12th NCERT/CBSE Matrices Exercise 3.4 Questions 18
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Using elementry transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17

Question (1)

\[\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\2&3\end{array}} \right]\]

Solution

\[\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\2&3\end{array}} \right]\] \[A = IA\] \[\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\2&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A\] \[{R_2} \to {R_2} - 2{R_1}\] \[\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{ - 2}&1\end{array}} \right]A\] \[{R_2} \to \frac{{{R_2}}}{5}\] \[\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{ - \frac{2}{5}}&{\frac{1}{5}}\end{array}} \right]A\] \[{R_1} \to {R_1} + {R_2}\] \[\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{3}{5}}&{\frac{1}{5}}\\{ - \frac{2}{5}}&{\frac{1}{5}}\end{array}} \right]A\] \[I = \left[ {\begin{array}{*{20}{c}}{\frac{3}{5}}&{\frac{1}{5}}\\{ - \frac{2}{5}}&{\frac{1}{5}}\end{array}} \right]A\] \[AA' = I\] \[\therefore \quad {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{\frac{3}{5}}&{\frac{1}{5}}\\{\frac{{ - 2}}{5}}&{\frac{1}{5}}\end{array}} \right]\]

Question (2)

\[\left[ {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right]\]

Solution

\[\left[ {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right]\] \[A = \left[ {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right]\] \[IA = A\] \[A\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right]\] \[{R_1} \to {R_1} - {R_2}\] \[A\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right]\] \[{R_2} \to {R_2} - {R_1}\] \[A\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = I\] \[\ text{But} \quad A{A^{ - 1}} = I\] \[\therefore {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&2\end{array}} \right]\]

Question (3)

\[\left[ {\begin{array}{*{20}{c}}1&3\\2&7\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}1&3\\2&7\end{array}} \right]\] \[IA = A\] \[A\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&3\\2&7\end{array}} \right]\] \[{R_2} \to {R_2} - 2{R_1}\] \[A\left[ {\begin{array}{*{20}{c}}1&0\\{ - 2}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&3\\0&1\end{array}} \right]\] \[{R_1} \to {R_1} - 3{R_2}\] \[A\left[ {\begin{array}{*{20}{c}}7&{ - 3}\\{ - 2}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[\text{But} \quad A{A^{ - 1}} = I\] \[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}7&{ - 3}\\{ - 2}&1\end{array}} \right]\] \[{R_1} \to \frac{{{R_1}}}{2}\]

Question (4)

\[\left[ {\begin{array}{*{20}{c}}2&3\\5&7\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}2&3\\5&7\end{array}} \right]\] \[IA = A\] \[A\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&3\\5&7\end{array}} \right]\] \[{R_1} \to \frac{{{R_1}}}{2}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\{ - 2}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{3}{2}}\\5&7\end{array}} \right]\] \[{R_2} \to {R_2} - 5{R_1}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\{ - \frac{5}{2}}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{3}{2}}\\0&{\frac{{ - 1}}{2}}\end{array}} \right]\] \[{R_2} \to - 2{R_2}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\5&{ - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{3}{2}}\\0&1\end{array}} \right]\] \[{R_1} \to {R_1} - \frac{3}{2}{R_2}\] \[A\left[ {\begin{array}{*{20}{c}}{ - 7}&3\\5&{ - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[\text{But} \quad A{A^{ - 1}} = I\] \[ \therefore {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{ - 7}&3\\5&{ - 2}\end{array}} \right]\]

Question (5)

\[ \left[ {\begin{array}{*{20}{c}}2&1\\7&4\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}2&1\\7&4\end{array}} \right]\] \[IA = A\] \[A\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&1\\7&4\end{array}} \right]\] \[{R_1} \to \frac{{{R_1}}}{2}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\7&4\end{array}} \right]\] \[{R_2} \to {R_2} - 7{R_1}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\{ - \frac{7}{2}}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\0&{\frac{1}{2}}\end{array}} \right]\] \[{R_2} \to 2{R_2}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\{ - 7}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\0&1\end{array}} \right]\] \[{R_1} \to {R_1} - \frac{1}{2}{R_2}\] \[A\left[ {\begin{array}{*{20}{c}}4&{ - 1}\\{ - 7}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[\text{But} A{A^{ - 1}} = I\] \[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}4&{ - 1}\\{ - 7}&2\end{array}} \right]\]

Question (6)

\[ \left[ {\begin{array}{*{20}{c}}2&5\\1&3\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}2&5\\1&3\end{array}} \right]\] \[IA = A\] \[A\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&5\\1&3\end{array}} \right]\] \[{R_1} \to \frac{{{R_1}}}{2}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{5}{2}}\\1&3\end{array}} \right]\] \[{R_2} \to {R_2} - {R_1}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\{ - \frac{1}{2}}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{5}{2}}\\0&{\frac{1}{2}}\end{array}} \right]\] \[{R_2} \to 2{R_2}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\{ - 1}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{5}{2}}\\0&1\end{array}} \right]\] \[{R_1} \to {R_1} - \frac{5}{2}{R_2}\] \[A\left[ {\begin{array}{*{20}{c}}3&{ - 5}\\{ - 1}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[A{A^{ - 1}} = I\] \[ \therefore {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}3&{ - 5}\\{ - 1}&2\end{array}} \right]\]

Question (7)

\[ \left[ {\begin{array}{*{20}{c}}3&1\\5&2\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}3&1\\5&2\end{array}} \right]\] \[IA = A\] \[A\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&1\\5&2\end{array}} \right]\] \[{R_1} \to \frac{{{R_1}}}{3}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{3}}&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{1}{3}}\\5&2\end{array}} \right]\] \[{R_2} \to {R_2} - 5{R_1}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{3}}&0\\{ - \frac{5}{3}}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{1}{3}}\\0&{\frac{1}{3}}\end{array}} \right]\] \[{R_2} \to 3{R_2}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{3}}&0\\{ - 5}&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{1}{3}}\\0&1\end{array}} \right]\] \[{R_1} \to {R_1} - \frac{1}{3}{R_2}\] \[A\left[ {\begin{array}{*{20}{c}}2&{ - 1}\\{ - 5}&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[A{A^{ - 1}} = I\] \[\threfore \quad {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}2&{ - 1}\\{ - 5}&3\end{array}} \right]\]

Question (8)

\[ \left[ {\begin{array}{*{20}{c}}4&5\\3&4\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}4&5\\3&4\end{array}} \right]\] \[IA = A\] \[A\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4&5\\3&4\end{array}} \right]\] \[{R_1} \to {R_1} - {R_2}\] \[A\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1\\3&4\end{array}} \right]\] \[{R_2} \to {R_2} - 3{R_1}\] \[A\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 3}&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]\] \[{R_1} \to {R_1} - {R_2}\] \[A\left[ {\begin{array}{*{20}{c}}4&{ - 5}\\{ - 3}&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[\text{But} \quad A{A^{ - 1}} = I\] \[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}4&{ - 5}\\{ - 3}&4\end{array}} \right]\]

Question (9)

\[\left[ {\begin{array}{*{20}{c}}3&{10}\\2&7\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}3&{10}\\2&7\end{array}} \right]\] \[IA = A\] \[A\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&{10}\\2&7\end{array}} \right]\] \[{R_1} \to {R_1} - {R_2}\] \[A\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&3\\2&7\end{array}} \right]\] \[{R_2} \to {R_2} - 2{R_1}\] \[A\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 2}&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&3\\0&1\end{array}} \right]\] \[{R_1} \to {R_1} - 3{R_2}\] \[A\left[ {\begin{array}{*{20}{c}}7&{ - 10}\\{ - 2}&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}7&{ - 10}\\{ - 2}&3\end{array}} \right]\]

Question (10)

\[\left[ {\begin{array}{*{20}{c}}3&{ - 1}\\4&2\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}3&{ - 1}\\4&2\end{array}} \right]\] \[IA = A\] \[A\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&{ - 1}\\{ - 4}&2\end{array}} \right]\] \[{R_1} \to \frac{{{R_1}}}{3}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{3}}&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{{ - 1}}{3}}\\{ - 4}&2\end{array}} \right]\] \[{R_2} \to {R_2} + 4{R_1}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{3}}&0\\{\frac{4}{3}}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{{ - 1}}{3}}\\0&{\frac{2}{3}}\end{array}} \right]\] \[{R_2} \to \frac{3}{2}{R_2}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{3}}&0\\2&{\frac{3}{2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{{ - 1}}{3}}\\0&1\end{array}} \right]\] \[{R_1} \to {R_1} + \frac{1}{3}{R_2}\] \[A\left[ {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\2&{\frac{3}{2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[A{A^{ - 1}} = I\] \[\therefore \quad {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\2&{\frac{3}{2}}\end{array}} \right]\]

Question (11)

\[ \left[ {\begin{array}{*{20}{c}}2&{ - 6}\\1&{ - 2}\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}2&{ - 6}\\1&{ - 2}\end{array}} \right]\] \[IA = A\] \[A\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&{ - 6}\\1&{ - 2}\end{array}} \right]\] \[{R_1} \to {R_1} - {R_2}\] \[A\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ - 4}\\1&{ - 2}\end{array}} \right]\] \[{R_2} \to {R_2} - {R_1}\] \[A\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ - 4}\\0&2\end{array}} \right]\] \[{R_2} \to \frac{1}{2}{R_2}\] \[A\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\{\frac{{ - 1}}{2}}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ - 4}\\0&1\end{array}} \right]\] \[{R_1} \to {R_1} + 4{R_2}\] \[A\left[ {\begin{array}{*{20}{c}}{ - 1}&3\\{\frac{{ - 1}}{2}}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[A{A^{ - 1}} = I\] \[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{ - 1}&3\\{\frac{{ - 1}}{2}}&1\end{array}} \right]\]

Question (12)

\[\left[ {\begin{array}{*{20}{c}}6&{ - 3}\\{ - 2}&1\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}6&{ - 3}\\{ - 2}&1\end{array}} \right]\] \[AI = A\] \[A\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}6&{ - 3}\\{ - 2}&1\end{array}} \right]\] \[{R_1} \to \frac{{{R_1}}}{6}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{6}}&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{{ - 1}}{2}}\\{ - 2}&1\end{array}} \right]\] \[{R_2} \to {R_2} + 2{R_1}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{6}}&0\\{\frac{1}{3}}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{{ - 1}}{2}}\\0&0\end{array}} \right]\] Since element of 2nd row become zero, inverse does not exists

Question (13)

\[\left[ {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 1}&2\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 1}&2\end{array}} \right]\] \[AI = A\] \[A\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 1}&2\end{array}} \right]\] \[{R_1} \to {R_1} + {R_2}\] \[A\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&2\end{array}} \right]\] \[{R_2} \to {R_2} + {R_1}\] \[A\left[ {\begin{array}{*{20}{c}}1&0\\1&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right]\] \[{R_1} \to {R_1} + {R_2}\] \[A\left[ {\begin{array}{*{20}{c}}2&3\\1&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[A{A^{ - 1}} = I\] \[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}2&3\\1&2\end{array}} \right]\]

Question (14)

\[\left[ {\begin{array}{*{20}{c}}2&1\\4&2\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}2&1\\4&2\end{array}} \right]\] \[AI = A\] \[A\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&1\\4&2\end{array}} \right]\] \[{R_1} \to \frac{{{R_1}}}{2}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&1\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\4&2\end{array}} \right]\] \[{R_2} \to {R_2} - 4{R_1}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\{ - 2}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\0&1\end{array}} \right]\] All elements of 2nd row is zero
So inverse does not exists for A

Question (15)

\[\left[ {\begin{array}{*{20}{c}}2&{ - 3}&3\\2&2&3\\3&{ - 2}&2\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}2&{ - 3}&3\\2&2&3\\3&{ - 2}&2\end{array}} \right]\] \[AI = A\] \[A\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&{ - 3}&3\\2&2&3\\3&{ - 2}&2\end{array}} \right]\] \[{R_1} \to \frac{{{R_1}}}{2}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{{ - 3}}{2}}&{\frac{3}{2}}\\2&2&3\\3&{ - 2}&2\end{array}} \right]\] \[{R_2} \to {R_2} - 2{R_1},{R_3} \to {R_3} - 3{R_1}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0&0\\{ - 1}&1&0\\{\frac{{ - 3}}{2}}&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{{ - 3}}{2}}&{\frac{3}{2}}\\0&5&0\\0&{\frac{5}{2}}&{\frac{{ - 5}}{2}}\end{array}} \right]\] \[{R_2} \to \frac{{{R_2}}}{5}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0&0\\{\frac{{ - 1}}{5}}&{\frac{1}{5}}&0\\{\frac{{ - 3}}{2}}&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{{ - 3}}{2}}&{\frac{3}{2}}\\0&1&0\\0&{\frac{5}{2}}&{\frac{{ - 5}}{2}}\end{array}} \right]\] \[{R_1} \to {R_1} + \frac{3}{2}{R_2},{R_3} \to {R_3} - \frac{5}{2}{R_2}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{5}}&{\frac{3}{{10}}}&0\\{\frac{{ - 1}}{5}}&{\frac{1}{5}}&0\\{ - 1}&{\frac{{ - 1}}{2}}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&{\frac{3}{2}}\\0&1&0\\0&0&{\frac{{ - 5}}{2}}\end{array}} \right]\] \[{R_3} \to - \frac{2}{5}{R_3}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{5}}&{\frac{3}{{10}}}&0\\{\frac{{ - 1}}{5}}&{\frac{1}{5}}&0\\{\frac{2}{5}}&{\frac{1}{5}}&{ - \frac{2}{5}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&{\frac{3}{2}}\\0&1&0\\0&0&1\end{array}} \right]\] \[{R_1} \to {R_1} - \frac{3}{2}{R_3}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{{ - 2}}{5}}&0&{\frac{3}{5}}\\{\frac{{ - 1}}{5}}&{\frac{1}{5}}&0\\{\frac{2}{5}}&{\frac{1}{5}}&{ - \frac{2}{5}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\] \[\text{But} \quad A{A^{ - 1}} = I\] \[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{\frac{{ - 2}}{5}}&0&{\frac{3}{5}}\\{\frac{{ - 1}}{5}}&{\frac{1}{5}}&0\\{\frac{2}{5}}&{\frac{1}{5}}&{ - \frac{2}{5}}\end{array}} \right]\]

Question (16)

\[\left[ {\begin{array}{*{20}{c}}1&3&{ - 2}\\{ - 3}&0&{ - 5}\\2&5&0\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}1&3&{ - 2}\\{ - 3}&0&{ - 5}\\2&5&0\end{array}} \right]\] \[AI = A\] \[A\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&3&{ - 2}\\{ - 3}&0&{ - 5}\\2&5&0\end{array}} \right]\] \[{R_2} \to {R_2} + 3{R_1},{R_3} \to {R_3} - 2{R_1}\] \[A\left[ {\begin{array}{*{20}{c}}1&0&0\\3&1&0\\{ - 2}&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&3&{ - 2}\\0&9&{ - 11}\\0&{ - 1}&4\end{array}} \right]\] \[{R_2} \to \frac{{{R_2}}}{9}\] \[A\left[ {\begin{array}{*{20}{c}}1&0&0\\{\frac{1}{3}}&{\frac{1}{9}}&0\\{ - 2}&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&3&{ - 2}\\0&1&{\frac{{ - 11}}{9}}\\0&{ - 1}&4\end{array}} \right]\] \[{R_1} \to {R_1} - 3{R_2},{R_3} \to {R_3} + {R_2}\] \[A\left[ {\begin{array}{*{20}{c}}0&{\frac{{ - 1}}{3}}&0\\{\frac{1}{3}}&{\frac{1}{9}}&0\\{\frac{{ - 5}}{3}}&{\frac{1}{9}}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&{\frac{5}{3}}\\0&1&{\frac{{ - 11}}{9}}\\0&0&{\frac{{25}}{9}}\end{array}} \right]\] \[{R_3} \to \frac{9}{{25}}{R_3}\] \[A\left[ {\begin{array}{*{20}{c}}0&{\frac{{ - 1}}{3}}&0\\{\frac{1}{3}}&{\frac{1}{9}}&0\\{\frac{{ - 3}}{5}}&{\frac{1}{{25}}}&{\frac{9}{{25}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&{\frac{5}{3}}\\0&1&{\frac{{ - 11}}{9}}\\0&0&1\end{array}} \right]\] \[{R_1} \to {R_1} - \frac{5}{3}{R_3},{R_2} \to {R_2} + \frac{{11}}{9}{R_3}\] \[A\left[ {\begin{array}{*{20}{c}}{ - 1}&{\frac{{ - 2}}{5}}&{\frac{{ - 3}}{5}}\\{\frac{{ - 2}}{5}}&{\frac{4}{{25}}}&{\frac{{11}}{{25}}}\\{\frac{{ - 3}}{5}}&{\frac{1}{{25}}}&{\frac{9}{{25}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\] \[\text{But} \quad A{A^{ - 1}} = I\] \[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{ - 1}&{\frac{{ - 2}}{5}}&{\frac{{ - 3}}{5}}\\{\frac{{ - 2}}{5}}&{\frac{4}{{25}}}&{\frac{{11}}{{25}}}\\{\frac{{ - 3}}{5}}&{\frac{1}{{25}}}&{\frac{9}{{25}}}\end{array}} \right]\]

Question (17)

\[ \left[ {\begin{array}{*{20}{c}}2&0&{ - 1}\\5&1&0\\0&1&3\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}2&0&{ - 1}\\5&1&0\\0&1&3\end{array}} \right]\] \[AI = A\] \[A\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&0&{ - 1}\\5&1&0\\0&1&3\end{array}} \right]\] \[{R_1} \to \frac{{{R_1}}}{2}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&{\frac{{ - 1}}{2}}\\5&1&0\\0&1&3\end{array}} \right]\] \[{R_2} \to {R_2} - 5{R_1}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0&0\\{\frac{{ - 5}}{2}}&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&{\frac{{ - 1}}{2}}\\0&1&{\frac{5}{2}}\\0&1&3\end{array}} \right]\] \[{R_3} \to {R_3} - {R_2}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0&0\\{\frac{{ - 5}}{2}}&1&0\\{\frac{5}{2}}&{ - 1}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&{\frac{{ - 1}}{2}}\\0&1&{\frac{5}{2}}\\0&0&{\frac{1}{2}}\end{array}} \right]\] \[{R_3} \to 2{R_3}\] \[A\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&0&0\\{\frac{{ - 5}}{2}}&1&0\\5&{ - 2}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&{\frac{{ - 1}}{2}}\\0&1&{\frac{5}{2}}\\0&0&1\end{array}} \right]\] \[{R_1} \to {R_1} + \frac{1}{2}{R_3},{R_2} \to {R_2} - \frac{5}{2}{R_3}\] \[A\left[ {\begin{array}{*{20}{c}}3&{ - 1}&1\\{ - 15}&6&{ - 5}\\5&{ - 2}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\] \[\text{But} \quad A{A^{ - 1}} = I\] \[\therefore {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 15}&6&{ - 5}\\5&{ - 2}&2\end{array}} \right]\]

Question (18)

Matrices A and B will be inverse of each other only if
(A) AB = BA     (B) AB = BA = 0
(C) BA = 0, BA = 1     (D) AB = BA = 1

Solution

A and B are inverse of each other
AB = BA = I
So D is correct option
Exercise 3.3 ⇐
⇒Miscellaneous Exercise