Question

The inverse of \[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&0&1\end{array}} \right]\] is
A. \[\left[ {\begin{array}{*{20}{c}}1&{ - 2}&3\\0&1&{ - 2}\\0&0&0\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\0&1&{ - 2}\\0&0&1\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}1&2&1\\0&1&2\\0&0&1\end{array}} \right]\]
D. None of these

Answer 1

Hint:We have to find \[{A^{ - 1}}\]of this 3×3 matrix\[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&0&1\end{array}} \right]\], so for that firstly we will solve adjA and the formula of this is mentioned below and then |A|which has also the formula below. And then dividing both of them will get our required solution. As for finding \[{A^{ - 1}}\] the formula suggests\[{A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}}\].



Formula Used:\[{A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}}\]
If A= \[\left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right]\]
Then signs of adjA for this matrix will be \[\left[ {\begin{array}{*{20}{c}} + & - & + \\ - & + & - \\ + & - & + \end{array}} \right]\]as in \[{a_{11}}\]the sum of this two numbers are 2 which is even so we will put + sign and like that for \[{a_{12}}\]The sum of these two numbers are 3 which is odd so we will put – sign. Like that it will be applicable for all elements in the matrix.
Now after knowing the signs of each element we will find cofactors of each eleme\[adjA = \left[ {\begin{array}{*{20}{c}}{{a_{22}}{a_{33}} - {a_{23}}{a_{32}}}&{{a_{21}}{a_{33}} - {a_{23}}{a_{31}}}&{{a_{21}}{a_{32}} - {a_{22}}{a_{31}}}\\{{a_{12}}{a_{33}} - {a_{13}}{a_{32}}}&{{a_{11}}{a_{33}} - {a_{13}}{a_{31}}}&{{a_{11}}{a_{32}} - {a_{12}}{a_{31}}}\\{{a_{12}}{a_{23}} - {a_{13}}{a_{22}}}&{{a_{11}}{a_{23}} - {a_{13}}{a_{21}}}&{{a_{11}}{a_{22}} - {a_{12}}{a_{21}}}\end{array}} \right]\]nt in this matrix.
\[adjA = \left[ {\begin{array}{*{20}{c}}{{a_{22}}{a_{33}} - {a_{23}}{a_{32}}}&{{a_{21}}{a_{33}} - {a_{23}}{a_{31}}}&{{a_{21}}{a_{32}} - {a_{22}}{a_{31}}}\\{{a_{12}}{a_{33}} - {a_{13}}{a_{32}}}&{{a_{11}}{a_{33}} - {a_{13}}{a_{31}}}&{{a_{11}}{a_{32}} - {a_{12}}{a_{31}}}\\{{a_{12}}{a_{23}} - {a_{13}}{a_{22}}}&{{a_{11}}{a_{23}} - {a_{13}}{a_{21}}}&{{a_{11}}{a_{22}} - {a_{12}}{a_{21}}}\end{array}} \right]\]
And if A= \[\left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right]\]
So for finding |A| we will hide that row and that column of that element and make a determinant of the other 2×2 matrix which will be left.
Then |A|=\[{a_{11}}({a_{22}}{a_{33}} - {a_{23}}{a_{32}}) - {a_{12}}({a_{21}}{a_{33}} - {a_{23}}{a_{31}}) + {a_{13}}({a_{21}}{a_{32}} - {a_{22}}{a_{31}})\]



Complete step by step solution:we are given matrix= \[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&0&1\end{array}} \right]\]and we have to find \[{A^{ - 1}}\]
so first of all as the formula suggests we have to find adjA
so adjA of this matrix\[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&0&1\end{array}} \right]\] is

\[adjA = \left[ {\begin{array}{*{20}{c}}{1(1) - 2(0)}&{0(1) - 2(0)}&{0(0) - 1(0)}\\{2(1) - 3(0)}&{1(1) - 3(0)}&{1(0) - 2(0)}\\{2(2) - 3(1)}&{1(2) - 3(0)}&{1(1) - 2(0)}\end{array}} \right]\]
\[adjA = \left[ {\begin{array}{*{20}{c}}1&0&0\\2&1&0\\1&2&1\end{array}} \right]\]
Now as we have done above that the signs of the adj will change as per the formula
\[adjA = \left[ {\begin{array}{*{20}{c}}1&0&0\\{ - 2}&1&0\\1&{ - 2}&1\end{array}} \right]\]
Now |A| of matrix \[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&0&1\end{array}} \right]\]is
\[\begin{array}{l}\left| A \right| = 1(1 - 0) - 2(0 - 0) + 3(0 - 0)\\\left| A \right| = 1(1) - 0 + 0\\\left| A \right| = 1 - 0\\\left| A \right| = 1\end{array}\]
Now as \[{A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}}\]
\[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&0&0\\{ - 2}&1&0\\1&{ - 2}&1\end{array}} \right]\]
Now by taking transpose of this matrix we will get
\[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&0&0\\{ - 2}&1&0\\1&{ - 2}&1\end{array}} \right]\]


So, Option ‘B’ is correct

Note:We have to remember that if our final answer is not matching with the options we can try taking the transpose of that matrix and then we can get our required solution. \[{A^{ - 1}}\] exists only when\[\left| A \right| \ne 0\]. We have to remember the formula for\[{A^{ - 1}}\]. Sometimes students make mistakes while solving adjA and |A| for a 3×3 matrix. If the inverse of a matrix exists, we can find the adjoint of the given matrix and divide it by the determinant of the matrix.