Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 312 Question 312 of 472 Medium MCQ
Let \[A = \left( { ( 1 − 1 1 2 1 − 3 1 1 1 ) \begin{pmatrix} 1 & { - 1} & 1 \\
2 & 1 & { - 3} \\
1 & 1 & 1 \\ \end{pmatrix} 1 2 1 − 1 1 1 1 − 3 1 } \right).\] and 10 10 10 \[B = \left( { ( 4 2 2 − 5 0 α 1 − 2 3 ) \begin{pmatrix} 4 & 2 & 2 \\
{ - 5} & 0 & \alpha \\
1 & { - 2} & 3 \\ \end{pmatrix} 4 − 5 1 2 0 − 2 2 α 3 } \right)\]. if B B B is
the inverse of matrix A A A , then \[\alpha \] is
Solutions · 1 Post your solution Official solution
Given that 10 B 10B 10 B \[\,\,\, = \left[ { ( 4 2 2 − 5 0 α 1 − 2 3 ) \begin{pmatrix} 4 & 2 & 2 \\
{ - 5} & 0 & \alpha \\
1 & { - 2} & 3 \\
\end{pmatrix} 4 − 5 1 2 0 − 2 2 α 3 } \right]\]
\[ \Rightarrow B = {1 \over {10}}\left[ { ( 4 2 2 − 5 0 α 1 − 2 3 ) \begin{pmatrix} 4 & 2 & 2 \\
{ - 5} & 0 & \alpha \\
1 & { - 2} & 3 \\
\end{pmatrix} 4 − 5 1 2 0 − 2 2 α 3 } \right]\]
Also since, \[B = {A^{ - 1}} \Rightarrow AB = I\]
\[ \Rightarrow {1 \over {10}}\left[ { ( 1 − 1 1 2 1 − 3 1 1 1 ) \begin{pmatrix} 1 & { - 1} & 1 \\
2 & 1 & { - 3} \\
1 & 1 & 1 \\
\end{pmatrix} 1 2 1 − 1 1 1 1 − 3 1 } \right]\left[ { ( 4 2 2 − 5 0 α 1 − 2 3 ) \begin{pmatrix} 4 & 2 & 2 \\
{ - 5} & 0 & \alpha \\
1 & { - 2} & 3 \\
\end{pmatrix} 4 − 5 1 2 0 − 2 2 α 3 } \right] = \left[ { ( 1 0 0 0 1 0 0 0 1 ) \begin{pmatrix} 1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix} 1 0 0 0 1 0 0 0 1 } \right]\]
\[ \Rightarrow {1 \over {10}}\left[ { ( 10 0 5 − 2 0 10 − 5 + α 0 0 5 + α ) \begin{pmatrix} {10} & 0 & {5 - 2} \\
0 & {10} & { - 5 + \alpha } \\
0 & 0 & {5 + \alpha } \\
\end{pmatrix} 10 0 0 0 10 0 5 − 2 − 5 + α 5 + α } \right] = \left[ { ( 1 0 0 0 1 0 0 0 1 ) \begin{pmatrix} 1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix} 1 0 0 0 1 0 0 0 1 } \right]\]
\[ \Rightarrow {{5 - \alpha } \over {10}} = 0\]
\[ \Rightarrow \alpha = 5\] Answer the question to unlock