
How do you find \[{{B}^{-1}}\]?; We know that \[{{B}^{2}}=B+2{{I}_{3}}\].
Answer
528k+ views
Hint: In this problem, we have to find the inverse of B with the given condition \[{{B}^{2}}=B+2{{I}_{3}}\]. We can first multiply \[{{B}^{-1}}\] 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 \[{{B}^{-1}}\], we can take the remaining terms to the other side to find the value of \[{{B}^{-1}}\].
Complete step by step answer:
We know that the given equation is,
\[{{B}^{2}}=B+2{{I}_{3}}\]….. (1)
We have to find \[{{B}^{-1}}\].
We can multiply \[{{B}^{-1}}\] on on both the left-hand side and the right-hand side of the equation (1), we get
\[\Rightarrow {{B}^{2}}{{B}^{-1}}=B{{B}^{-1}}+2{{I}_{3}}{{B}^{-1}}\]
We can now write the left-hand side as,
\[\Rightarrow \dfrac{{{B}^{2}}}{B}=B{{B}^{-1}}+2{{I}_{3}}{{B}^{-1}}\]
We can now cancel the terms in the left-hand side, we get
\[\Rightarrow B=B{{B}^{-1}}+2{{I}_{3}}{{B}^{-1}}\]…… (2)
We can now simplify the right-hand side.
We can write \[B{{B}^{-1}}={{I}_{3}}\] and \[{{I}_{3}}{{B}^{-1}}={{B}^{-1}}\].
We can now substitute the above values in the equation (2), we get
\[\Rightarrow B={{I}_{3}}+2{{B}^{-1}}\]
We can now subtract \[{{I}_{3}}\] on both the left-hand side and the right-hand side.
\[\Rightarrow B-{{I}_{3}}=2{{B}^{-1}}\]
Now we can divide the number 2 on both right-hand side and the left-hand side, we get
\[\Rightarrow \dfrac{1}{2}\left( B-{{I}_{3}} \right)={{B}^{-1}}\]
Therefore, the value of \[{{B}^{-1}}=\dfrac{1}{2}\left( B-{{I}_{3}} \right)\].
Note: Students make mistakes while substituting the equivalent values to simplify the steps, like substituting \[B{{B}^{-1}}={{I}_{3}}\] and \[{{I}_{3}}{{B}^{-1}}={{B}^{-1}}\]. 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 \[{{B}^{-1}}\].
Complete step by step answer:
We know that the given equation is,
\[{{B}^{2}}=B+2{{I}_{3}}\]….. (1)
We have to find \[{{B}^{-1}}\].
We can multiply \[{{B}^{-1}}\] on on both the left-hand side and the right-hand side of the equation (1), we get
\[\Rightarrow {{B}^{2}}{{B}^{-1}}=B{{B}^{-1}}+2{{I}_{3}}{{B}^{-1}}\]
We can now write the left-hand side as,
\[\Rightarrow \dfrac{{{B}^{2}}}{B}=B{{B}^{-1}}+2{{I}_{3}}{{B}^{-1}}\]
We can now cancel the terms in the left-hand side, we get
\[\Rightarrow B=B{{B}^{-1}}+2{{I}_{3}}{{B}^{-1}}\]…… (2)
We can now simplify the right-hand side.
We can write \[B{{B}^{-1}}={{I}_{3}}\] and \[{{I}_{3}}{{B}^{-1}}={{B}^{-1}}\].
We can now substitute the above values in the equation (2), we get
\[\Rightarrow B={{I}_{3}}+2{{B}^{-1}}\]
We can now subtract \[{{I}_{3}}\] on both the left-hand side and the right-hand side.
\[\Rightarrow B-{{I}_{3}}=2{{B}^{-1}}\]
Now we can divide the number 2 on both right-hand side and the left-hand side, we get
\[\Rightarrow \dfrac{1}{2}\left( B-{{I}_{3}} \right)={{B}^{-1}}\]
Therefore, the value of \[{{B}^{-1}}=\dfrac{1}{2}\left( B-{{I}_{3}} \right)\].
Note: Students make mistakes while substituting the equivalent values to simplify the steps, like substituting \[B{{B}^{-1}}={{I}_{3}}\] and \[{{I}_{3}}{{B}^{-1}}={{B}^{-1}}\]. 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 \[{{B}^{-1}}\].
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Differentiate between action potential and resting class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

