
How do you find \[{{B}^{-1}}\]?; We know that \[{{B}^{2}}=B+2{{I}_{3}}\].
Answer
541.2k+ 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
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

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

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

Why cannot DNA pass through cell membranes class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

