NCERT Solutions For Class 12 Maths Chapter 3 Exercise 3.4

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Chapter 3 Ex.3.4 Question 1

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}} 1&{ - 1} \\ 2&3 \end{array}} \right)\), if exists.

Solution

Let \(A = \left( {\begin{array}{*{20}{c}} 1&{ - 1} \\ 2&3 \end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\[\begin{align} &\Rightarrow \; \left( {\begin{array}{*{20}{c}}1&{ - 1}\\2&3\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)A\\ &\Rightarrow \;\left( {\begin{array}{*{20}{c}}1&{ - 1}\\0&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\{ - 2}&1\end{array}} \right)A \qquad \quad \left( {{R_2} \to {R_2} - 2{R_1}} \right)\\&\Rightarrow \;\left( {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\{\frac{{ - 2}}{5}}&{\frac{1}{5}}\end{array}} \right)A \qquad \quad \left( {{R_2} \to \frac{1}{5}{R_2}} \right)\\& \Rightarrow\; \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{3}{5}}&{\frac{1}{5}}\\{\frac{{ - 2}}{5}}&{\frac{1}{5}}\end{array}} \right)A \qquad \quad \left( {{R_1} \to {R_1} + {R_2}} \right)\\ &\Rightarrow\; {A^{ - 1}} = \left( {\begin{array}{*{20}{c}}{\frac{3}{5}}&{\frac{1}{5}}\\{\frac{{ - 2}}{5}}&{\frac{1}{5}}\end{array}} \right)\end{align}\]

Chapter 3 Ex.3.4 Question 2

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right)\), if exists.

Solution

Let \(A = \left( {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\[\begin{align}& \Rightarrow \;\left( {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)A\\ &\Rightarrow \;\left( {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right)A \qquad \qquad \left( {{R_1} \to {R_1} - {R_2}} \right)\\ &\Rightarrow \;\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&2\end{array}} \right)A \qquad \qquad \left( {{R_2} \to {R_2} - {R_1}} \right)\\& \Rightarrow\; {A^{ - 1}} = \left( {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&2\end{array}} \right)\end{align}\]

Chapter 3 Ex.3.4 Question 3

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}}1&3\\2&7\end{array}} \right)\), if exists.

Solution

Let \(A = \left( {\begin{array}{*{20}{c}}1&3\\2&7\end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\[\begin{align} &\Rightarrow\; \left( {\begin{array}{*{20}{c}}1&3\\2&7\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)A\\ &\Rightarrow \;\left( {\begin{array}{*{20}{c}}1&3\\0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\{ - 2}&1\end{array}} \right)A \qquad \qquad \left( {{R_2} \to {R_2} - 2{R_1}} \right)\\& \Rightarrow \;\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}7&{ - 3}\\{ - 2}&1\end{array}} \right)A \qquad \qquad \left( {{R_1} \to {R_1} - 3{R_2}} \right)\\ &\Rightarrow \;{A^{ - 1}} = \left( {\begin{array}{*{20}{c}}7&{ - 3}\\{ - 2}&1\end{array}} \right)\end{align}\]

Chapter 3 Ex.3.4 Question 4

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}}2&3\\5&7\end{array}} \right)\), if exists.

Solution

Let \(A = \left( {\begin{array}{*{20}{c}}2&3\\5&7\end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\[\begin{align}& \Rightarrow \; \left( {\begin{array}{*{20}{c}}2&3\\5&7\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)A\\& \Rightarrow \; \left( {\begin{array}{*{20}{c}}1&{\frac{3}{2}}\\5&7\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\0&1\end{array}} \right)A \qquad \qquad \left( {{R_1} \to \frac{1}{2}{R_1}} \right)\\& \Rightarrow\; \left( {\begin{array}{*{20}{c}}1&{\frac{3}{2}}\\0&{\frac{{ - 1}}{2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\{\frac{{ - 5}}{2}}&1\end{array}} \right)A\qquad \qquad\left( {{R_2} \to {R_2} - 5{R_1}} \right)\\& \Rightarrow \; \left( {\begin{array}{*{20}{c}}1&0\\0&{\frac{{ - 1}}{2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 7}&3\\{\frac{{ - 5}}{2}}&1\end{array}} \right)A\qquad \qquad\left( {{R_1} \to {R_1} + 3{R_2}} \right)\\& \Rightarrow\; \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 7}&3\\5&{ - 2}\end{array}} \right)A\qquad \qquad\left( {{R_2} \to - 2{R_1}} \right)\\& \Rightarrow \;{A^{ - 1}} = \left( {\begin{array}{*{20}{c}}{ - 7}&3\\5&{ - 2}\end{array}} \right)\end{align}\]

Chapter 3 Ex.3.4 Question 5

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}}2&1\\7&4\end{array}} \right)\), if exists.

Solution

Let \(A = \left( {\begin{array}{*{20}{c}}2&1\\7&4\end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\[\begin{align} &\Rightarrow \; \left( {\begin{array}{*{20}{c}}2&1\\7&4\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)A\\& \Rightarrow\; \left( {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\7&4\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\0&1\end{array}} \right)A \qquad \qquad \left( {{R_1} \to \frac{1}{2}{R_1}} \right)\\ &\Rightarrow \; \left( {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\0&{\frac{1}{2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\{\frac{{ - 7}}{2}}&1\end{array}} \right)A\qquad \qquad \left( {{R_2} \to {R_2} - 7{R_1}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}1&0\\0&{\frac{1}{2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}4&{ - 1}\\{\frac{{ - 7}}{2}}&1\end{array}} \right)A\qquad \qquad \left( {{R_1} \to {R_1} - {R_2}} \right)\\& \Rightarrow \; \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}4&{ - 1}\\{ - 7}&2\end{array}} \right)A\qquad \qquad \left( {{R_2} \to 2{R_1}} \right)\\ &\Rightarrow \; {A^{ - 1}} = \left( {\begin{array}{*{20}{c}}4&{ - 1}\\{ - 7}&2\end{array}} \right)\end{align}\]

Chapter 3 Ex.3.4 Question 6

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}}2&5\\1&3\end{array}} \right)\), if exists.

Solution

Let \(A = \left( {\begin{array}{*{20}{c}}2&5\\1&3\end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\[\begin{align}&\Rightarrow \; \left( {\begin{array}{*{20}{c}}2&5\\1&3\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)A\\& \Rightarrow \; \left( {\begin{array}{*{20}{c}}1&{\frac{5}{2}}\\1&3\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\0&1\end{array}} \right)A \qquad \qquad \left( {{R_1} \to \frac{1}{2}{R_1}} \right)\\& \Rightarrow\; \left( {\begin{array}{*{20}{c}}1&{\frac{5}{2}}\\0&{\frac{1}{2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\{\frac{{ - 1}}{2}}&1\end{array}} \right)A \qquad \qquad \left( {{R_2} \to {R_2} - {R_1}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}1&0\\0&{\frac{1}{2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&{ - 5}\\{\frac{{ - 1}}{2}}&1\end{array}} \right)A \qquad \qquad \left( {{R_1} \to {R_1} - 5{R_2}} \right)\\ &\Rightarrow \;\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&{ - 5}\\{ - 1}&2\end{array}} \right)A \qquad \qquad \left( {{R_2} \to 2{R_2}} \right)\\& \Rightarrow \; {A^{ - 1}} = \left( {\begin{array}{*{20}{c}}3&{ - 5}\\{ - 1}&2\end{array}} \right)\end{align}\]

Chapter 3 Ex.3.4 Question 7

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}}3&1\\5&2\end{array}} \right)\), if exists.

Solution

Let \(A = \left( {\begin{array}{*{20}{c}}3&1\\5&2\end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\[\begin{align}& \Rightarrow \; \left( {\begin{array}{*{20}{c}}3&1\\5&2\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\ &\Rightarrow \; \left( {\begin{array}{*{20}{c}}1&1\\1&2\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}1&0\\{ - 2}&1\end{array}} \right) \qquad \qquad \left( {{C_1} \to {C_1} - 2{C_2}} \right)\\&\Rightarrow \; \left( {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 2}&3\end{array}} \right)\qquad \qquad\left( {{C_2} \to {C_2} - {C_1}} \right)\\& \Rightarrow\; \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}2&{ - 1}\\{ - 5}&3\end{array}} \right)\qquad \qquad\left( {{C_1} \to {C_1} - {C_2}} \right)\\ &\Rightarrow \; {A^{ - 1}} = \left( {\begin{array}{*{20}{c}}2&{ - 1}\\{ - 5}&3\end{array}} \right)\end{align}\]

Chapter 3 Ex.3.4 Question 8

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}}4&5\\3&4\end{array}} \right)\), if exists.

Solution

Let\(A = \left( {\begin{array}{*{20}{c}}4&5\\3&4\end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\[\begin{align}& \Rightarrow \; \left( {\begin{array}{*{20}{c}}4&5\\3&4\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)A\\& \Rightarrow \; \left( {\begin{array}{*{20}{c}}1&1\\3&4\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right)A \qquad \quad \left( {{R_1} \to {R_1} - {R_2}} \right)\\& \Rightarrow \; \left( {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 3}&4\end{array}} \right)A\qquad \quad \left( {{R_2} \to {R_2} - 3{R_1}} \right)\\& \Rightarrow \; \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}4&{ - 5}\\{ - 3}&4\end{array}} \right)A\qquad \quad \left( {{R_1} \to {R_1} - {R_2}} \right)\\& \Rightarrow \;{A^{ - 1}} = \left( {\begin{array}{*{20}{c}}4&{ - 5}\\{ - 3}&4\end{array}} \right)\end{align}\]

Chapter 3 Ex.3.4 Question 9

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}}3&{10}\\2&7\end{array}} \right)\), if exists.

Solution

Let\(A = \left( {\begin{array}{*{20}{c}}3&{10}\\2&7\end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\[\begin{align} &\Rightarrow \; \left( {\begin{array}{*{20}{c}}3&{10}\\2&7\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)A\\ &\Rightarrow \; \left( {\begin{array}{*{20}{c}}1&3\\2&7\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right)A \qquad \qquad \left( {{R_1} \to {R_1} - {R_2}} \right)\\ &\Rightarrow \; \left( {\begin{array}{*{20}{c}}1&3\\0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 2}&3\end{array}} \right)A \qquad \qquad \left( {{R_2} \to {R_2} - 2{R_1}} \right)\\ &\Rightarrow \; \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}7&{ - 10}\\{ - 2}&3\end{array}} \right)A \qquad \qquad \left( {{R_1} \to {R_1} - 3{R_2}} \right)\\& \Rightarrow \; {A^{ - 1}} = \left( {\begin{array}{*{20}{c}}7&{ - 10}\\{ - 2}&3\end{array}} \right)\end{align}\]

Chapter 3 Ex.3.4 Question 10

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}}3&{ - 1}\\{ - 4}&2\end{array}} \right)\), if exists.

Solution

Let\(A = \left( {\begin{array}{*{20}{c}}3&{ - 1}\\{ - 4}&2\end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\[\begin{align}&\Rightarrow \; \left( {\begin{array}{*{20}{c}}3&{ - 1}\\{ - 4}&2\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\ &\Rightarrow \; \left( {\begin{array}{*{20}{c}}1&{ - 1}\\0&2\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}1&0\\2&1\end{array}} \right) \qquad \qquad \left( {{C_1} \to {C_1} + 2{C_2}} \right)\\ &\Rightarrow \; \left( {\begin{array}{*{20}{c}}1&0\\0&2\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}1&1\\2&3\end{array}} \right) \qquad \qquad \left( {{C_2} \to {C_2} + {C_1}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\2&{\frac{3}{2}}\end{array}} \right) \qquad \qquad \left( {{C_2} \to \frac{1}{2}{C_2}} \right)\\& \Rightarrow \; {A^{ - 1}} = \left( {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\2&{\frac{3}{2}}\end{array}} \right)\end{align}\]

Chapter 3 Ex.3.4 Question 11

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}}2&{ - 6}\\1&{ - 2}\end{array}} \right)\), if exists.

Solution

Let \(A = \left( {\begin{array}{*{20}{c}}2&{ - 6}\\1&{ - 2}\end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\[\begin{align}& \Rightarrow \;\left( {\begin{array}{*{20}{c}}2&{ - 6}\\1&{ - 2}\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\& \Rightarrow \;\left( {\begin{array}{*{20}{c}}2&0\\1&1\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}1&3\\0&1\end{array}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {{C_2} \to {C_2} + 3{C_1}} \right)\\& \Rightarrow \;\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}{ - 2}&3\\{ - 1}&1\end{array}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {{C_1} \to {C_1} - {C_2}} \right)\\& \Rightarrow \;\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}{ - 1}&3\\{ - \frac{1}{2}}&1\end{array}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {{C_1} \to \frac{1}{2}{C_1}} \right)\\ &\Rightarrow \; {A^{ - 1}} = \left( {\begin{array}{*{20}{c}}{ - 1}&3\\{ - \frac{1}{2}}&1\end{array}} \right)\end{align}\]

Chapter 3 Ex.3.4 Question 12

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}}6&{ - 3}\\{ - 2}&1\end{array}} \right)\), if exists.

Solution

Let \(A = \left( {\begin{array}{*{20}{c}}6&{ - 3}\\{ - 2}&1\end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\[\begin{align} &\Rightarrow \; \left( {\begin{array}{*{20}{c}}6&{ - 3}\\{ - 2}&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)A\\ &\Rightarrow\; \left( {\begin{array}{*{20}{c}}1&{\frac{{ - 1}}{2}}\\{ - 2}&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{1}{6}}&0\\0&1\end{array}} \right)A\qquad \quad \left( {{R_1} \to \frac{1}{6}{R_1}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}1&{\frac{{ - 1}}{2}}\\0&0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{1}{6}}&0\\{\frac{1}{3}}&1\end{array}} \right)A \qquad \qquad \left( {{R_2} \to {R_2} + 2{R_1}} \right)\end{align}\]

In the above equation, we can see all the zeros in the second row of the matrix on the L.H.S.

Thus, \({A^{ - 1}}\)does not exist.

Chapter 3 Ex.3.4 Question 13

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 1}&2\end{array}} \right)\), if exists.

Solution

Let \(A = \left( {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 1}&2\end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\[\begin{align}& \Rightarrow \; \left( {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 1}&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)A\\& \Rightarrow \; \left( {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right)A \qquad \quad \left( {{R_1} \to {R_1} + {R_2}} \right)\\& \Rightarrow\; \left( {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&1\\1&2\end{array}} \right)A\qquad \qquad \left( {{R_2} \to {R_2} + {R_1}} \right)\\ &\Rightarrow \; \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}2&3\\1&2\end{array}} \right)A\qquad \qquad \quad\left( {{R_1} \to {R_1} + {R_2}} \right)\\ &\Rightarrow\; {A^{ - 1}} = \left( {\begin{array}{*{20}{c}}2&3\\1&2\end{array}} \right)\end{align}\]

Chapter 3 Ex.3.4 Question 14

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}}2&1\\4&2\end{array}} \right)\), if exists.

Solution

Let \(A = \left( {\begin{array}{*{20}{c}}2&1\\4&2\end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\[\begin{align} &\Rightarrow \; \left( {\begin{array}{*{20}{c}}2&1\\4&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)A\\& \Rightarrow\; \left( {\begin{array}{*{20}{c}}0&0\\4&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{\frac{{ - 1}}{2}}\\0&1\end{array}} \right)A \qquad \quad \left( {{R_1} \to {R_1} - \frac{1}{2}{R_2}} \right)\end{align}\]

In the above equation, we can see all the zeros in the first row of the matrix on the L.H.S.

Thus, \({A^{ - 1}}\) does not exist.

Chapter 3 Ex.3.4 Question 15

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}}2&{ - 3}&3\\2&2&3\\3&{ - 2}&2\end{array}} \right)\), if exists.

Solution

Let \(A = \left( {\begin{array}{*{20}{c}}2&{ - 3}&3\\2&2&3\\3&{ - 2}&2\end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\[\begin{align} &\Rightarrow \; \left( {\begin{array}{*{20}{c}}2&{ - 3}&3\\2&2&3\\3&{ - 2}&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)A\\& \Rightarrow\; \left( {\begin{array}{*{20}{c}}2&{ - 3}&3\\0&5&0\\3&{ - 2}&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&0\\{ - 1}&1&0\\0&0&1\end{array}} \right)A \qquad \quad \left( {{R_2} \to {R_2} - {R_1}} \right)\\ &\Rightarrow\; \left( {\begin{array}{*{20}{c}}2&{ - 3}&3\\0&1&0\\3&{ - 2}&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&0\\{\frac{{ - 1}}{5}}&{\frac{1}{5}}&0\\0&0&1\end{array}} \right)A \qquad \quad \left( {{R_2} \to \frac{1}{5}{R_2}} \right)\\& \Rightarrow\; \left( {\begin{array}{*{20}{c}}{ - 1}&{ - 1}&1\\0&1&0\\3&{ - 2}&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&{ - 1}\\{\frac{{ - 1}}{5}}&{\frac{1}{5}}&0\\0&0&1\end{array}} \right)A \quad \quad \left( {{R_1} \to {R_1} - {R_3}} \right)\\ &\Rightarrow\; \left( {\begin{array}{*{20}{c}}{ - 1}&0&1\\0&1&0\\3&0&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{4}{5}}&{\frac{1}{5}}&{ - 1}\\{\frac{{ - 1}}{5}}&{\frac{1}{5}}&0\\{\frac{{ - 2}}{5}}&{\frac{2}{5}}&1\end{array}} \right)A \qquad \quad \left( {{R_1} \to {R_1} + {R_2}{\rm{ and }}{R_3} \to {R_3} + 2{R_2}} \right)\\ &\Rightarrow\; \left( {\begin{array}{*{20}{c}}{ - 1}&0&1\\0&1&0\\0&0&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{4}{5}}&{\frac{1}{5}}&{ - 1}\\{\frac{{ - 1}}{5}}&{\frac{1}{5}}&0\\2&1&{ - 2}\end{array}} \right)A \qquad \quad \left( {{\rm{ }}{R_3} \to {R_3} + 3{R_1}} \right)\end{align}\]

\[\begin{align} &\Rightarrow\; \left( {\begin{array}{*{20}{c}}{ - 1}&0&1\\0&1&0\\0&0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{4}{5}}&{\frac{1}{5}}&{ - 1}\\{\frac{{ - 1}}{5}}&{\frac{1}{5}}&0\\{\frac{2}{5}}&{\frac{1}{5}}&{\frac{{ - 2}}{5}}\end{array}} \right)A \qquad \quad \left( {{\rm{ }}{R_3} \to \frac{1}{5}{R_3}} \right)\\& \Rightarrow \; \left( {\begin{array}{*{20}{c}}{ - 1}&0&0\\0&1&0\\0&0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{2}{5}}&0&{\frac{{ - 3}}{5}}\\{\frac{{ - 1}}{5}}&{\frac{1}{5}}&0\\{\frac{2}{5}}&{\frac{1}{5}}&{\frac{{ - 2}}{5}}\end{array}} \right)A \qquad \quad \left( {{\rm{ }}{R_1} \to {R_1} - {R_3}} \right)\\ &\Rightarrow \; \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{{ - 2}}{5}}&0&{\frac{3}{5}}\\{\frac{{ - 1}}{5}}&{\frac{1}{5}}&0\\{\frac{2}{5}}&{\frac{1}{5}}&{\frac{{ - 2}}{5}}\end{array}} \right)A \qquad \qquad \left( {{\rm{ }}{R_1} \to \left( { - 1} \right){R_1}} \right)\\& \Rightarrow\; {A^{ - 1}} = \left( {\begin{array}{*{20}{c}}{\frac{{ - 2}}{5}}&0&{\frac{3}{5}}\\{\frac{{ - 1}}{5}}&{\frac{1}{5}}&0\\{\frac{2}{5}}&{\frac{1}{5}}&{\frac{{ - 2}}{5}}\end{array}} \right)\end{align}\]

Chapter 3 Ex.3.4 Question 16

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}}1&3&{ - 2}\\{ - 3}&0&{ - 5}\\2&5&0\end{array}} \right)\), if exists.

Solution

Let \(A = \left( {\begin{array}{*{20}{c}}1&3&{ - 2}\\{ - 3}&0&{ - 5}\\2&5&0\end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\( \Rightarrow \left( {\begin{array}{*{20}{c}}1&3&{ - 2}\\{ - 3}&0&{ - 5}\\2&5&0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)A\)

\[\begin{align}& \Rightarrow \; \left( {\begin{array}{*{20}{c}}1&3&{ - 2}\\0&9&{ - 11}\\0&{ - 1}&4\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&0\\3&1&0\\{ - 2}&0&1\end{array}} \right)A \qquad \qquad \left( {{R_2} \to {R_2} + 3{R_1}{\text{ and }}{R_3} \to {R_3} - 2{R_1}{\rm{ }}} \right)\\ &\Rightarrow \; \left( {\begin{array}{*{20}{c}}1&0&{10}\\0&1&{21}\\0&{ - 1}&4\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 5}&0&3\\{ - 13}&1&8\\{ - 2}&0&1\end{array}} \right)A\qquad \qquad \left( {{R_1} \to {R_1} + 3{R_3}{\text{ and }}{R_2} \to {R_2} + 8{R_3}} \right)\\& \Rightarrow \left( {\begin{array}{*{20}{c}}1&0&{10}\\0&1&{21}\\0&0&{25}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 5}&0&3\\{ - 13}&1&8\\{ - 15}&1&9\end{array}} \right)A\qquad \qquad \quad \left( {{R_3} \to {R_3} + {R_2}} \right)\\& \Rightarrow \; \left( {\begin{array}{*{20}{c}}1&0&{10}\\0&1&{21}\\0&0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 5}&0&3\\{ - 13}&1&8\\{\frac{{ - 3}}{5}}&{\frac{1}{{25}}}&{\frac{9}{{25}}}\end{array}} \right)A\qquad \quad \left( {{R_3} \to \frac{1}{{25}}{R_3}} \right)\\& \Rightarrow \; \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{\frac{{ - 2}}{5}}&{\frac{{ - 3}}{5}}\\{\frac{{ - 2}}{5}}&{\frac{4}{{25}}}&{\frac{{11}}{{25}}}\\{\frac{{ - 3}}{5}}&{\frac{1}{{25}}}&{\frac{9}{{25}}}\end{array}} \right)A\qquad \qquad \left( {{\rm{ }}{R_1} \to {R_1} - 10{R_3}{\text{ and }}{R_2} \to {R_2} - 21{R_3}{\rm{ }}} \right)\\ &\Rightarrow \; {A^{ - 1}} = \left( {\begin{array}{*{20}{c}}1&{\frac{{ - 2}}{5}}&{\frac{{ - 3}}{5}}\\{\frac{{ - 2}}{5}}&{\frac{4}{{25}}}&{\frac{{11}}{{25}}}\\{\frac{{ - 3}}{5}}&{\frac{1}{{25}}}&{\frac{9}{{25}}}\end{array}} \right)\end{align}\]

Chapter 3 Ex.3.4 Question 17

Using elementary transformation, Find the inverse of the matrix \(\left( {\begin{array}{*{20}{c}}2&0&{ - 1}\\5&1&0\\0&1&3\end{array}} \right)\), if exists.

Solution

Let \(A = \left( {\begin{array}{*{20}{c}}2&0&{ - 1}\\5&1&0\\0&1&3\end{array}} \right)\)

We know that \(A = IA\)

Therefore,

\[\begin{align} &\Rightarrow \; \left( {\begin{array}{*{20}{c}}2&0&{ - 1}\\5&1&0\\0&1&3\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)A\\& \Rightarrow \; \left( {\begin{array}{*{20}{c}}1&0&{\frac{{ - 1}}{2}}\\5&1&0\\0&1&3\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{1}{2}}&0&0\\0&1&0\\0&0&1\end{array}} \right)A \qquad \quad \left( {{R_1} \to \frac{1}{2}{R_1}} \right)\\ & \Rightarrow \; \left( {\begin{array}{*{20}{c}}1&0&{\frac{{ - 1}}{2}}\\0&1&{\frac{5}{2}}\\0&1&3\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{1}{2}}&0&0\\{\frac{{ - 5}}{2}}&1&0\\0&0&1\end{array}} \right)A \qquad \quad \left( {{R_2} \to {R_2} - 5{R_1}} \right)\\ &\Rightarrow \; \left( {\begin{array}{*{20}{c}}1&0&{\frac{{ - 1}}{2}}\\0&1&{\frac{5}{2}}\\0&0&{\frac{1}{2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{1}{2}}&0&0\\{\frac{{ - 5}}{2}}&1&0\\{\frac{5}{2}}&{ - 1}&1\end{array}} \right)A \qquad \quad \left( {{R_3} \to {R_3} - {R_2}} \right)\\ &\Rightarrow \; \left( {\begin{array}{*{20}{c}}1&0&{\frac{{ - 1}}{2}}\\0&1&{\frac{5}{2}}\\0&0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{1}{2}}&0&0\\{\frac{{ - 5}}{2}}&1&0\\5&{ - 2}&2\end{array}} \right)A \qquad \quad \left( {{R_3} \to 2{R_3}} \right)\\ &\Rightarrow \; \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&{ - 1}&1\\{ - 15}&6&{ - 5}\\5&{ - 2}&2\end{array}} \right)A \qquad \quad \left( {{\rm{ }}{R_1} \to {R_1} + \frac{1}{2}{R_3}{\text{ and }}{R_2} \to {R_2} - \frac{5}{2}{R_3}{\rm{ }}} \right)\\& \Rightarrow \;{A^{ - 1}} = \left( {\begin{array}{*{20}{c}}3&{ - 1}&1\\{ - 15}&6&{ - 5}\\5&{ - 2}&2\end{array}} \right)\end{align}\]

Chapter 3 Ex.3.4 Question 18

Matrices \(A\) and \(B\) will be the inverse of each other only if:

(A) \(AB = BA\)

(B) \(AB = BA = 0\)

(C) \(AB = 0,\;BA = I\)

(D) \(AB = BA = I\)

Solution

We know that if \(A\) is a square matrix of order \(m\), and

if there exists another square matrix \(B\) of the same order \(m\),

such that \(AB = BA = I\), then \(B\) is said to be the inverse of \(A\).

In this case, it is clear that \(A\) is the inverse of \(B\).

Thus, matrices \(A\)and \(B\) will be inverses of each other only if \(AB = BA = I\).

The correct option is D.