12th NCERT Matrices Exercise 4.5 Questions 18
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Find the adjoint of each of the matrices in Exercise 1 and 2

Question (1)

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

Solution

\[A = \left[ {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right]\] cofactor A
A11 = 4; A12; A21 =2 ; A22 = 1
\[\text{cofactor} \;A = \left[ {\begin{array}{*{20}{c}}4&{ - 3}\\{ - 2}&1\end{array}} \right]\] adj A = (cofactor A)'
\[adj\;A = \left[ {\begin{array}{*{20}{c}}4&{ - 3}\\{ - 2}&1\end{array}} \right]\]

Question (2)

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

Solution

\[A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&2\\2&3&5\\{ - 2}&0&1\end{array}} \right]\] \[cofactor\;A = \left[ {\begin{array}{*{20}{c}}3&{ - 12}&6\\1&5&2\\{ - 11}&{ - 1}&{15}\end{array}} \right]\] adj A = (cofactor A)' \[adj\;A = \left[ {\begin{array}{*{20}{c}}3&1&{ - 11}\\{ - 12}&5&{ - 1}\\6&2&5\end{array}} \right]\]

Verify A(adj A) = (adj A) A = |A|I in Exercise 3 and 4

Question (3)

\[\left[ {\begin{array}{*{20}{c}}2&3\\{ - 4}&{ - 6}\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}2&3\\{ - 4}&{ - 6}\end{array}} \right]\] \[\left| A \right| = - 12 + 12 = 0\] \[cofactor\;A = \left[ {\begin{array}{*{20}{c}}{ - 6}&4\\{ - 3}&2\end{array}} \right]\] adj A = (cofactor A)' \[adj\;A = \left[ {\begin{array}{*{20}{c}}{ - 6}&{ - 3}\\4&2\end{array}} \right]\] \[LHS\; = A\left( {adj\;A} \right)\] \[RHS\; = \left[ {\begin{array}{*{20}{c}}2&3\\{ - 4}&{ - 6}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 6}&{ - 3}\\4&2\end{array}} \right]\] \[RHS\; = \left[ {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right] = \left| A \right|I \] \[RHS\; = \left[ {\begin{array}{*{20}{c}}{ - 6}&{ - 3}\\4&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2&3\\{ - 4}&{ - 6}\end{array}} \right]\] \[RHS\; = \left[ {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right] = \left| A \right|I = LHS\]

Question (4)

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

Solution

\[A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&2\\3&0&{ - 2}\\1&0&3\end{array}} \right]\] \[\left| A \right| = 1\left( {0 - 0} \right) + 1\left( {9 + 2} \right) + 2\left( {0 - 0} \right) = 11\] \[cofactorA = \left[ {\begin{array}{*{20}{c}}0&{ - 11}&0\\3&1&{ - 1}\\2&8&3\end{array}} \right]\] adj A = (cofactor A)' \[adjA = \left[ {\begin{array}{*{20}{c}}0&3&2\\{ - 11}&1&8\\0&{ - 1}&3\end{array}} \right]\] LHS = A adjA
\[LHS = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&2\\3&0&{ - 2}\\1&0&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0&3&2\\{ - 11}&1&8\\0&{ - 1}&3\end{array}} \right]\] \[LHS = \left[ {\begin{array}{*{20}{c}}{11}&0&0\\0&{11}&0\\0&0&{11}\end{array}} \right]\] \[LHS = 11\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\] \[LHS = 11I = \left| A \right|I\] RHS = adj A. A
\[RHS = \left[ {\begin{array}{*{20}{c}}0&3&2\\{ - 11}&1&8\\0&{ - 1}&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ - 1}&2\\3&0&{ - 2}\\1&0&3\end{array}} \right]\] \[RHS = \left[ {\begin{array}{*{20}{c}}{11}&0&0\\0&{11}&0\\0&0&{11}\end{array}} \right]\] \[RHS = 11\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\] \[RHS = \left| A \right|I\]

Find the inverse of each of the matrices (if it exists) given in Exercise 5 to 11

Question (5)

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

Solution

\[A = \left[ {\begin{array}{*{20}{c}}2&{ - 2}\\4&3\end{array}} \right]\] \[A = 6 + 8 = 14\] \[cofactorA = \left[ {\begin{array}{*{20}{c}}3&{ - 4}\\2&2\end{array}} \right]\] \[adjA = \left[ {\begin{array}{*{20}{c}}3&2\\{ - 4}&2\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{{14}}\left[ {\begin{array}{*{20}{c}}3&2\\{ - 4}&2\end{array}} \right]\]

Question (6)

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

Solution

\[A = \left[ {\begin{array}{*{20}{c}}{ - 1}&5\\{ - 3}&1\end{array}} \right]\] \[\left| A \right| = - 2 + 15 = 13\] \[cofactorA = \left[ {\begin{array}{*{20}{c}}2&3\\{ - 5}&{ - 1}\end{array}} \right]\] \[adjA = \left[ {\begin{array}{*{20}{c}}2&{ - 5}\\3&{ - 1}\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{{13}}\left[ {\begin{array}{*{20}{c}}2&{ - 5}\\3&{ - 1}\end{array}} \right]\]

Question (7)

\[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&2&4\\0&0&5\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}1&2&3\\0&2&4\\0&0&5\end{array}} \right]\] \[\left| A \right| = 1\left( {10} \right) - 2\left( 0 \right) + 3\left( 0 \right) = 10\] \[cofactorA = \left[ {\begin{array}{*{20}{c}}{10}&0&0\\{ - 10}&5&0\\2&{ - 4}&2\end{array}} \right]\] \[adjA = \left[ {\begin{array}{*{20}{c}}{10}&{ - 10}&2\\0&5&{ - 4}\\0&0&2\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}}{10}&{ - 10}&2\\0&5&{ - 4}\\0&0&2\end{array}} \right]\]

Question (8)

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

Solution

\[A = \left[ {\begin{array}{*{20}{c}}1&0&0\\3&3&0\\5&2&{ - 1}\end{array}} \right]\] \[\left| A \right| = 1\left( { - 3 - 0} \right) - 0 + 0 = - 3\] \[cofactorA = \left[ {\begin{array}{*{20}{c}}{ - 3}&{ + 3}&{ - 9}\\0&{ - 1}&{ - 2}\\0&0&3\end{array}} \right]\] \[adjA = \left[ {\begin{array}{*{20}{c}}{ - 3}&0&0\\3&{ - 1}&0\\{ - 9}&{ - 2}&3\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{{ - 3}}\left[ {\begin{array}{*{20}{c}}{ - 3}&0&0\\3&{ - 1}&0\\{ - 9}&{ - 2}&3\end{array}} \right]\]

Question (9)

\[\left[ {\begin{array}{*{20}{c}}2&1&3\\4&{ - 1}&0\\{ - 7}&2&1\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}2&1&3\\4&{ - 1}&0\\{ - 7}&2&1\end{array}} \right]\] \[\left| A \right| = 2\left( { - 1 - 0} \right) - 1\left( {4 - 0} \right) + 3\left( {8 - 7} \right)\] \[\left| A \right| = - 2 - 4 + 3 = - 3\] \[cofactorA = \left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 4}&1\\5&{23}&{ - 11}\\3&{12}&{ - 6}\end{array}} \right]\] adj A = (cofactor A)'
\[adjA = \left[ {\begin{array}{*{20}{c}}{ - 1}&5&3\\{ - 4}&{23}&{12}\\1&{ - 11}&{ - 6}\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{{ - 3}}\left[ {\begin{array}{*{20}{c}}{ - 1}&5&3\\{ - 4}&{23}&{12}\\1&{ - 11}&{ - 6}\end{array}} \right]\]

Question (10)

\[\left[ {\begin{array}{*{20}{c}}1&{ - 1}&2\\0&2&{ - 3}\\3&{ - 2}&4\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&2\\0&2&{ - 3}\\3&{ - 2}&4\end{array}} \right]\] \[\left| A \right| = 1\left( {8 - 6} \right) + \left( {0 + 9} \right) + \left( {0 - 6} \right)\] \[\left| A \right| = 2 + 9 - 12 = - 1\] \[cofactorA = \left[ {\begin{array}{*{20}{c}}2&{ - 9}&{ - 6}\\0&{ - 2}&{ - 1}\\{ - 1}&{ + 3}&2\end{array}} \right]\] adj A = (cofactor A)'
\[adjA = \left[ {\begin{array}{*{20}{c}}2&0&{ - 1}\\{ - 9}&{ - 2}&3\\{ - 6}&{ - 1}&2\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}}2&0&{ - 1}\\{ - 9}&{ - 2}&3\\{ - 6}&{ - 1}&2\end{array}} \right]\] \[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{ - 2}&0&1\\9&2&{ - 3}\\6&1&{ - 2}\end{array}} \right]\]

Question (11)

\[\left[ {\begin{array}{*{20}{c}}1&0&0\\0&{\cos \alpha }&{\sin \alpha }\\0&{\sin \alpha }&{ - \cos \alpha }\end{array}} \right]\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&{\cos \alpha }&{\sin \alpha }\\0&{\sin \alpha }&{ - \cos \alpha }\end{array}} \right]\] \[\left| A \right| = - {\cos ^2}\alpha - si{n^2}\alpha = - 1\] \[cofactorA = \left[ {\begin{array}{*{20}{c}}{ - 1}&0&0\\0&{ - \cos \alpha }&{ - \sin \alpha }\\0&{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]\] adj A = (cofactor A)'
\[AdjA = \left[ {\begin{array}{*{20}{c}}{ - 1}&0&0\\0&{ - \cos \alpha }&{ - \sin \alpha }\\0&{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}}{ - 1}&0&0\\0&{ - \cos \alpha }&{ - \sin \alpha }\\0&{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]\] \[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&{\cos \alpha }&{\sin \alpha }\\0&{\sin \alpha }&{ - \cos \alpha }\end{array}} \right]\]

Question (12)

Let $A = \left[ {\begin{array}{*{20}{c}}3&7\\2&5\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}6&8\\7&9\end{array}} \right]$.
Verify that (AB)-1 = B-1 A-1

Solution

\[A = \left[ {\begin{array}{*{20}{c}}3&7\\2&5\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}6&8\\7&9\end{array}} \right]\] \[AB = \left[ {\begin{array}{*{20}{c}}3&7\\2&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}6&8\\7&9\end{array}} \right]\] \[AB = \left[ {\begin{array}{*{20}{c}}{18 + 49}&{24 + 63}\\{12 + 35}&{16 + 45}\end{array}} \right]\] \[AB = \left[ {\begin{array}{*{20}{c}}{67}&{87}\\{47}&{61}\end{array}} \right]\] \[\left| {AB} \right| = 4087 - 4089 = - 2\] \[cofactor\;AB = \left[ {\begin{array}{*{20}{c}}{61}&{ - 47}\\{ - 87}&{67}\end{array}} \right]\] \[adj\;AB = \left[ {\begin{array}{*{20}{c}}{61}&{ - 47}\\{ - 47}&{67}\end{array}} \right]\] \[{\left( {AB} \right)^{ - 1}} = \frac{1}{{\left| {AB} \right|}}adj\;AB\] \[{\left( {AB} \right)^{ - 1}} = \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}}{61}&{ - 47}\\{ - 47}&{67}\end{array}} \right]\] \[A = \left[ {\begin{array}{*{20}{c}}3&7\\2&5\end{array}} \right]\] \[\left| A \right| = 15 - 14 = 1\] \[cofactorA = \left[ {\begin{array}{*{20}{c}}5&{ - 2}\\{ - 7}&3\end{array}} \right]\] \[adjA = \left[ {\begin{array}{*{20}{c}}5&{ - 7}\\{ - 2}&3\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{1}\left[ {\begin{array}{*{20}{c}}5&{ - 7}\\{ - 2}&3\end{array}} \right]\] \[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}5&{ - 7}\\{ - 2}&3\end{array}} \right]\] \[B = \left[ {\begin{array}{*{20}{c}}6&8\\7&9\end{array}} \right]\] \[\left| B \right| = 54 - 56 = - 2\] \[cofactorB = \left[ {\begin{array}{*{20}{c}}9&{ - 7}\\{ - 8}&6\end{array}} \right]\] \[adjB = \left[ {\begin{array}{*{20}{c}}9&{ - 7}\\{ - 8}&6\end{array}} \right]\] \[{B^{ - 1}} = \frac{1}{{\left| B \right|}}adjB\] \[{B^{ - 1}} = \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}}9&{ - 8}\\{ - 7}&6\end{array}} \right]\] \[RHS = {B^{ - 1}}{A^{ - 1}}\] \[RHS = \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}}9&{ - 8}\\{ - 7}&6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}5&{ - 7}\\{ - 2}&3\end{array}} \right]\] \[RHS = \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}}{45 + 16}&{ - 63 - 24}\\{ - 35 - 12}&{49 + 18}\end{array}} \right]\] \[RHS = \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}}{61}&{ - 87}\\{ - 47}&{67}\end{array}} \right]\] \[RHS = {\left( {AB} \right)^{ - 1}}\] \[ \therefore {\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}\]

Question (13)

If $A = \left[ {\begin{array}{*{20}{c}}3&2\\{ - 1}&1\end{array}} \right]$, show that A2 -5A + 7I = 0
Hence find A-1

Solution

\[A = \left[ {\begin{array}{*{20}{c}}3&2\\{ - 1}&1\end{array}} \right]\] \[{A^2} = \left[ {\begin{array}{*{20}{c}}3&2\\{ - 1}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3&2\\{ - 1}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}8&5\\{ - 5}&3\end{array}} \right]\] \[LHS = {A^2} - 5A + 7I\] \[LHS = \left[ {\begin{array}{*{20}{c}}8&5\\{ - 5}&3\end{array}} \right] - 5\left[ {\begin{array}{*{20}{c}}3&2\\{ - 1}&1\end{array}} \right] + 7\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[LHS = \left[ {\begin{array}{*{20}{c}}8&5\\{ - 5}&3\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{15}&{10}\\{ - 5}&5\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}7&0\\0&7\end{array}} \right]\] \[LHS = \left[ {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right] = 0 = RHS\] \[{A^2} - 5A + 7I = 0\] \[{A^2}{A^{ - 1}} - 5A{A^{ - 1}} + 7I{A^{ - 1}} = 0\] \[A - 5I + 7{A^{ - 1}} = 0\] \[{A^{ - 1}} = \frac{1}{7}\left[ { - A + 5I} \right]\] \[{A^{ - 1}} = \frac{1}{7}\left[ {\left[ {\begin{array}{*{20}{c}}5&0\\0&5\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right]} \right]\] \[{A^{ - 1}} = \frac{1}{7}\left[ {\begin{array}{*{20}{c}}2&{ - 1}\\1&3\end{array}} \right]\]

Question (14)

For the matrix $A = \left[ {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right]$, find the number a and b such that A2 + aA + bI = 0

Solution

\[A = \left[ {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right]\] \[{A^2} = \left[ {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right]\] \[{A^2} = \left[ {\begin{array}{*{20}{c}}{11}&8\\4&3\end{array}} \right]\] \[{A^2} + aA + bI = 0\] \[\left[ {\begin{array}{*{20}{c}}{11}&8\\4&3\end{array}} \right] + a\left[ {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right] + b\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = 0\] \[\left[ {\begin{array}{*{20}{c}}{11 + 3a + b}&{8 + 2a}\\{4 + a}&{3 + a + b}\end{array}} \right] = 0\] \[ \Rightarrow 11 + 3a + b = 0 - - - \left( 1 \right)\] \[4 + a = 0\] \[a = - 4\] Replace a = -4 in equation(1)
\[11 + 3\left( { - 4} \right) + b = 0\] \[11 - 12 + b = 0\] \[b = 1\] \[ \therefore a = - 4\; \text{and}\; b = 1\]

Question (15)

For the matrix $A = \left[ {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right]$
Show that A3 - 6A2 + 5A +11I = 0, Hence find A-1

Solution

\[A = \left[ {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right]\] \[{A^2} = \left[ {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right]\] \[{A^2} = \left[ {\begin{array}{*{20}{c}}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right]\] \[{A^3} = \left[ {\begin{array}{*{20}{c}}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right]\] \[{A^3} = \left[ {\begin{array}{*{20}{c}}8&{27}&1\\{ - 23}&{27}&{ - 69}\\{32}&{ - 13}&{58}\end{array}} \right]\] \[LHS = {A^3} - 6{A^2} + 5A + 11I\] \[ = \left[ {\begin{array}{*{20}{c}}8&{27}&1\\{ - 23}&{27}&{ - 69}\\{32}&{ - 13}&{58}\end{array}} \right] - 6\left[ {\begin{array}{*{20}{c}}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right] + 5\left[ {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right] + 11\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\] \[ = \left[ {\begin{array}{*{20}{c}}8&{27}&1\\{ - 23}&{27}&{ - 69}\\{32}&{ - 13}&{58}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{24}&{12}&6\\{ - 18}&{48}&{ - 84}\\{42}&{ - 18}&{84}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}5&5&5\\5&{10}&{ - 15}\\{10}&{ - 5}&{15}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{11}&0&0\\0&{11}&0\\0&0&{11}\end{array}} \right]\] \[ = \left[ {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] = 0 = RHS\] \[{A^3} - 6{A^2} + 5A + 11I = 0\] Multiply by A-1
\[{A^3}{A^{ - 1}} - 6{A^2}{A^{ - 1}} + 5A{A^{ - 1}} + 11I{A^{ - 1}} = 0\] \[{A^2} - 6A + 5I + 11{A^{ - 1}} = 0\] \[11{A^{ - 1}} = - {A^2} + 6A - 5I\] \[{A^{ - 1}} = \frac{1}{{11}}\left[ { - {A^2} + 6A - 5I} \right]\] \[ = \frac{1}{{11}}\left[ { - \left[ {\begin{array}{*{20}{c}}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right] + 6\left[ {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right] - 5\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]} \right]\] \[ = \frac{1}{{11}}\left[ {\left[ {\begin{array}{*{20}{c}}{ - 4}&{ - 2}&{ - 1}\\3&{ - 8}&{14}\\{ - 7}&3&{ - 14}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}6&6&6\\6&{12}&{ - 18}\\{12}&{ - 6}&{18}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}5&0&0\\0&5&0\\0&0&5\end{array}} \right]} \right]\] \[ = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}}{ - 3}&4&5\\9&{ - 1}&{ - 4}\\5&{ - 3}&9\end{array}} \right]\]

Question (16)

If $A = \left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right]$
Verify that A3 - 6A2 + 9A - 4I = 0, Hence find A-1

Solution

\[A = \left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right]\] \[{A^2} = \left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right]\] \[{A^2} = \left[ {\begin{array}{*{20}{c}}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right]\] \[{A^3} = {A^2}A\] \[{A^3} = \left[ {\begin{array}{*{20}{c}}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right]\] \[{A^3} = \left[ {\begin{array}{*{20}{c}}{22}&{ - 21}&{21}\\{ - 21}&{22}&{ - 21}\\{21}&{ - 21}&{22}\end{array}} \right]\] \[LHS = {A^3} - 6{A^2} + 9A - 4I\] \[LHS = \left[ {\begin{array}{*{20}{c}}{22}&{ - 21}&{21}\\{ - 21}&{22}&{ - 21}\\{21}&{ - 21}&{22}\end{array}} \right] - 6\left[ {\begin{array}{*{20}{c}}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right] + 9\left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right] - 4\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\] \[LHS = \left[ {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] = 0 = RHS\] \[{A^3} - 6{A^2} + 9A - 4I = 0\] Multiply by A-1
\[{A^3}{A^{ - 1}} - 6{A^2}{A^{ - 1}} + 9A{A^{ - 1}} - 4I{A^{ - 1}} = 0\] \[{A^2} - 6A + 9I - 4{A^{ - 1}} = 0\] \[{A^2} - 6A + 9I = 4{A^{ - 1}}\] \[{A^{ - 1}} = \frac{1}{4}\left[ {{A^2} - 6A + 9I} \right]\] \[{A^{ - 1}} = \frac{1}{4}\left[ {\left[ {\begin{array}{*{20}{c}}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right] - 6\left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right] + 9\left[ {\begin{array}{*{20}{c}}9&0&0\\0&9&0\\0&0&9\end{array}} \right]} \right]\] \[{A^{ - 1}} = \frac{1}{4}\left[ {\begin{array}{*{20}{c}}3&1&{ - 1}\\1&3&1\\{ - 1}&1&3\end{array}} \right]\]

Question (17)

Let A be a nonsingular square matrix of order 3×3. The |adj A| is equal to
(A) |A|   (B) |A|2   (C) |A|3   (D) 3|A|

Solution

\[{\left[ A \right]_{3 \times 3}}\] \[\left| {adjA} \right| = {\left| A \right|^{n - 1}}\] \[\left| {adjA} \right| = {\left| A \right|^2}\] Option (B) is correct

Question (18)

If A is an invertible matrix of order 2, then det(A-1) is equal to
(A) det (A)   (B) $\frac{1}{{\det \left( A \right)}}$   (C) 1   (D) 0

Solution

A is an invertible matrix of order 2 \[\det \left( {{A^{ - 1}}} \right) = \frac{1}{{\det \left( A \right)}}\] ∴ Option (B) is correct
Exercise 4.4⇐
⇒ Exercise 4.6