
How do you find \[{{B}^{-1}}\]?; We know that \[{{B}^{2}}=B+2{{I}_{3}}\].
Answer
454.5k+ 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 \[{{B}^{-1}}\].
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