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Class 12 RD Sharma Solutions – Chapter 7 Adjoint and Inverse of a Matrix – Exercise 7.2

  • Last Updated : 28 Mar, 2021

Find the inverse of each of the following matrices by using elementary row transformation(Questions 1- 16):

Question 1. \begin{bmatrix}7&1\\4&-3\end{bmatrix}        

Solution:

Here, A = \begin{bmatrix}7&1\\4&-3\end{bmatrix}

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A = AI 



Using elementary row operation

\begin{bmatrix}7&1\\4&-3\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A

R1 -> 1/7R1

\begin{bmatrix}1&1/7\\4&-3\end{bmatrix}=\begin{bmatrix}1/7&0\\0&1\end{bmatrix}A

R2 -> R2 – 4R1

\begin{bmatrix}1&1/7\\0&-25/7\end{bmatrix}=\begin{bmatrix}1/7&0\\-4/7&1\end{bmatrix}A

R2 -> (-7/25)R2

\begin{bmatrix}1&1/7\\0&1\end{bmatrix}=\begin{bmatrix}1/7&0\\4/25&-7/25\end{bmatrix}A



R1 -> R1 – 1/7R2

\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}21/175&1/25\\4/25&-7/25\end{bmatrix}A

Therefore, A-1 = \begin{bmatrix}3/25&1/25\\4/25&-7/25\end{bmatrix}

Question 2. \begin{bmatrix}5&2\\2&1\end{bmatrix}

Solution:

Here, A = \begin{bmatrix}5&2\\2&1\end{bmatrix}

A = AI

Using elementary row operation

\begin{bmatrix}5&2\\2&1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A

R1 -> 1/5R1

\begin{bmatrix}1&2/5\\2&1\end{bmatrix}=\begin{bmatrix}1/5&0\\0&1\end{bmatrix}A



R1 -> R2 – 2R1

\begin{bmatrix}1&2/5\\0&1/5\end{bmatrix}=\begin{bmatrix}1/5&0\\-2/5&1\end{bmatrix}A

R2 -> 5R2

\begin{bmatrix}1&2/5\\0&1\end{bmatrix}=\begin{bmatrix}1/5&0\\-2&5\end{bmatrix}A

R1 -> R1 – 2/5R2

\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}1&-2\\-2&5\end{bmatrix}A

Therefore, A-1\begin{bmatrix}1&-2\\-2&5\end{bmatrix}

Question 3. \begin{bmatrix}1&6\\-3&5\end{bmatrix}

Solution:

Here, A = \begin{bmatrix}1&6\\-3&5\end{bmatrix}

A = AI



Using elementary row operation

 ⇒\begin{bmatrix}1&6\\-3&5\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A

R2 -> R2 + 3R1

\begin{bmatrix}1&6\\0&23\end{bmatrix}=\begin{bmatrix}1&0\\3&1\end{bmatrix}A

R2 -> 1/23R2

\begin{bmatrix}1&6\\0&1\end{bmatrix}=\begin{bmatrix}1&0\\3/23&1/23\end{bmatrix}A

R1 -> R1 – 6R1

\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}5/23&-6/23\\3/23&1/23\end{bmatrix}A

Therefore, A-11/23\begin{bmatrix}5&-6\\3&1\end{bmatrix}     

Question 4. \begin{bmatrix}2&5\\1&3\end{bmatrix}

Solution:



Here, \begin{bmatrix}2&5\\1&3\end{bmatrix}

A = AI

Using elementary row operation

\begin{bmatrix}2&5\\1&3\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A

R1 -> 1/2R1

\begin{bmatrix}1&5/2\\1&3\end{bmatrix}=\begin{bmatrix}1/2&0\\0&1\end{bmatrix}A

R2 -> R2 – R1

\begin{bmatrix}1&5/2\\0&1/2\end{bmatrix}=\begin{bmatrix}1/2&0\\-1/2&1\end{bmatrix}A

R2 -> 2R2

\begin{bmatrix}1&5/2\\0&1\end{bmatrix}=\begin{bmatrix}1/2&0\\-1&2\end{bmatrix}A



R1 -> R1 – 5/2R2

\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}3&-5\\-1&2\end{bmatrix}A

Therefore, A-1 \begin{bmatrix}3&-5\\-1&2\end{bmatrix}     

Question 5. \begin{bmatrix}3&10\\2&7\end{bmatrix}

Solution:

Here, A = \begin{bmatrix}3&10\\2&7\end{bmatrix}

A = AI

\begin{bmatrix}3&10\\2&7\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A

R1 -> 1/3R1

\begin{bmatrix}1&10/3\\2&7\end{bmatrix}=\begin{bmatrix}1/3&0\\0&1\end{bmatrix}A

R2 -> R2 – 2R1



\begin{bmatrix}1&10/3\\0&1/3\end{bmatrix}=\begin{bmatrix}1/3&0\\-2/3&1\end{bmatrix}A

R2 -> 3R2

\begin{bmatrix}1&10/3\\0&1\end{bmatrix}=\begin{bmatrix}1/3&0\\-2&3\end{bmatrix}A

R1 -> R1 – 10/3R2

\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}7&-10\\-2&3\end{bmatrix}A

Therefore, A-1 \begin{bmatrix}7&-10\\-2&3\end{bmatrix}     

Question 6. \begin{bmatrix}0&1&2\\1&2&3\\3&1&1\end{bmatrix}

Solution:

Here, A = \begin{bmatrix}0&1&2\\1&2&3\\3&1&1\end{bmatrix}

A = IA

\begin{bmatrix}0&1&2\\1&2&3\\3&1&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A



R1 ↔ R2

\begin{bmatrix}1&2&2\\0&1&2\\3&1&1\end{bmatrix}=\begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix}A

R3 -> R3 – 3R1

\begin{bmatrix}1&2&2\\0&1&2\\0&-5&-8\end{bmatrix}=\begin{bmatrix}0&1&0\\1&0&0\\0&-3&1\end{bmatrix}A

R1 -> R1 – 2R2, R3 -> R3 + 5R2

\begin{bmatrix}1&0&-1\\0&1&2\\0&0&2\end{bmatrix}=\begin{bmatrix}-2&1&0\\1&0&0\\5&-3&1\end{bmatrix}A

R3 -> R3/2

\begin{bmatrix}1&0&-1\\0&1&2\\0&0&1\end{bmatrix}=\begin{bmatrix}-2&1&0\\1&0&0\\5/2&-3/2&1/2\end{bmatrix}A

R1 -> R1 + R3, R2 -> R2 – 2R3

\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1/2&-1/2&1/2\\-4&3&1\\5/2&-3/2&1/2\end{bmatrix}A



Therefore, A-1 = \begin{bmatrix}1/2&-1/2&1/2\\-4&3&1\\5/2&-3/2&1/2\end{bmatrix}

Question 7. \begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}

Solution:

Here, A = \begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}

A = IA

\begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A

R1 -> R1/2

\begin{bmatrix}1&0&-1/2\\5&1&0\\0&1&3\end{bmatrix}=\begin{bmatrix}1/2&0&0\\0&1&0\\0&0&1\end{bmatrix}A

R2 -> R2 – 5R1

\begin{bmatrix}1&0&-1/2\\0&1&5/2\\0&1&3\end{bmatrix}=\begin{bmatrix}1/2&0&0\\-5/2&1&0\\0&0&1\end{bmatrix}A

R3 -> R3 – R2



\begin{bmatrix}1&0&-1/2\\0&1&5/2\\0&0&1/2\end{bmatrix}=\begin{bmatrix}1/2&0&0\\-5/2&1&0\\5/2&-1&1\end{bmatrix}A

R3 -> 2R3

\begin{bmatrix}1&0&-1/2\\0&1&5/2\\0&0&1\end{bmatrix}=\begin{bmatrix}1/2&0&0\\-5/2&1&0\\5&-2&2\end{bmatrix}A

R1 -> R1 + 1/2R3, R2 -> R2 – 5/2R3

\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}A

Therefore, A-1 \begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}     

Question 8. \begin{bmatrix}2&3&1\\2&4&1\\3&7&2\end{bmatrix}

Solution:

Here, A = \begin{bmatrix}2&3&1\\2&4&1\\3&7&2\end{bmatrix}

A = IA

\begin{bmatrix}2&3&1\\2&4&1\\3&7&2\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A



R1 -> 1/2R1

\begin{bmatrix}1&3/2&1/2\\2&4&1\\3&7&2\end{bmatrix}=\begin{bmatrix}1/2&0&0\\0&1&0\\0&0&1\end{bmatrix}A

R2 -> R2 – 2R1, R3 -> R3 – 3R1

 ⇒\begin{bmatrix}1&3/2&1/2\\0&1&0\\0&5/2&1/2\end{bmatrix}=\begin{bmatrix}1/2&0&0\\-1&1&0\\-3/2&0&1\end{bmatrix}A

R1 -> R1 – 3/2R2, R3 -> R3 – 5/2R2

\begin{bmatrix}1&0&1/2\\0&1&0\\0&0&1/2\end{bmatrix}=\begin{bmatrix}2&-3/2&0\\-1&1&0\\1&-5/2&1\end{bmatrix}A

R3 -> 2R3

\begin{bmatrix}1&0&1/2\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}2&-3/2&0\\-1&1&0\\2&-5&2\end{bmatrix}A

R1 -> R1 – 1/2R3

\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&1&-1\\-1&1&0\\2&-5&2\end{bmatrix}A

Therefore, A-1 \begin{bmatrix}1&1&-1\\-1&1&0\\2&-5&2\end{bmatrix}      

Question 9. \begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix}

Solution:

Here, A = \begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix}

A = IA

\begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A

R1 -> 1/3R1

\begin{bmatrix}1&-1&4/3\\2&-3&4\\0&-1&1\end{bmatrix}=\begin{bmatrix}1/3&0&0\\0&1&0\\0&0&1\end{bmatrix}A

R2 -> R2 – 2R1

\begin{bmatrix}1&-1&4/3\\0&-1&4/3\\0&-1&1\end{bmatrix}=\begin{bmatrix}1/3&0&0\\-2/3&1&0\\0&0&1\end{bmatrix}A

R2 -> (-1)R2



\begin{bmatrix}1&-1&4/3\\0&1&-4/3\\0&-1&1\end{bmatrix}=\begin{bmatrix}1/3&0&0\\2/3&-1&0\\0&0&1\end{bmatrix}A

R1 -> R1 + R2, R3 -> R3 + R2

\begin{bmatrix}1&0&0\\0&1&-4/3\\0&0&-1/3\end{bmatrix}=\begin{bmatrix}1&-1&0\\2/3&-1&0\\2/3&-1&1\end{bmatrix}A

R3 -> (-3)R3

\begin{bmatrix}1&0&0\\0&1&-4/3\\0&0&1\end{bmatrix}=\begin{bmatrix}1&-1&0\\2/3&-1&0\\2&-3&3\end{bmatrix}A

R2 -> R2 + 4/3R3

\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&-1&0\\-2&3&-4\\-2&3&3\end{bmatrix}A

Therefore, A-1 \begin{bmatrix}1&-1&0\\-2&3&-4\\-2&3&-3\end{bmatrix}    

Question 10. \begin{bmatrix}1&2&0\\2&3&-1\\1&-1&3\end{bmatrix}

Solution:

Here, A = \begin{bmatrix}1&2&0\\2&3&-1\\1&-1&3\end{bmatrix}

\begin{bmatrix}1&2&0\\2&3&-1\\1&-1&3\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A

R2 -> R2 – 2R1, R3 -> R3 – R1

\begin{bmatrix}1&2&0\\0&-1&-1\\0&-3&3\end{bmatrix}=\begin{bmatrix}1&0&0\\-2&1&0\\-1&0&1\end{bmatrix}A

R2 -> (-1)R2

\begin{bmatrix}1&2&0\\0&1&1\\0&-3&3\end{bmatrix}=\begin{bmatrix}1&0&0\\2&-1&0\\-1&0&1\end{bmatrix}A

R1 -> R1 – 2R2, R3 -> R3 + 3R2

\begin{bmatrix}1&0&-2\\0&1&1\\0&0&6\end{bmatrix}=\begin{bmatrix}-3&2&0\\2&-1&0\\5&-3&1\end{bmatrix}A

R3 -> R3/6

\begin{bmatrix}1&0&-2\\0&1&1\\0&0&1\end{bmatrix}=\begin{bmatrix}-3&2&0\\2&-1&0\\5/6&-1/2&1/6\end{bmatrix}A

R1 -> R1 + 2R3, R2 -> R2 – R3



\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}-4/3&1&1/3\\7/6&-1/2&-1/6\\5/6&-1/2&1/6\end{bmatrix}A

Therefore, A-1 \begin{bmatrix}-4/3&1&1/3\\7/6&-1/2&-1/6\\5/6&-1/2&1/6\end{bmatrix}    

Question 11. \begin{bmatrix}2&-1&3\\1&2&4\\3&1&1\end{bmatrix}

Solution:

Here, A = \begin{bmatrix}2&-1&3\\1&2&4\\3&1&1\end{bmatrix}

A = IA

\begin{bmatrix}2&-1&3\\1&2&4\\3&1&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A

R1 -> R1/2

\begin{bmatrix}1&-1/2&3/2\\1&2&4\\3&1&1\end{bmatrix}=\begin{bmatrix}1/2&0&0\\0&1&0\\0&0&1\end{bmatrix}A

R2 -> R2 – R1, R3 -> R3 – 3R1

\begin{bmatrix}1&-1/2&3/2\\0&5/2&5/2\\0&5/2&-7/2\end{bmatrix}=\begin{bmatrix}1/2&0&0\\-1/2&1&0\\-3/2&0&1\end{bmatrix}A

R2 -> (2/5)R2

\begin{bmatrix}1&-1/2&3/2\\0&1&1\\0&5/2&-7/2\end{bmatrix}=\begin{bmatrix}1/2&0&0\\-1/5&2/5&0\\-3/2&-1&1\end{bmatrix}A

R1 -> R1 + 1/2 R2, R3 -> R3 – 5/2R2

\begin{bmatrix}1&0&2\\0&1&1\\0&0&-6\end{bmatrix}=\begin{bmatrix}2/5&1/5&0\\-1/5&2/5&0\\-1&-1&1\end{bmatrix}A

R3 -> R3/-6

\begin{bmatrix}1&0&2\\0&1&1\\0&0&-6\end{bmatrix}=\begin{bmatrix}2/5&1/5&0\\-1/5&2/5&0\\1/6&1/6&-1/6\end{bmatrix}A

R2 -> R2 – R3, R1 -> R1 – 2R3

\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1/15&-2/15&-1/3\\-11/30&7/30&1/6\\1/6&1/6&-1/6\end{bmatrix}A

Therefore, A-1\begin{bmatrix}1/15&-2/15&-1/3\\-11/30&7/30&1/6\\1/6&1/6&-1/6\end{bmatrix}

Question 12. \begin{bmatrix}1&1&2\\3&1&1\\2&3&1\end{bmatrix}

Solution:



Here, A = \begin{bmatrix}1&1&2\\3&1&1\\2&3&1\end{bmatrix}

A = IA

\begin{bmatrix}1&1&2\\3&1&1\\2&3&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A

R2 -> R2 – 3R1, R3 -> R3 – 2R1

\begin{bmatrix}1&1&2\\0&-2&-5\\0&1&-3\end{bmatrix}=\begin{bmatrix}1&0&0\\-3&1&0\\-2&0&1\end{bmatrix}A

R2 -> R2/(-2)

\begin{bmatrix}1&1&2\\0&1&5/2\\0&1&-3\end{bmatrix}=\begin{bmatrix}1&0&0\\3/2&-1/2&0\\-2&0&1\end{bmatrix}A

R1 -> R1 – R2, R3 -> R3 – R2

\begin{bmatrix}1&0&-1/2\\0&1&5/2\\0&0&-11/2\end{bmatrix}=\begin{bmatrix}-1/2&1/2&0\\3/2&-1/2&0\\-7/2&1/2&-2/11\end{bmatrix}A

R3 -> (-2/11)R3

\begin{bmatrix}1&0&-1/2\\0&1&5/2\\0&0&1\end{bmatrix}=\begin{bmatrix}-1/2&1/2&0\\3/2&-1/2&0\\7/11&-1/11&-2/11\end{bmatrix}A

R1 -> R1 + 1/2R3, R2 -> R2 – 5/2R3

\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}-2/11&5/11&-1/11\\-1/11&-3/11&5/11\\7/11&-1/11&-2/11\end{bmatrix}A

Therefore, A-1 \begin{bmatrix}-2/11&5/11&-1/11\\-1/11&-3/11&5/11\\7/11&-1/11&-2/11\end{bmatrix}

Question 13. \begin{bmatrix}2&-1&4\\4&0&2\\3&-2&7\end{bmatrix}

Solution:

Here, A = \begin{bmatrix}2&-1&4\\4&0&2\\3&-2&7\end{bmatrix}

A = IA

\begin{bmatrix}2&-1&4\\4&0&2\\3&-2&7\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A

R1 -> 1/2R1

\begin{bmatrix}1&-1/2&2\\4&0&2\\3&-2&7\end{bmatrix}=\begin{bmatrix}1/2&0&0\\0&1&0\\0&0&1\end{bmatrix}A



R2 -> R2 – 4R1, R3 -> R3 – 3R1

\begin{bmatrix}1&-1/2&2\\0&2&-6\\0&-1/2&1\end{bmatrix}=\begin{bmatrix}1/2&0&0\\-2&1&0\\-3/2&0&1\end{bmatrix}A

R2 -> 1/2R2

\begin{bmatrix}1&-1/2&2\\0&1&-3\\0&-1/2&1\end{bmatrix}=\begin{bmatrix}1/2&0&0\\-1&1/2&0\\-3/2&0&1\end{bmatrix}A

R1 -> R1 + 1/2R2, R3 -> R3 + 1/2R2

\begin{bmatrix}1&0&1/2\\0&1&-3\\0&0&-1/2\end{bmatrix}=\begin{bmatrix}0&1/4&0\\-1&1/2&0\\-2&1/4&1\end{bmatrix}A

R3 -> (-2)R3

\begin{bmatrix}1&0&1/2\\0&1&-3\\0&0&1\end{bmatrix}=\begin{bmatrix}0&1/4&0\\-1&1/2&0\\4&-1/2&-2\end{bmatrix}A

R1 -> R1 – 1/2R3, R2 -> R2 + 3R3

\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}-2&1/2&1\\11&-1&-6\\4&-1/2&-2\end{bmatrix}A

Therefore, A-1 \begin{bmatrix}-2&1/2&1\\11&-1&-6\\4&-1/2&-2\end{bmatrix}

Question 14. \begin{bmatrix}3&0&-1\\2&3&0\\0&4&1\end{bmatrix}

Solution:

Here, A = \begin{bmatrix}3&0&-1\\2&3&0\\0&4&1\end{bmatrix}

A = IA

\begin{bmatrix}3&0&-1\\2&3&0\\0&4&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A

R1 -> (1/3)R1

\begin{bmatrix}1&0&-1/3\\2&3&0\\0&4&1\end{bmatrix}=\begin{bmatrix}1/3&0&0\\0&1&0\\0&0&1\end{bmatrix}A

R2 -> R2 – 2R1

\begin{bmatrix}1&0&-1/3\\0&3&2/3\\0&4&1\end{bmatrix}=\begin{bmatrix}1/3&0&0\\-2/3&1&0\\0&0&1\end{bmatrix}A

R2 -> (1/3)R2



\begin{bmatrix}1&0&-1/3\\0&1&2/9\\0&4&1\end{bmatrix}=\begin{bmatrix}1/3&0&0\\-2/9&1/3&0\\0&0&1\end{bmatrix}A

R3 -> R3 – 4R2

\begin{bmatrix}1&0&-1/3\\0&1&2/9\\0&0&1/9\end{bmatrix}=\begin{bmatrix}1/3&0&0\\-2/9&1/3&0\\8/9&-4/3&1\end{bmatrix}A

R3 -> 9R3

\begin{bmatrix}1&0&-1/3\\0&1&2/9\\0&0&1\end{bmatrix}=\begin{bmatrix}1/3&0&0\\-2/9&1/3&0\\8&-12&9\end{bmatrix}A

R1 -> R1 + 1/3R3, R2 -> R2 – 2/9R3

\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}3&-4&3\\-2&3&-2\\8&-12&9\end{bmatrix}A

Therefore, A-1 \begin{bmatrix}3&-4&3\\-2&3&-2\\8&-12&9\end{bmatrix}

Question 15. \begin{bmatrix}1&3&-2\\-3&0&1\\2&1&0\end{bmatrix}

Solution:

Here, A = \begin{bmatrix}1&3&-2\\-3&0&1\\2&1&0\end{bmatrix}

A = IA

\begin{bmatrix}1&3&-2\\-3&0&1\\2&1&0\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A

R2 -> 3R1 + R2, R3 -> R3 – 2R1

\begin{bmatrix}1&3&-2\\0&9&-5\\0&-5&4\end{bmatrix}=\begin{bmatrix}1&0&0\\3&1&0\\-2&0&1\end{bmatrix}A

R1 -> R1 – 3R2, R3 -> R3 + 5R2

\begin{bmatrix}1&0&-1/3\\0&1&-5/9\\0&0&11/9\end{bmatrix}=\begin{bmatrix}0&-1/3&0\\1/3&1/9&0\\-3/11&5/11&9/11\end{bmatrix}A

R2 -> R2 + 5/9R3, R1 -> R1 + 1/3R3

\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}-1/11&-2/11&3/11\\2/11&4/11&5/11\\-3/11&5/11&9/11\end{bmatrix}A

Therefore, A-1 \begin{bmatrix}-1/11&-2/11&3/11\\2/11&4/11&5/11\\-3/11&5/11&9/11\end{bmatrix}

Question 16. \begin{bmatrix}-1&1&2\\1&2&3\\3&1&1\end{bmatrix}

Solution:



Here, A= \begin{bmatrix}-1&1&2\\1&2&3\\3&1&1\end{bmatrix}

A = IA

\begin{bmatrix}-1&1&2\\1&2&3\\3&1&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A

R1 -> (-1)R1

\begin{bmatrix}1&-1&-2\\1&2&3\\3&1&1\end{bmatrix}=\begin{bmatrix}-1&0&0\\0&1&0\\0&0&1\end{bmatrix}A

R2 -> R2 – R1, R3 -> R3 – 3R1

\begin{bmatrix}1&-1&-2\\0&3&5\\0&4&7\end{bmatrix}=\begin{bmatrix}-1&0&0\\1&1&0\\3&0&1\end{bmatrix}A

R2 -> R2/3

\begin{bmatrix}1&-1&-2\\0&1&5/3\\0&4&7\end{bmatrix}=\begin{bmatrix}-1&0&0\\1/3&1/3&0\\3&0&1\end{bmatrix}A

R1 -> R1 + R2, R3 -> R3 – 4R2

\begin{bmatrix}1&0&-1/3\\0&1&5/3\\0&0&1/3\end{bmatrix}=\begin{bmatrix}-2/3&1/3&0\\1/3&1/3&0\\5/3&-4/3&1\end{bmatrix}A

R3 -> R3/3

\begin{bmatrix}1&0&-1/3\\0&1&5/3\\0&0&1\end{bmatrix}=\begin{bmatrix}-2/3&1/3&0\\1/3&1/3&0\\5&-4&3\end{bmatrix}A

R1 -> R1 + 1/3R3, R2 -> R2 – 5/3R3

\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&-1&1\\-8&7&-5\\5&-4&3\end{bmatrix}A

Therefore, A-1\begin{bmatrix}1&-1&1\\-8&7&-5\\5&-4&3\end{bmatrix}   




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