Ncert Solutions For Class 12 Maths Ex 4.6

Ncert Solutions For Class 12 Maths Chapter 4 Ex 4.6

Q-1: Check the consistency for the system of two equations given below:

a + 3b = 2

2a + 4b = 3

 

Sol:

As per the data given in the question,

The given system of the two equations is:

a + 3b = 2

2a + 4b = 3

The given system of equations will be written as in the form of MX = N, where

\(M = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}, X = \begin{bmatrix} a \\ b \end{bmatrix} \; and \; N = \begin{bmatrix} 2 \\ 3 \end{bmatrix}\)

Thus,

|M| = 1( 4 ) – 3( 2 ) = 3 – 4 = -1 ≠ 0

Hence, M is non- singular.

Thus, M-1 exists.

Therefore, the given system of two equations will be consistent.

 

Q-2. Check the consistency of the system of two equations given below:

a + 4b = 5

2a + 7b = 8

Sol:

The given system of the two equations is:

a + 4b = 5

2a + 7b = 8

The given system of equations will be written as in the form of MX = N, where

\(M = \begin{bmatrix} 1 & 4 \\ 2 & 7 \end{bmatrix}, X = \begin{bmatrix} a \\ b \end{bmatrix} \; and \; N = \begin{bmatrix} 5 \\ 8 \end{bmatrix}\)

Thus,

|M| = 1(7) – 4(2) = 7 – 8 = -1 ≠ 0

Hence, M is a singular matrix.

Thus,

\(\left ( adj M \right ) = \begin{bmatrix} 7 & -4 \\ -2 & 1 \end{bmatrix}\) \(\left ( adj M \right )N = \begin{bmatrix} 7 & -4 \\ -2 & 1 \end{bmatrix}\begin{bmatrix} 5 \\ 8 \end{bmatrix}\)

= \(\begin{bmatrix} 35 – 32 \\ -10 + 8 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \end{bmatrix} \neq 0\)

Therefore, the solution for the given system of the equation does not exist. Thus, the given system of the equations will be inconsistent.

 

Q-3. Check the consistency of the system of three equations given below:

a + b + c = 1

2a + 3b + 2c = 2

pa + pb + 2pc = 4

 

Sol:

The given system of the two equations is:

a + b + c = 1

2a + 3b + 2c = 2

pa + pb + 2pc = 4

The given system of three equations will be written as in the form MX = N, where

\(M = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ p & p & 2p \end{bmatrix}, X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}\)

|M| = 1( 6p – 2p ) – 1( 4p – 2p ) + 1( 2p – 3p )

= 1( 4p ) – 1( 2p ) + 1( -p )

= 4p – 2p – p

= p ≠ 0

Hence, M is non- singular.

Thus, M-1 exists.

Therefore, the given system of two equations will be consistent.

 

 

Q-4. Check the consistency of the system of three equations given below:

3a – b – 2c = 2

2b – c = -1

3a – 5b = 3

Sol:

The given system of the two equations is:

3a – b – 2c = 2

2b – c = -1

3a – 5b = 3

The given system of three equations will be written as in the form MX = N, where

\(M = \begin{bmatrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}, X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}\)

|M| = 3[ 2 × 0 – ( -1 ) × ( -5 ) ] – 0[ ( -1 ) × 0 – ( -2 ) × ( -5 ) ] + 3[ ( -1 ) × ( -1 ) – ( -1 ) × ( 4 ) ]

= 3( 0 – 5 ) – 0 + 3(1 + 4)

= 3( -5 ) – 0 + 3( 5 )

= -15 + 15 = 0

Thus,

M is a singular matrix.

Then,

\(\left ( adj M \right ) = \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix}\) \(\left ( adj M \right )N = \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix}\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}\)

= \(\begin{bmatrix} -10 – 10 + 15 \\ -6 -6 + 9 \\ -12 – 12 + 18 \end{bmatrix} = \begin{bmatrix} -5 \\ -3 \\ -6 \end{bmatrix} \neq 0\)

Therefore, the solution for the given system of the equation does not exist. Thus, the given system of the equations will be inconsistent.

 

 

Q-5. Solve the following system of the linear equations by suing the matrix method:

5a + 2b = 4

7a + 3b = 5

Sol:

The given system of linear equations will be written as in the form of MX = N, where

\(M = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}, X = \begin{bmatrix} a \\ b \end{bmatrix} \; and \; N = \begin{bmatrix} 4 \\ 5 \end{bmatrix}\)

|M| = ( 5 × 3 ) – ( 7 × 2 ) = 15 – 14 = 1 ≠ 0

Hence, M is non- singular.

Thus, M-1 exists.

Now,

M-1 = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\)

⟹ M-1 = \(\frac{ 1 }{ 1 }\begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}\)

Thus,

X = M-1 N = \(\begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4 \\ 5 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4 \\ 5 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 3 \times 4 -2 \times 5 \\ -7\times 4 + 5 \times 5 \end{bmatrix} = \begin{bmatrix} 12 – 10 \\ -28 + 25 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}\)

Hence, a = 2 and b = -3.

 

 

Q-6. Solve the following system of the linear equations by suing the matrix method:

4a – 3b = 3

3a – 5b = 7

 

Sol:

The given system of linear equations will be written as in the form of MX = N, where

\(M = \begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}, X = \begin{bmatrix} a \\ b \end{bmatrix} \; and \; N = \begin{bmatrix} 3 \\ 7 \end{bmatrix}\)

|M| = [ 4 × (-5) ] – [ ( -3 ) × 3 ] = -20 + 9= -11 ≠ 0

Hence, M is non- singular.

Thus, M-1 exists.

Now,

M-1 = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\)

⟹ M-1 = \(– \frac{ 1 }{ 11 }\begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix}\) = \(\frac{ 1 }{ 11 }\begin{bmatrix} 5 & -3 \\ 3 & -4 \end{bmatrix}\)

Thus,

X = M-1 N = \( \frac{ 1 }{ 11 }\begin{bmatrix} 5 & -3 \\ 3 & -4 \end{bmatrix}\begin{bmatrix} 3 \\ 7 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \end{bmatrix} = \frac{ 1 }{ 11 } \begin{bmatrix} 5 & -3 \\ 3 & -4 \end{bmatrix}\begin{bmatrix} 3 \\ 7 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \end{bmatrix} = \frac{ 1 }{ 11 } \begin{bmatrix} 5 \times 3 + \left( -3 \right )\times 7 \\ 3\times 3 + \left( -4 \right ) \times 7 \end{bmatrix} = \frac{ 1 }{ 11 } \begin{bmatrix} 15 – 21 \\ 9 – 28 \end{bmatrix} = \frac{ 1 }{ 11 } \begin{bmatrix} -6 \\ -19 \end{bmatrix} = \begin{bmatrix} \frac{ -6 }{ 11 } \\ -\frac{ 19 }{ 11 } \end{bmatrix} \)

⟹ \(\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} \frac{ -6 }{ 11 } \\ -\frac{ 19 }{ 11 }\end{bmatrix}\)

Hence, a = \(– \frac{ 6 }{ 11 }\) and b = \(– \frac{ 19 }{ 11 }\).

 

Q-7. Solve the following system of the linear equations by suing the matrix method:

2a + b + c = 1

a – 2b – c = \(\frac{ 3 }{ 2 }\)

3b – 5c = 9

Sol:

The given system of linear equations will be written as in the form of MX = N, where

\(M = \begin{bmatrix} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{bmatrix}, \; X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 1 \\ \frac{ 3 }{ 2 }\\ 9 \end{bmatrix}\)

Thus,

|M| = 2[ (-2) × (-5) – (-1) × 3 ] – 1[ 1 × (-5) – 3 × 1 ] + 0[ (-2) × 1 – 1 × (-1) ] = 2( 10 + 3 ) – 1( -5 – 3 ) + 0

= 2 × 13 – (-8)

= 26 + 8 = 34 ≠ 0

Hence, M is non- singular.

Thus, M-1 exists.

Now,

M11 = 13, M12 = 5, M13 = 3

M21 = 8, M22 = -10, M23 = -6

M31 = 1, M32 = 3, M33 = -5

Thus,

M-1 = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\) = \(\frac{ 1 }{ 34 } \begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}\)

Thus,

X = M-1 N = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\) = \(\frac{ 1 }{ 34 } \begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix} \begin{bmatrix} 1 \\ \frac{ 3 }{ 2 }\\ 9 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 34 } \begin{bmatrix} 13\times 1 + 8 \times \frac{ 3 }{ 2 } + 1 \times 9 \\ 5\times 1 + \left ( -10 \right ) \times \frac{ 3 }{ 2 } + 3 \times 9 \\ 3\times 1 + \left ( -6 \right ) \times \frac{ 3 }{ 2 } + \left ( -5 \right ) \times 9 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 34 } \begin{bmatrix} 13 + 12 + 9 \\ 5 – 15 + 27 \\ 3 – 9 – 45 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 34 } \begin{bmatrix} 34 \\ 17 \\ -51 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 1 \\ \frac{ 1 }{ 2 } \\ -\frac{ 3 }{ 2 } \end{bmatrix}\)

Therefore, a = 1, b = \(\frac{ 1 }{ 2 }\) and c = –\(\frac{ 3 }{ 2 }\)

 

Q-8. Solve the following system of the linear equations by suing the matrix method:

a – b + c = 4

2a + b – 3c = 0

a + b + c = 2

Sol:

The given system of linear equations will be written as in the form of MX = N, where

\(M = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}, \; X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}\)

Thus,

|M| = 1[ 1 × 1 – (-3) × 1 ] – ( -1 )[ 2 × 1 – ( -3 ) × 1 ] + 1[ 2 × 1 – 1 × 1 ] = 2( 1 + 3 ) + 1( 2 + 3 ) + 1( 2 – 1 )

= 1 × 4 + 1 × 5 + 1 × 1

= 4 + 5 + 1 = 10 ≠ 0

Hence, M is non- singular.

Thus, M-1 exists.

Now,

M11 = 4, M12 = -5, M13 = 1

M21 = 2, M22 = 0, M23 = -2

M31 = 2, M32 = 5, M33 = 3

Thus,

M-1 = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\) = \(\frac{ 1 }{ 10 } \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}\)

Thus,

X = M-1 N = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\) = \(\frac{ 1 }{ 10 } \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 10 } \begin{bmatrix} 4\times 4 + 2 \times 0 + 2 \times 2 \\ -5\times 4 + 0 \times 0 + 5 \times 2 \\ 1 \times 4 + \left ( -2 \right ) \times 0 + 3 \times 2 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 10 } \begin{bmatrix} 16 + 0 + 4 \\ -20 + 0 + 10 \\ 4 + 0 + 6 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 10 } \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}\)

Therefore, a = 2, b = -1 and c = 1

 

 

Q-9. If from equations,

2a – 3b + 5c = 11, 3a + 2b – 4c = -5 and a + b – 2c = -3

M = \(\begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}\)

Find M-1. By using M-1, solve the system of the linear equations.

Sol:

M = \(\begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}\)

|M| = 2[ 2 × (-2) – 1 × (-4) ] + 3[ 3 × (-2) – 1 × (-4) ] + 5[ 3 × 1 – 2 × 1 ]

= 2( -4 + 4 ) + 3( -6 + 4 ) + 5( 3 – 2 )

= 2 × 0 + 3 × (-2) + 5 × 1

= 0 – 6 + 5 = -1 ≠ 0

Hence, M is non- singular.

Thus, M-1 exists.

Now,

M11 = 0, M12 = 2, M13 = 1

M21 = -1, M22 = -9, M23 = -5

M31 = 2, M32 = 23, M33 = 13

Thus,

M-1 = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\) = \(\frac{ 1 }{ -1 } \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}\)

The given system of linear equations will be written as in the form of MX = N, where

\(M = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}, \; X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}\)

The solution of the given system of linear equations will be given by X = M-1 N.

⟹ X = M-1 N = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\) = \(\begin{bmatrix} 4 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 0\times 11 + 1 \times \left( -5 \right ) + \left( -2 \right ) \times \left( -3 \right ) \\ -2\times 11 + 9 \times \left( -5 \right ) + \left( -23 \right ) \times \left( -3 \right ) \\ \left( -1 \right )\times 11 + 5 \times \left( -5 \right ) + \left( -13 \right ) \times \left( -3 \right ) \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 0 – 5 + 6 \\ -22 – 45 + 69 \\ -11 – 25 + 39 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\)

Therefore, a = 1, b = 2 and c = 3.

 

Q-10. The price of 4 kg of onion, 3 kg of wheat and 2 kg of rice is Rs 60. The price of 2 kg of onion, 4 kg of wheat and 6 kg of rice is Rs 90. The price of 6 kg of onion, 2 kg of wheat and 3 kg of rice is Rs 70. What is the price of each of the items (per kg)? Use matrix method to find the price.

 

Sol:

Let us consider the cost of onions, wheat, and rice per kg be given by Rs a, Rs b,and Rs c, respectively.

Thus, the given situation will be represented by the system of the equations as:

4a + 3b + 2c = 60

2a + 4b + 6c = 90

6a + 2b + 3c = 70

The given system of the equations will be written as in the form of MX = N, where

\(M = \begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix}, \; X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix}\)

Thus,

|M| = 4[ 4 × 3 – 6 × 2 ] – 3[ 2 × 3 – 6 × 6 ] + 2[ 2 × 2 – 6 × 4 ] = 2( 12 – 12 ) – 3( 6 – 36 ) + 2( 4 – 24 )

= 0 + 90 – 40 = 50 ≠ 0

Hence, M is non- singular.

Thus, M-1 exists.

Now,

M11 = 0, M12 = 30, M13 = -20

M21 = -5, M22 = 0, M23 = 10

M31 = 10, M32 = -20, M33 = 10

M-1 = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\) = \(\frac{ 1 }{ 50 } \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}\)

Thus,

X = M-1 N = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\) = \(\frac{ 1 }{ 50 } \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix} \begin{bmatrix} 60 \\90 \\ 70 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 50 } \begin{bmatrix} 0\times 60 + \left( -5 \right ) \times 90 + 10 \times 70 \\ 30 \times 60 + 0 \times 90 + \left( -20 \right ) \times 70 \\ \left( -20 \right ) \times 60 + 10 \times 90 + 10 \times 70 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 50 } \begin{bmatrix} 0 – 450 + 700 \\ 1800 + 0 – 1400 \\ -1200 + 900 + 700 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 50 } \begin{bmatrix} 250 \\ 400 \\ 400 \end{bmatrix}\)

⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix}\)

Therefore, a = 5, b = 8 and c = 8

Thus, the price for onions is Rs. 5 per kg, the price for wheat is Rs. 8 per kg and the price for the rice is Rs. 8 per kg.

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