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Question

Question: How do you find \[{{B}^{-1}}\]?; We know that \[{{B}^{2}}=B+2{{I}_{3}}\]....

How do you find B1{{B}^{-1}}?; We know that B2=B+2I3{{B}^{2}}=B+2{{I}_{3}}.

Explanation

Solution

In this problem, we have to find the inverse of B with the given condition B2=B+2I3{{B}^{2}}=B+2{{I}_{3}}. We can first multiply B1{{B}^{-1}} on both the left-hand side and the right-hand side of the given equation. We can then substitute equivalent values for each term, we will get a term B1{{B}^{-1}}, we can take the remaining terms to the other side to find the value of B1{{B}^{-1}}.

Complete step by step answer:
We know that the given equation is,
B2=B+2I3{{B}^{2}}=B+2{{I}_{3}}….. (1)
We have to find B1{{B}^{-1}}.
We can multiply B1{{B}^{-1}} on on both the left-hand side and the right-hand side of the equation (1), we get
B2B1=BB1+2I3B1\Rightarrow {{B}^{2}}{{B}^{-1}}=B{{B}^{-1}}+2{{I}_{3}}{{B}^{-1}}
We can now write the left-hand side as,
B2B=BB1+2I3B1\Rightarrow \dfrac{{{B}^{2}}}{B}=B{{B}^{-1}}+2{{I}_{3}}{{B}^{-1}}
We can now cancel the terms in the left-hand side, we get
B=BB1+2I3B1\Rightarrow B=B{{B}^{-1}}+2{{I}_{3}}{{B}^{-1}}…… (2)
We can now simplify the right-hand side.
We can write BB1=I3B{{B}^{-1}}={{I}_{3}} and I3B1=B1{{I}_{3}}{{B}^{-1}}={{B}^{-1}}.
We can now substitute the above values in the equation (2), we get
B=I3+2B1\Rightarrow B={{I}_{3}}+2{{B}^{-1}}
We can now subtract I3{{I}_{3}} on both the left-hand side and the right-hand side.
BI3=2B1\Rightarrow B-{{I}_{3}}=2{{B}^{-1}}
Now we can divide the number 2 on both right-hand side and the left-hand side, we get
12(BI3)=B1\Rightarrow \dfrac{1}{2}\left( B-{{I}_{3}} \right)={{B}^{-1}}

Therefore, the value of B1=12(BI3){{B}^{-1}}=\dfrac{1}{2}\left( B-{{I}_{3}} \right).

Note: Students make mistakes while substituting the equivalent values to simplify the steps, like substituting BB1=I3B{{B}^{-1}}={{I}_{3}} and I3B1=B1{{I}_{3}}{{B}^{-1}}={{B}^{-1}}. We had divided the number 2 in the above step on the both right-hand side and left-hand side, as we have to find the value of B1{{B}^{-1}}.