These NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-4-ex-4-6/
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.6
Ex 4.6 Class 12 NCERT Solutions Question 1.
x + 2y = 2
2x + 3y = 3
Solution:
x + 2y = 2,
2x + 3y = 3
⇒ \(\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right] \)
⇒ AX = B
Now |A| = \(\begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix}\)
= 3 – 4
= – 1 ≠ 0.
Hence, equations are consistent.
Exercise 4.6 Class 12 NCERT Solutions Question 2.
2x – y = 5
x + y = 4
Solution:
2x – y = 5,
x + y = 4
⇒ \(\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 4 \end{matrix} \right] \)
⇒ AX = B
Now |A| = \(\begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix}\)
= 2 + 1
= 3 ≠ 0.
Hence, equations are consistent.
Exercise 4.6 Maths Class 12 Question 3.
x + 3y = 5,
2x + 6y = 8
Solution:
x + 3y = 5,
2x + 6y = 8
Hence solution does not exists and the system is inconsistent.
Exercise 4.6 Class 12 Maths Question 4.
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Solution:
The solution can be written as AX = B where A = \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
2 & 3 & 2 \\
a & a & 2 a
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\); B = \(\left[\begin{array}{l}
1 \\
2 \\
4
\end{array}\right]\)
|A| = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & 3 & 2 \\
a & a & 2 a
\end{array}\right|\)
= 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a) = 4a – 2a – a = a ≠ 0
∴ A is non singular and has a unique solution.
Hence the system is consistent (if a ≠ 0)
4.6 Class 12 NCERT Solutions Question 5.
3x – y – 2z = 2
2y – z = – 1
3x – 5y = 3
Solution:
The system can be written as AX = B where A = \(\left[\begin{array}{ccc}
3 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\); B = \(\left[\begin{array}{l}
2 \\
-1 \\
3
\end{array}\right]\)
∴ The solution does not exists and the system is inconsistent.
Class 12 Maths Ex 4.6 Solutions Question 6.
5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = – 1
Solution:
Given
5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = -1
\(\left[ \begin{matrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 2 \\ -1 \end{matrix} \right] \)
\(AX=B|A|=\left[ \begin{matrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{matrix} \right] \)
= 5(18 + 10)+1(12 – 25)+4(-4-15)
= 140-13-76
= 51 ≠ 0
∴ Hence equations are consistent with a unique solution.
Ex 4.6 Class 12 Maths NCERT Solutions Question 7.
5x + 2y = 4
7x + 3y = 5
Solution:
The given system of equations can be written as
Class 12 Maths Chapter 4 Exercise 4.6 Solutions Question 8.
2x – y = – 2
3x + 3y = 3
Solution:
The given system of equations can be written
Ex4.6 Class 12 NCERT Solutions Question 9.
4x – 3y = 3
3x – 5y = 7
Solution:
Ex 4.6 Class 12 Maths Ncert Solutions Question 10.
5x + 2y = 3
3x + 2y = 5
Solution:
4.6 Maths Class 12 NCERT Solutions Question 11.
2x + y + z = 1,
x – 2y – z = 3/2
3y – 5z = 9
Solution:
The system can be written as AX = B
Exercise 4.6 Class 12 Maths Ncert Solutions Question 12.
x – y + z = 4
2x + y – 3z = 0
x + y + z = 2.
Solution:
The system can be written as AX = B where A = \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\); B = \(\left[\begin{array}{l}
4 \\
0 \\
2
\end{array}\right]\)
|A| = \(\left|\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right|\) = 1(1 + 3) +1(2 + 3) + 1(2 – 1) = 10 ≠ 0
∴ A is non singular and has a unique solution.
Class 12 Ex 4.6 NCERT Solutions Question 13.
2x + 3y + 3z = 5
x – 2y + z = – 4
3x – y – 2z = 3
Solution:
The system can be written as AX = B where A = \(\left[\begin{array}{ccc}
2 & 3 & 3 \\
1 & -2 & 1 \\
3 & -1 & -2
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\); B = \(\left[\begin{array}{l}
5 \\
– 4 \\
3
\end{array}\right]\)
|A| = \(\left|\begin{array}{ccc}
2 & 3 & 3 \\
1 & -2 & 1 \\
3 & -1 & -2
\end{array}\right|\) = 2(5) – 3(- 5) + 2(5) = 40 ≠ 0
∴ A is non singular and has a unique solution.
Ncert Solutions For Class 12 Maths Chapter 4 Exercise 4.6 Question 14.
x – y + 2z = 7
3x + 4y – 5z = – 5
2x – y + 3z = 12.
Solution:
The system can be written as AX = B where A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\); B = \(\left[\begin{array}{l}
7 \\
– 5 \\
12
\end{array}\right]\)
|A| = \(\left|\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right|\) = 1(7) + 1(19) + 2(-11) = 4 ≠ 0
∴ A is non singular and has a unique solution.
Class 12 Maths Ex 4.6 NCERT Solutions Question 15.
If A = \(\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{matrix} \right] \) Find A-1. Using A-1. Solve the following system of linear equations 2x – 3y + 5z = 11, 3x + 2y – 4z = – 5, x + y – 2z = – 3
Solution:
Solution Of Exercise 4.6 Class 12 Question 16.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 69. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs. 90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is Rs. 70. Find the cost of each item per kg by matrix method
Solution:
Let cost of 1 kg onion = Rs x
and cost of 1 kg wheat = Rs y
and cost of 1 kg rice = Rs z
4x + 3y + 2z = 60
2x + 4y + 6z = 90
6x + 2y + 3z = 70
i.e., x = 5; y = 8; z = 8
i.e., Price of onion = ₹ 5/kg
Price of wheat = ₹ 8/kg
Price of rice = ₹ 8/kg