NCERT Solutions For Class 12 Maths Chapter 4 Exercise 4.6


Chapter 4 Ex.4.6 Question 1

Examine the consistency of the system of equations:

\[\begin{align}x + 2y &= 2\\2x + 3y &= 3\end{align}\]

 

Solution

 

The given system of equations is: \(\begin{align}x + 2y = 2\\2x + 3y = 3\end{align}\)

The given system of equations can be written in the form of \(AX = B\), where

\(A = \left( {\begin{array}{*{20}{c}}1&2\\2&3\end{array}} \right),X = \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right]\)

Hence,

\[\begin{align}\left| A \right|& = 1\left( 3 \right) - 2\left( 2 \right)\\ &= 3 - 4\\ &= - 1\\ &\ne 0\end{align}\]

So, \(A\) is non-singular.

Therefore, \({A^{ - 1}}\) exists.

Thus, the given system of equations is consistent.

Chapter 4 Ex.4.6 Question 2

Examine the consistency of the system of equations:

\[\begin{align}2x - y = 5\\x + y = 4\end{align}\]

 

Solution

 

The given system of equations is: \(\begin{align}2x - y &= 5\\x + y &= 4\end{align}\)

The given system of equations can be written in the form of \(AX = B\), where

\(A = \left( {\begin{array}{*{20}{c}}2&{ - 1}\\1&1\end{array}} \right),X = \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right]\)

Hence,

\[\begin{align}\left| A \right| &= 2\left( 1 \right) - 1\left( { - 1} \right)\\& = 2 + 1\\ &= 3\\ &\ne 0\end{align}\]

So, \(A\) is non-singular.

Therefore, \({A^{ - 1}}\) exists.

Hence, the given system of equations is consistent.

Chapter 4 Ex.4.6 Question 3

Examine the consistency of the system of equations:

\(\begin{align}x + 3y = 5\\2x + 6y = 8\end{align}\)

 

Solution

 

The given system of equations is: \(\begin{align}&x + 3y = 5\\&2x + 6y = 8\end{align}\)

The given system of equations can be written in the form of \(AX = B\), where

\(A = \left( {\begin{array}{*{20}{c}}1&3\\2&6\end{array}} \right),X = \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}5\\8\end{array}} \right]\)

Hence,

\[\begin{align}\left| A \right|& = 1\left( 6 \right) - 3\left( 2 \right)\\ &= 6 - 6\\& = 0\end{align}\]

So, \(A\) is a singular matrix.

Now,

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

Therefore,

\[\begin{align}\left( {adjA} \right)B &= \left( {\begin{array}{*{20}{c}}6&{ - 5}\\{ - 2}&1\end{array}} \right)\left[ {\begin{array}{*{20}{c}}5\\8\end{array}} \right]\\& = \left( {\begin{array}{*{20}{c}}{30 - 24}\\{ - 10 + 8}\end{array}} \right)\\& = \left[ {\begin{array}{*{20}{c}}6\\{ - 2}\end{array}} \right]\\& \ne 0\end{align}\]

Thus, the solution of the given system of equations does not exist.

Hence, the system of equations is inconsistent.

Chapter 4 Ex.4.6 Question 4

Examine the consistency of the system of equations:

\(\begin{align}x + y + z &= 1\\2x + 3y + 2z &= 2\\ax + ay + 2az &= 4\end{align}\)

 

Solution

 

The given system of equations is: \(\begin{align}&x + y + z = 1\\&2x + 3y + 2z = 2\\&ax + ay + 2az = 4\end{align}\)

The given system of equations can be written in the form of \(AX = B\), where

\(A = \left( {\begin{array}{*{20}{c}}1&1&1\\2&3&2\\a&a&{2a}\end{array}} \right),X = \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}1\\2\\4\end{array}} \right]\)

Hence,

\[\begin{align}\left| A \right|& = 1\left( {6a - 2a} \right) - 1\left( {4a - 2a} \right) + 1\left( {2a - 3a} \right)\\ &= 4a - 2a - a\\& = 4a - 3a\\ &= a \ne 0\end{align}\]

So,\(A\) is non-singular.

Therefore, \({A^{ - 1}}\) exists.

Thus, the given system of equations is consistent.

Chapter 4 Ex.4.6 Question 5

Examine the consistency of the system of equations:

\(\begin{align}&3x - y - 2z = 2\\&2y - z = - 1\\&3x - 5y = 3\end{align}\)

 

Solution

 

The given system of equations is: \(\begin{align}&3x - y - 2z = 2\\&2y - z = - 1\\&3x - 5y = 3\end{align}\)

The given system of equations can be written in the form of \(AX = B\), where

\(A = \left( {\begin{array}{*{20}{c}}3&{ - 1}&{ - 2}\\0&2&{ - 1}\\3&{ - 5}&0\end{array}} \right),X = \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\3\end{array}} \right]\)

Hence,

\[\begin{align}\left| A \right| &= 3\left( {0 - 5} \right) - 0 + 3\left( {1 + 4} \right)\\& = - 15 + 15\\& = 0\end{align}\]

So, \(A\) is a singular matrix.

Now,

\[\left( {adjA} \right) = \left( {\begin{array}{*{20}{c}}{ - 5}&{10}&5\\{ - 3}&6&3\\{ - 6}&{12}&6\end{array}} \right)\]

Therefore,

\[\begin{align}\left( {adjA} \right)B &= \left( {\begin{array}{*{20}{c}}{ - 5}&{10}&5\\{ - 3}&6&3\\{ - 6}&{12}&6\end{array}} \right)\left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\3\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 10 - 10 + 15}\\{ - 6 - 6 + 9}\\{ - 12 - 12 + 18}\end{array}} \right]\\& = \left[ {\begin{array}{*{20}{c}}{ - 5}\\{ - 3}\\{ - 6}\end{array}} \right]\\& \ne 0\end{align}\]

Thus, the solution of the given system of equations does not exist.

Hence, the system of equations is inconsistent.

Chapter 4 Ex.4.6 Question 6

Examine the consistency of the system of equations:

\(\begin{align}&5x - y + 4z = 5\\&2x + 3y + 5z = 2\\&5x - 2y + 6z = - 1\end{align}\)

 

Solution

 

The given system of equations is: \(\begin{align}&5x - y + 4z = 5\\&2x + 3y + 5z = 2\\&5x - 2y + 6z = - 1\end{align}\)

The given system of equations can be written in the form of \(AX = B\), where

\(A = \left( {\begin{array}{*{20}{c}}5&{ - 1}&4\\2&3&5\\5&{ - 2}&6\end{array}} \right),\,X = \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right]\,\,{\rm{and}}\,\,B = \left[ {\begin{array}{*{20}{c}}5\\2\\{ - 1}\end{array}} \right]\)

Hence,

\[\begin{align}\left| A \right| &= 5\left( {18 + 10} \right) + 1\left( {12 - 25} \right) + 4\left( { - 4 - 15} \right)\\ &= 5\left( {28} \right) + 1\left( { - 13} \right) + 4\left( { - 19} \right)\\ &= 140 - 13 - 76\\& = 51 \ne 0\end{align}\]

So, \(A\) is nonsingular.

Therefore, \({A^{ - 1}}\) exists.

Hence, the given system of equations is consistent.

Chapter 4 Ex.4.6 Question 7

Solve system of linear equations, using matrix method.

\(\begin{align}5x + 2y &= 4\\7x + 3y &= 5\end{align}\)

 

Solution

 

The given system of equations is: \(\begin{align}5x + 2y &= 4\\7x + 3y& = 5\end{align}\)

The given system of equations can be written in the form of \(AX = B\), where

\(A = \left( {\begin{array}{*{20}{c}}5&2\\7&3\end{array}} \right),X = \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}4\\5\end{array}} \right]\)

Hence,

\[\begin{align}\left| A \right| &= 15 - 14\\& = 1\\& \ne 0\end{align}\]

So,\(A\) is non-singular.

Therefore, \({A^{ - 1}}\) exists.

Now,

\[\begin{align}{A^{ - 1}} &= \frac{1}{{\left| A \right|}}\left( {adjA} \right)\\& = \left( {\begin{array}{*{20}{c}}3&{ - 2}\\&{ - 7}&5\end{array}} \right)\end{align}\]

Then,

\[\begin{align}& \Rightarrow X = {A^{ - 1}}B\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left( {\begin{array}{*{20}{c}}3&{ - 2}\\{ - 7}&5\end{array}} \right)\left[ {\begin{array}{*{20}{c}}4\\5\end{array}} \right]\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{12 - 10}\\{ - 28 + 25}\end{array}} \right]\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2\\{ - 3}\end{array}} \right]\end{align}\]

Hence, \(x = 2\) and \(y = - 3\)

Chapter 4 Ex.4.6 Question 8

Solve system of linear equations, using matrix method.

\(\begin{align}2x - y &= - 2\\3x + 4y &= 3\end{align}\)

 

Solution

 

The given system of equations is: \(\begin{align}2x - y &= - 2\\3x + 4y &= 3\end{align}\)

The given system of equations can be written in the form of \(AX = B\), where

\(A = \left( {\begin{array}{*{20}{c}}2&{ - 1}\\3&4\end{array}} \right),X = \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{ - 2}\\3\end{array}} \right]\)

Hence,

\[\begin{align}\left| A \right| &= 8 + 3\\ &= 11\\ &\ne 0\end{align}\]

So, \(A\) is non-singular.

Therefore, \({A^{ - 1}}\) exists.

Now,

\[\begin{align}{A^{ - 1}} &= \frac{1}{{\left| A \right|}}\left( {adjA} \right)\\ &= \frac{1}{{11}}\left( {\begin{array}{*{20}{c}}4&1\\{ - 3}&2\end{array}} \right)\end{align}\]

Therefore,

\[\begin{align}& \Rightarrow X = {A^{ - 1}}B\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \frac{1}{{11}}\left( {\begin{array}{*{20}{c}}4&1\\{ - 3}&2\end{array}} \right)\left[ {\begin{array}{*{20}{c}}{ - 2}\\3\end{array}} \right]\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}}{ - 8 + 3}\\{6 + 6}\end{array}} \right]\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}}{ - 5}\\{12}\end{array}} \right]\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{{ - 5}}{{11}}}\\{\frac{{12}}{{11}}}\end{array}} \right]\end{align}\]

Hence, \(x = \frac{{ - 5}}{{11}}\) and \(y = \frac{{12}}{{11}}\)

Chapter 4 Ex.4.6 Question 9

Solve system of linear equations, using matrix method.

\(\begin{align}4x - 3y &= 3\\3x - 5y &= 7\end{align}\)

 

Solution

 

The given system of equations is: \(\begin{align}4x - 3y &= 3\\3x - 5y &= 7\end{align}\)

The given system of equations can be written in the form of \(AX = B\), where

\(A = \left( {\begin{array}{*{20}{c}}4&{ - 3}\\3&{ - 5}\end{array}} \right),X = \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}3\\7\end{array}} \right]\)

Hence,

\[\begin{align}\left| A \right| &= - 20 + 9\\ &= - 11\\& \ne 0\end{align}\]

So, \(A\) is nonsingular.

Therefore, \({A^{ - 1}}\) exists.

Now,

\[\begin{align}{A^{ - 1}} &= \frac{1}{{\left| A \right|}}\left( {adjA} \right)\\& = - \frac{1}{{11}}\left( {\begin{array}{*{20}{c}}{ - 5}&3\\{ - 3}&4\end{array}} \right)\\& = \frac{1}{{11}}\left( {\begin{array}{*{20}{c}}5&{ - 3}\\3&{ - 4}\end{array}} \right)\end{align}\]

Therefore,

\[\begin{align} &\Rightarrow \; X = {A^{ - 1}}B\\& \Rightarrow \; \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \frac{1}{{11}}\left( {\begin{array}{*{20}{c}}5&{ - 3}\\3&{ - 4}\end{array}} \right)\left[ {\begin{array}{*{20}{c}}3\\7\end{array}} \right]\\& \Rightarrow \;\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \frac{1}{{11}}\left( {\begin{array}{*{20}{c}}5&{ - 3}\\3&{ - 4}\end{array}} \right)\left[ {\begin{array}{*{20}{c}}3\\7\end{array}} \right]\\& \Rightarrow \; \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}}{15 - 21}\\{9 - 28}\end{array}} \right]\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}}{ - 6}\\{ - 19}\end{array}} \right]\\&\Rightarrow \; \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{{ - 6}}{{11}}}\\{\frac{{ - 19}}{{11}}}\end{array}} \right]\end{align}\]

Hence, \(x = \frac{{ - 6}}{{11}}\) and \(y = \frac{{ - 19}}{{11}}\)

Chapter 4 Ex.4.6 Question 10

Solve system of linear equations, using matrix method.

\(\begin{align}5x + 2y &= 3\\3x + 2y &= 5\end{align}\)

 

Solution

 

The given system of equations is: \(\begin{align}5x + 2y &= 3\\3x + 2y &= 5\end{align}\)

The given system of equations can be written in the form of \(AX = B\), where

\(A = \left( {\begin{array}{*{20}{c}}5&2\\3&2\end{array}} \right),X = \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}3\\5\end{array}} \right]\)

Hence,

\[\begin{align}\left| A \right| &= 10 - 6\\ &= 4\\ &\ne 0\end{align}\]

So, \(A\) is non-singular.

Therefore, \({A^{ - 1}}\) exists.

Now,

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

Therefore,

\[\begin{align}& \Rightarrow \;X = {A^{ - 1}}B\\ &\Rightarrow\; \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \frac{1}{4}\left( {\begin{array}{*{20}{c}}2&{ - 2}\\{ - 3}&5\end{array}} \right)\left[ {\begin{array}{*{20}{c}}3\\5\end{array}} \right]\\& \Rightarrow \;\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \frac{1}{4}\left( {\begin{array}{*{20}{c}}2&{ - 2}\\{ - 3}&5\end{array}} \right)\left[ {\begin{array}{*{20}{c}}3\\5\end{array}} \right]\\& \Rightarrow \; \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{c}}{6 - 10}\\{ - 9 + 25}\end{array}} \right]\\& \Rightarrow \;\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{c}}{ - 4}\\{16}\end{array}} \right]\\& \Rightarrow \;\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 1}\\4\end{array}} \right]\end{align}\]

Hence, \(x = - 1\) and \(y = 4\)

Chapter 4 Ex.4.6 Question 11

Solve system of linear equations, using matrix method.

\(\begin{align}2x + y + z &= 1\\x - 2y - z& = \frac{3}{2}\\3y - 5z &= 9\end{align}\)

 

Solution

 

The given system of equations is: \(\begin{align}2x + y + z &= 1\\x - 2y - z &= \frac{3}{2}\\3y - 5z &= 9\end{align}\)

The given system of equations can be written in the form of \(AX = B\), where

\(A = \left( {\begin{array}{*{20}{c}}2&1&1\\1&{ - 2}&{ - 1}\\0&3&{ - 5}\end{array}} \right),X = \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}1\\{\frac{3}{2}}\\9\end{array}} \right]\)

Hence,

\[\begin{align}\left| A \right| &= 2\left( {10 + 3} \right) - 1\left( { - 5 - 3} \right) + 0\\ &= 2\left( {13} \right) - 1\left( { - 8} \right)\\& = 26 + 8\\& = 34\\ &\ne 0\end{align}\]

So, \(A\) is non-singular.

Therefore, \({A^{ - 1}}\) exists.

Now,

\[\begin{align}&{A_{11}} = 13\;\;\;\;\;\;\;\;\;\;{A_{12}} = 5\;\;\;\;\;\;\;\;\;\;{A_{13}} = 3\\&{A_{21}} = 8\;\;\;\;\;\;\;\;\;\;\;{A_{22}} = - 10\;\;\;\;\;\;{A_{23}} = - 6\\&{a_{31}} = 1\;\;\;\;\;\;\;\;\;\;\;\;{A_{32}} = 3\;\;\;\;\;\;\;\;\;\;{A_{33}} = - 5\end{align}\]

Hence,

\[\begin{align}{A^{ - 1}} &= \frac{1}{{\left| A \right|}}\left( {adjA} \right)\\ &= \frac{1}{{34}}\left( {\begin{array}{*{20}{c}}{13}&8&1\\5&{ - 10}&3\\3&{ - 6}&{ - 5}\end{array}} \right)\end{align}\]

Therefore,

\[\begin{align} &\Rightarrow \; X = {A^{ - 1}}B\\& \Rightarrow \; \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{34}}\left( {\begin{array}{*{20}{c}}{13}&8&1\\5&{ - 10}&3\\3&{ - 6}&{ - 5}\end{array}} \right)\left[ {\begin{array}{*{20}{c}}1\\{\frac{3}{2}}\\9\end{array}} \right]\\& \Rightarrow \; \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{34}}\left( {\begin{array}{*{20}{c}}{13}&8&1\\5&{ - 10}&3\\3&{ - 6}&{ - 5}\end{array}} \right)\left[ {\begin{array}{*{20}{c}}1\\{\frac{3}{2}}\\9\end{array}} \right]\\ & \Rightarrow\; \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}}{13 + 12 + 9}\\{5 - 15 + 27}\\{3 - 9 - 45}\end{array}} \right]\\ &\Rightarrow \;\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}}{34}\\{17}\\{ - 51}\end{array}} \right]\\& \Rightarrow \;\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{\frac{1}{2}}\\{ - \frac{3}{2}}\end{array}} \right]\end{align}\]

Hence, \(x = 1,y = \frac{1}{2}\) and \(z = \frac{{ - 3}}{2}\)

Chapter 4 Ex.4.6 Question 12

Solve system of linear equations, using matrix method.

\(\begin{align}x - y + z &= 4\\2x + y - 3z &= 0\\x + y + z &= 2\end{align}\)

 

Solution

 

The given system of equations is: \(\begin{align}x - y + z &= 4\\2x + y - 3z &= 0\\x + y + z &= 2\end{align}\)

The given system of equations can be written in the form of \(AX = B\), where

\(A = \left( {\begin{array}{*{20}{c}}1&{ - 1}&1\\2&1&{ - 3}\\1&1&1\end{array}} \right),X = \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}4\\0\\2\end{array}} \right]\)

Hence,

\[\begin{align}\left| A \right| &= 1\left( {1 + 3} \right) + 1\left( {2 + 3} \right) + 1\left( {2 - 1} \right)\\ &= 4 + 5 + 1\\ &= 10\\ &\ne 0\end{align}\]

So, \(A\) is nonsingular.

Therefore, \({A^{ - 1}}\) exists.

Now,

\[\begin{align}&{A_{11}} = 4\;\;\;\;\;\;\;\;\;{A_{12}} = - 5\;\;\;\;\;\;\;\;\;{A_{13}} = 1\\&{A_{21}} = 2\;\;\;\;\;\;\;\;\;{A_{22}} = 0\;\;\;\;\;\;\;\;\;\;\;{A_{23}} = - 2\\&{a_{31}} = 2\;\;\;\;\;\;\;\;\;{A_{32}} = 5\;\;\;\;\;\;\;\;\;\;\;\;{A_{33}} = 3\end{align}\]

Hence,

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

Therefore,

\[\begin{align} &\Rightarrow X = {A^{ - 1}}B\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{10}}\left( {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&5\\1&{ - 2}&3\end{array}} \right)\left[ {\begin{array}{*{20}{c}}4\\0\\2\end{array}} \right]\\ &\Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{10}}\left( {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&5\\1&{ - 2}&3\end{array}} \right)\left[ {\begin{array}{*{20}{c}}4\\0\\2\end{array}} \right]\\ &\Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}}{16 + 0 + 4}\\{ - 20 + 0 + 10}\\{4 + 0 + 6}\end{array}} \right]\\ &\Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}}{20}\\{ - 10}\\{10}\end{array}} \right]\\ &\Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\1\end{array}} \right]\end{align}\]

Hence, \(x = 2,y = - 1\) and \(z = 1\)

Chapter 4 Ex.4.6 Question 13

Solve system of linear equations, using matrix method.

\(\begin{align}2x + 3y + 3z &= 5\\x - 2y + z &= - 4\\3x - y - 2z& = 3\end{align}\)

 

Solution

 

The given system of equations is: \(\begin{align}2x + 3y + 3z &= 5\\x - 2y + z &= - 4\\3x - y - 2z &= 3\end{align}\)

The given system of equations can be written in the form of \(AX = B\), where

\(A = \left( {\begin{array}{*{20}{c}}2&3&3\\1&{ - 2}&1\\3&{ - 1}&{ - 2}\end{array}} \right),X = \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\3\end{array}} \right]\)

Hence,

\[\begin{align}\left| A \right| &= 2\left( {4 + 1} \right) - 3\left( { - 2 - 3} \right) + 3\left( { - 1 + 6} \right)\\ &= 10 + 15 + 15\\ &= 40\\& \ne 0\end{align}\]

So, \(A\) is non-singular.

Therefore, \({A^{ - 1}}\) exists.

Now,

\[\begin{align}&{A_{11}} = 5\;\;\;\;\;\;\;\;\;\;{A_{12}} = 5\;\;\;\;\;\;\;\;\;\;{A_{13}} = 5\\&{A_{21}} = 3\;\;\;\;\;\;\;\;\;\;{A_{22}} = - 13\;\;\;\;\;\;{A_{23}} = 11\\&{A_{31}} = 9\;\;\;\;\;\;\;\;\;\;{A_{32}} = 1\;\;\;\;\;\;\;\;\;\;{A_{33}} = - 7\end{align}\]

Hence,

\[\begin{align}{A^{ - 1}} &= \frac{1}{{\left| A \right|}}\left( {adjA} \right)\\ &= \frac{1}{{40}}\left( {\begin{array}{*{20}{c}}5&3&9\\5&{ - 13}&1\\5&{11}&{ - 7}\end{array}} \right)\end{align}\]

Therefore,

\[\begin{align}& \Rightarrow X = {A^{ - 1}}B\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{40}}\left( {\begin{array}{*{20}{c}}5&3&9\\5&{ - 13}&1\\5&{11}&{ - 7}\end{array}} \right)\left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\3\end{array}} \right]\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{40}}\left( {\begin{array}{*{20}{c}}5&3&9\\5&{ - 13}&1\\5&{11}&{ - 7}\end{array}} \right)\left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\3\end{array}} \right]\\ &\Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{40}}\left[ {\begin{array}{*{20}{c}}{25 - 12 + 27}\\{25 + 52 + 3}\\{25 - 44 - 21}\end{array}} \right]\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{40}}\left[ {\begin{array}{*{20}{c}}{40}\\{80}\\{ - 40}\end{array}} \right]\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\2\\{ - 1}\end{array}} \right]\end{align}\]

Hence, \(x = 1,y = 2\) and \(z = - 1\)

Chapter 4 Ex.4.6 Question 14

Solve system of linear equations, using matrix method.

\(\begin{align}x - y + 2z &= 7\\3x + 4y - 5z &= - 5\\2x - y + 3z& = 12\end{align}\)

 

Solution

 

The given system of equations is: \(\begin{align}x - y + 2z &= 7\\3x + 4y - 5z &= - 5\\2x - y + 3z &= 12\end{align}\)

The given system of equations can be written in the form of \(AX = B\), where

\(A = \left( {\begin{array}{*{20}{c}}1&{ - 1}&2\\3&4&{ - 5}\\2&{ - 1}&3\end{array}} \right),\,X = \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right]\,\,{\rm{and}}\,\,B = \left[ {\begin{array}{*{20}{c}}7\\{ - 5}\\{12}\end{array}} \right]\)

Hence,

\[\begin{align}\left| A \right| &= 1\left( {12 - 5} \right) + 1\left( {9 + 10} \right) + 2\left( { - 3 - 8} \right)\\& = 7 + 19 - 22\\& = 4\\ &\ne 0\end{align}\]

So, \(A\) is non-singular.

Therefore, \({A^{ - 1}}\) exists.

Now,

\[\begin{align}&{A_{11}} = 7\;\;\;\;\;\;\;\;\;\;\;{A_{12}} = - 19\;\;\;\;\;\;\;\;\;\;\;{A_{13}} = - 11\\&{A_{21}} = 1\;\;\;\;\;\;\;\;\;\;\;{A_{22}} = - 1\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{23}} = - 1\\&{a_{31}} = - 3\;\;\;\;\;\;\;\;\;{A_{32}} = 11\;\;\;\;\;\;\;\;\;\;\;\;{A_{33}} = 7\end{align}\]

Hence,

\[\begin{align}{A^{ - 1}} &= \frac{1}{{\left| A \right|}}\left( {adjA} \right)\\ &= \frac{1}{4}\left( {\begin{array}{*{20}{c}}7&1&{ - 3}\\{ - 19}&{ - 1}&{11}\\{ - 11}&{ - 1}&7\end{array}} \right)\end{align}\]

Therefore,

\[\begin{align}& \Rightarrow X = {A^{ - 1}}B\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{4}\left( {\begin{array}{*{20}{c}}7&1&{ - 3}\\{ - 19}&{ - 1}&{11}\\{ - 11}&{ - 1}&7\end{array}} \right)\left[ {\begin{array}{*{20}{c}}7\\{ - 5}\\{12}\end{array}} \right]\\ &\Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{4}\left( {\begin{array}{*{20}{c}}7&1&{ - 3}\\{ - 19}&{ - 1}&{11}\\{ - 11}&{ - 1}&7\end{array}} \right)\left[ {\begin{array}{*{20}{c}}7\\{ - 5}\\{12}\end{array}} \right]\\ &\Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{c}}{49 - 5 - 36}\\{ - 133 + 5 + 132}\\{ - 77 + 5 + 84}\end{array}} \right]\end{align}\]

\[\begin{align}& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{c}}{49 - 5 - 36}\\{ - 133 + 5 + 132}\\{ - 77 + 5 + 84}\end{array}} \right]\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{c}}8\\4\\{12}\end{array}} \right]\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2\\1\\3\end{array}} \right]\end{align}\]

Hence, \(x = 2,y = 1\) and \(z = 3\)

Chapter 4 Ex.4.6 Question 15

If \(A = \left( {\begin{array}{*{20}{c}}2&{ - 3}&5\\3&2&{ - 4}\\1&1&{ - 2}\end{array}} \right)\), find \({A^{ - 1}}\). Using \({A^{ - 1}}\) solve the system of equations

\(\begin{align}&2x - 3y + 5z = 11\\&3x + 2y - 4z = - 5\\&x + y - 2z = - 3\end{align}\)

 

Solution

 

It is given that \(A = \left( {\begin{array}{*{20}{c}}2&{ - 3}&5\\3&2&{ - 4}\\1&1&{ - 2}\end{array}} \right)\)

Therefore,

\[\begin{align}\left| A \right|& = 2\left( { - 4 + 4} \right) + 3\left( { - 6 + 4} \right) + 5\left( {3 - 2} \right)\\& = 0 - 6 + 5\\& = - 1\\ &\ne 0\end{align}\]

Now,

\[\begin{align}&{A_{11}} = 0\;\;\;\;\;\;\;\;\;\;{A_{12}} = 2\;\;\;\;\;\;\;\;\;\;{A_{13}} = 1\\&{A_{21}} = - 1\;\;\;\;\;\;\;\;{A_{22}} = - 9\;\;\;\;\;\;\;\;{A_{23}} = - 5\\&{a_{31}} = 2\;\;\;\;\;\;\;\;\;\;{A_{32}} = 23\;\;\;\;\;\;\;\;\;{A_{33}} = 13\end{align}\]

Hence,

\[\begin{align}{A^{ - 1}} &= \frac{1}{{\left| A \right|}}\left( {adjA} \right)\\ &= - \left( {\begin{array}{*{20}{c}}0&{ - 1}&2\\2&{ - 9}&{23}\\1&{ - 5}&{13}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0&1&{ - 2}\\{ - 2}&9&{ - 23}\\{ - 1}&5&{ - 13}\end{array}} \right)\end{align}\]

The given system of equations can be written in the form of \(AX = B\), where

\(A = \left( {\begin{array}{*{20}{c}}2&{ - 3}&5\\3&2&{ - 4}\\1&1&{ - 2}\end{array}} \right),X = \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{11}\\{ - 5}\\{ - 3}\end{array}} \right]\)

The solution of the system of equations is given by \(X = {A^{ - 1}}B\).

Therefore,

\[\begin{align}& \Rightarrow X = {A^{ - 1}}B\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left( {\begin{array}{*{20}{c}}0&1&{ - 2}\\{ - 2}&9&{ - 23}\\{ - 1}&5&{ - 13}\end{array}} \right)\left[ {\begin{array}{*{20}{c}}{11}\\{ - 5}\\{ - 3}\end{array}} \right]\\ &\Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left( {\begin{array}{*{20}{c}}0&1&{ - 2}\\{ - 2}&9&{ - 23}\\{ - 1}&5&{ - 13}\end{array}} \right)\left[ {\begin{array}{*{20}{c}}{11}\\{ - 5}\\{ - 3}\end{array}} \right]\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{0 - 5 + 6}\\{ - 22 - 45 + 69}\\{ - 11 - 25 + 39}\end{array}} \right]\\ &\Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\2\\3\end{array}} \right]\end{align}\]

Hence, \(x = 1,y = 2\) and \(z = 3\)

Chapter 4 Ex.4.6 Question 16

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹\({\rm{6}}0\). The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹\({\rm{9}}0\). The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹\({\rm{7}}0\). Find cost of each item per kg by matrix method.

 

Solution

 

Let the cost of onions, wheat, and rice per kg in ₹ be \(x,y\) and \(z\) respectively.

Then, the given situation can be represented by a system of equations as:

\[\begin{align}4x + 3y + 2z &= 60\\2x + 4y + 6z &= 90\\6x + 2y + 3z &= 70\end{align}\]

The given system of equations can be written in the form of \(AX = B\), where

\(A = \left( {\begin{array}{*{20}{c}}4&3&2\\2&4&6\\6&2&3\end{array}} \right),X = \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{60}\\{90}\\{70}\end{array}} \right]\)

Therefore,

\[\begin{align}\left| A \right| &= 4\left( {12 - 12} \right) - 3\left( {6 - 36} \right) + 2\left( {4 - 24} \right)\\ &= 0 + 90 - 40\\ &= 50\\ &\ne 0\end{align}\]

So, \(A\) is non-singular.

Therefore, \({A^{ - 1}}\) exists.

Now,

\[\begin{align}&{A_{11}} = 0\;\;\;\;\;\;\;\;\;\;\;{A_{12}} = 30\;\;\;\;\;\;\;\;\;\;\;{A_{13}} = - 20\\&{A_{21}} = - 5\;\;\;\;\;\;\;\;\;{A_{22}} = 0\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{23}} = 10\\&{A_{31}} = 10\;\;\;\;\;\;\;\;\;{A_{32}} = - 20\;\;\;\;\;\;\;\;\;\;{A_{33}} = 10\end{align}\]

Therefore,

\[\begin{align}{A^{ - 1}} &= \frac{1}{{\left| A \right|}}\left( {adjA} \right)\\& = \frac{1}{{50}}\left( {\begin{array}{*{20}{c}}0&{ - 5}&{10}\\{30}&0&{ - 20}\\{ - 20}&{10}&{10}\end{array}} \right)\end{align}\]

Hence,

\[\begin{align}& \Rightarrow X = {A^{ - 1}}B\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{50}}\left( {\begin{array}{*{20}{c}}0&{ - 5}&{10}\\{30}&0&{ - 20}\\{ - 20}&{10}&{10}\end{array}} \right)\left[ {\begin{array}{*{20}{c}}{60}\\{90}\\{70}\end{array}} \right]\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{50}}\left( {\begin{array}{*{20}{c}}0&{ - 5}&{10}\\{30}&0&{ - 20}\\{ - 20}&{10}&{10}\end{array}} \right)\left[ {\begin{array}{*{20}{c}}{60}\\{90}\\{70}\end{array}} \right]\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{50}}\left[ {\begin{array}{*{20}{c}}{0 - 450 + 700}\\{1800 + 0 - 1400}\\{ - 1200 + 900 + 700}\end{array}} \right]\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{50}}\left[ {\begin{array}{*{20}{c}}{250}\\{400}\\{400}\end{array}} \right]\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5\\8\\8\end{array}} \right]\end{align}\]

Thus, \(x = 5,y = 8\) and \(z = 8\)

Hence, the cost of onions is ₹ 5 per kg

the cost of wheat is ₹ 8 per kg, and

the cost of rice is ₹ 8 per kg.

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