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Question: If A and B are invertible matrices of order 3. \(\left| A \right|\)=2 and\(\left| {{{\left( {AB} \ri...

If A and B are invertible matrices of order 3. ∣A∣\left| A \right|=2 and∣(AB)−1∣=−16\left| {{{\left( {AB} \right)}^{ - 1}}} \right| = \dfrac{{ - 1}}{6}, find ∣B∣\left| B \right|

Explanation

Solution

Hint – In this question use the concept that since A and B are invertible hence ∣A∣∣A−1∣=1\left| A \right|\left| {{A^{ - 1}}} \right| = 1 and ∣B∣∣B−1∣=1\left| B \right|\left| {{B^{ - 1}}} \right| = 1. Then use the property of invertible matrix that (AB)−1=B−1A−1{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}and the property of determinant that ∣BA∣=∣B∣∣A∣\left| {BA} \right| = \left| B \right|\left| A \right|this will help getting the value of∣B∣\left| B \right|.
Complete step-by-step answer:
If A and B are invertible matrices then the inverse of A and B exist.
⇒∣A∣∣A−1∣=1\Rightarrow \left| A \right|\left| {{A^{ - 1}}} \right| = 1..................... (1)
And
⇒∣B∣∣B−1∣=1\Rightarrow \left| B \right|\left| {{B^{ - 1}}} \right| = 1.................... (2)
Now it is given that ∣A∣=2\left| A \right| = 2 ................ (3)
And∣(AB)−1∣=−16\left| {{{\left( {AB} \right)}^{ - 1}}} \right| = - \dfrac{1}{6}................... (4)
And if A and B are invertible then AB is invertible and, (AB)−1=B−1A−1{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}
⇒∣(AB)−1∣=∣B−1A−1∣\Rightarrow \left| {{{\left( {AB} \right)}^{ - 1}}} \right| = \left| {{B^{ - 1}}{A^{ - 1}}} \right|
And we all know that ∣BA∣=∣B∣∣A∣\left| {BA} \right| = \left| B \right|\left| A \right| and from equation (4) we have,
⇒∣(AB)−1∣=∣B−1A−1∣=∣B−1∣∣A−1∣=−16\Rightarrow \left| {{{\left( {AB} \right)}^{ - 1}}} \right| = \left| {{B^{ - 1}}{A^{ - 1}}} \right| = \left| {{B^{ - 1}}} \right|\left| {{A^{ - 1}}} \right| = - \dfrac{1}{6}............... (5)
Now from equation (1) and (3) we have,
⇒2∣A−1∣=1\Rightarrow 2\left| {{A^{ - 1}}} \right| = 1
⇒∣A−1∣=12\Rightarrow \left| {{A^{ - 1}}} \right| = \dfrac{1}{2}
Now from equation (5) we have,
⇒∣B−1∣12=−16\Rightarrow \left| {{B^{ - 1}}} \right|\dfrac{1}{2} = - \dfrac{1}{6}
⇒∣B−1∣=−26=−13\Rightarrow \left| {{B^{ - 1}}} \right| = - \dfrac{2}{6} = - \dfrac{1}{3}
Now from equation (2) we have,
⇒∣B∣(−13)=1\Rightarrow \left| B \right|\left( {\dfrac{{ - 1}}{3}} \right) = 1
⇒∣B∣=−3\Rightarrow \left| B \right| = - 3
So this is the required answer.

Note – A matrix (square matrix) is invertible matrix if and only if there exist another matrix B (square matrix) such that AB=BA=IAB = BA = I where I is the identity matrix of same order as that of order of A and B. If a square matrix has an invertible matrix then determinant value should be non-zero, or it must be non-singular.