Question

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

Answer 1

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