
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
512.4k+ 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 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
Give simple chemical tests to distinguish between the class 12 chemistry CBSE

How was the Civil Disobedience Movement different from class 12 social science CBSE

India is the secondlargest producer of AJute Bcotton class 12 biology CBSE

Define peptide linkage class 12 chemistry CBSE

How is democracy better than other forms of government class 12 social science CBSE

Differentiate between lanthanoids and actinoids class 12 chemistry CBSE
