
If \[B\] is a nonsingular matrix and \[A\] is a square matrix then \[\det \left( {{B^{ - 1}}AB} \right)\] is equal to
A. \[\det \left( A \right)\]
B. \[\det \left( B \right)\]
C. \[\det \left( {{B^{ - 1}}} \right)\]
D. \[\det \left( {{A^{ - 1}}} \right)\]
Answer
510.6k+ views
- Hint: First of all, split the matrices inside the determinants by using the multiplicative properties of determinants. Then make the product of two matrices equal to a unit matrix to obtain the required answer.
Complete step-by-step solution -
Given \[B\] is a nonsingular matrix and \[A\] is a square matrix.
Now, consider \[\det \left( {{B^{ - 1}}AB} \right)\]
We know that \[\det \left( {ABC} \right) = \det \left( A \right)\det \left( B \right)\det \left( C \right)\], so we have
\[\det \left( {{B^{ - 1}}AB} \right) = \det \left( {{B^{ - 1}}} \right)\det \left( A \right)\det \left( B \right)\]
As determinants obeys commutative property of multiplication, we have
\[\det \left( {{B^{ - 1}}AB} \right) = \det \left( {{B^{ - 1}}} \right)\det \left( B \right)\det \left( A \right)\]
We know that, \[\det \left( {AB} \right) = \det \left( A \right)\det \left( B \right)\]
\[\det \left( {{B^{ - 1}}AB} \right) = \det \left( {{B^{ - 1}}B} \right)\det \left( A \right)\]
As \[{B^{ - 1}}B = I\], we have
\[\det \left( {{B^{ - 1}}AB} \right) = \det \left( I \right)\det \left( A \right)\]
We know that, \[\det \left( I \right) = 1\]
\[\therefore \det \left( {{B^{ - 1}}AB} \right) = \det \left( A \right)\]
Thus, the correct option is A. \[\det \left( A \right)\]
Note: The multiplicative property of determinants tells us that the det of product of the given matrices is equal to their product of individual det of matrices. The product of a matrix and its inverse matrix gives us a unit matrix.
Complete step-by-step solution -
Given \[B\] is a nonsingular matrix and \[A\] is a square matrix.
Now, consider \[\det \left( {{B^{ - 1}}AB} \right)\]
We know that \[\det \left( {ABC} \right) = \det \left( A \right)\det \left( B \right)\det \left( C \right)\], so we have
\[\det \left( {{B^{ - 1}}AB} \right) = \det \left( {{B^{ - 1}}} \right)\det \left( A \right)\det \left( B \right)\]
As determinants obeys commutative property of multiplication, we have
\[\det \left( {{B^{ - 1}}AB} \right) = \det \left( {{B^{ - 1}}} \right)\det \left( B \right)\det \left( A \right)\]
We know that, \[\det \left( {AB} \right) = \det \left( A \right)\det \left( B \right)\]
\[\det \left( {{B^{ - 1}}AB} \right) = \det \left( {{B^{ - 1}}B} \right)\det \left( A \right)\]
As \[{B^{ - 1}}B = I\], we have
\[\det \left( {{B^{ - 1}}AB} \right) = \det \left( I \right)\det \left( A \right)\]
We know that, \[\det \left( I \right) = 1\]
\[\therefore \det \left( {{B^{ - 1}}AB} \right) = \det \left( A \right)\]
Thus, the correct option is A. \[\det \left( A \right)\]
Note: The multiplicative property of determinants tells us that the det of product of the given matrices is equal to their product of individual det of matrices. The product of a matrix and its inverse matrix gives us a unit matrix.
Recently Updated Pages
Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Questions & Answers - Ask your doubts

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Science: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Trending doubts
Types of lever in which effort is in between fulcrum class 12 physics CBSE

Distinguish between esterification and saponification class 12 chemistry CBSE

A two input XOR Gate produces a high output only when class 12 physics CBSE

Give five points to show the significance of varia class 12 biology CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

The Coordination number of bcc fcc ccp hcp is class 12 chemistry CBSE
