Answer
Verified
418.2k+ 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
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE