}_{1}=-1 \\ {AB}_{2}=[[|\begin{array}{|c|}2 \quad 0 \quad 1 \\ 1 \quad 1 \quad 0 \\ 1 \quad 0 \quad 1\end{array}]]|[[|\begin{array}{|c|}{x}_{2} \\ {y}_{2} \\ {z}_{2}\end{array}]]|=[[|\begin{array}{|c|}2 \\ 3 \\ 0\end{array}]]| \\ {x}_{2}=2,{y}_{2}=1,{z}_{2}=-2 \\ {AB}_{3}=[[|\begin{array}{|c|}2 \quad 0 \quad 1 \\ 1 \quad 1 \quad 0 \\ 1 \quad 0 \quad 1\end{array}]]|[[|\begin{array}{|c|}{x}_{3} \\ {y}_{3} \\ {z}_{3}\end{array}]]|=[[|\begin{array}{|c|}3 \\ 2 \\ 1\end{array}]]| \\ {x}_{3}=2,{y}_{3}=0,{z}_{3}=-1 \\ B=[[|\begin{array}{|c|}1 \quad 2 \quad 2 \\ -1 \quad 1 \quad 0 \\ -1 \quad -2 \quad -1\end{array}]]| \\ \alpha =|B|=3 \\ \beta =1 \\ {\alpha }^{3}+{\beta }^{3}=27+1=28 \end{gathered} x 1 = 1 , y 1 = − 1 , z 1 = − 1 A B 2 = [[ ∣ 2 0 1 1 1 0 1 0 1 ]] ∣ [[ ∣ x 2 y 2 z 2 ]] ∣ = [[ ∣ 2 3 0 ]] ∣ x 2 = 2 , y 2 = 1 , z 2 = − 2 A B 3 = [[ ∣ 2 0 1 1 1 0 1 0 1 ]] ∣ [[ ∣ x 3 y 3 z 3 ]] ∣ = [[ ∣ 3 2 1 ]] ∣ x 3 = 2 , y 3 = 0 , z 3 = − 1 B = [[ ∣ 1 2 2 − 1 1 0 − 1 − 2 − 1 ]] ∣ α = ∣ B ∣ = 3 β = 1 α 3 + β 3 = 27 + 1 = 28 Answer the question to unlock