NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths 4 Determinants Ex 4.6.
- Determinants Class 12 Ex 4.1
- Determinants Class 12 Ex 4.2
- Determinants Class 12 Ex 4.3
- Determinants Class 12 Ex 4.4
- Determinants Class 12 Ex 4.5
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 12 |
| Subject | Maths |
| Chapter | Chapter 4 |
| Chapter Name | Determinants |
| Exercise | Ex 4.6 |
| Number of Questions Solved | 16 |
| Category | NCERT Solutions |
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6
Examine the consistency of the system of equations in Questions 1 to 6:
Question 1.
x + 2y = 2
2x + 3y = 3
Solution:
x + 2y = 2,
2x + 3y = 3
=>
=> AX = B
Now |A| =
= 3 – 4
= – 1 ≠ 0.
Hence, equations are consistent.
![\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right]](http://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D+1+%26+2+%5C%5C+2+%26+3+%5Cend%7Bbmatrix%7D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+x+%5C%5C+y+%5Cend%7Bmatrix%7D+%5Cright%5D+%3D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+2+%5C%5C+3+%5Cend%7Bmatrix%7D+%5Cright%5D+&bg=ffffff&fg=000&s=2 "\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right]")
Question 2.
2x – y = 5
x + y = 4
Solution:
2x – y = 5,
x + y = 4
=>
=> AX = B
Now |A| =
= 2 + 1
= 3 ≠ 0.
Hence, equations are consistent.
![\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 4 \end{matrix} \right]](http://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D+2+%26+-1+%5C%5C+1+%26+1+%5Cend%7Bbmatrix%7D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+x+%5C%5C+y+%5Cend%7Bmatrix%7D+%5Cright%5D+%3D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+5+%5C%5C+4+%5Cend%7Bmatrix%7D+%5Cright%5D+&bg=ffffff&fg=000&s=2 "\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 4 \end{matrix} \right]")
Question 3.
x + 3y = 5,
2x + 6y = 8
Solution:
x + 3y = 5,
2x + 6y = 8
=>
=> AX = B
Now |A| =
= 6 – 6
= 0.
Hence, equations are consistent with no solution
![\begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 8 \end{matrix} \right]](http://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D+1+%26+3+%5C%5C+2+%26+6+%5Cend%7Bbmatrix%7D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+x+%5C%5C+y+%5Cend%7Bmatrix%7D+%5Cright%5D+%3D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+5+%5C%5C+8+%5Cend%7Bmatrix%7D+%5Cright%5D+&bg=ffffff&fg=000&s=2 "\begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 8 \end{matrix} \right]")
Question 4.
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Solution:
x + y + z = 1
2x + 3y + 2z = 2
x + y + z =
Question 5.
3x – y – 2z = 2
2y – z = – 1
3x – 5y = 3
Solution:
=> AX = B
![\left[ \begin{matrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right]](http://s0.wp.com/latex.php?latex=%5Cleft%5B+%5Cbegin%7Bmatrix%7D+3+%26+-1+%26+-2+%5C%5C+0+%26+2+%26+-1+%5C%5C+3+%26+-5+%26+0+%5Cend%7Bmatrix%7D+%5Cright%5D+%5Cleft%5B+%5Cbegin%7Bmatrix%7D+x+%5C%5C+y+%5C%5C+z+%5Cend%7Bmatrix%7D+%5Cright%5D+%3D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+2+%5C%5C+-1+%5C%5C+3+%5Cend%7Bmatrix%7D+%5Cright%5D+&bg=ffffff&fg=000&s=2 "\left[ \begin{matrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right]")
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
= 5(18 + 10)+1(12 – 25)+4(-4-15)
= 140-13-76
= 51 ≠ 0
Hence equations are consistent with a unique
solution.
![\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]](http://s0.wp.com/latex.php?latex=%5Cleft%5B+%5Cbegin%7Bmatrix%7D+5+%26+-1+%26+4+%5C%5C+2+%26+3+%26+5+%5C%5C+5+%26+-2+%26+6+%5Cend%7Bmatrix%7D+%5Cright%5D+%5Cleft%5B+%5Cbegin%7Bmatrix%7D+x+%5C%5C+y+%5C%5C+z+%5Cend%7Bmatrix%7D+%5Cright%5D+%3D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+5+%5C%5C+2+%5C%5C+-1+%5Cend%7Bmatrix%7D+%5Cright%5D+&bg=ffffff&fg=000&s=2 "\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]](http://s0.wp.com/latex.php?latex=AX%3DB%7CA%7C%3D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+5+%26+-1+%26+4+%5C%5C+2+%26+3+%26+5+%5C%5C+5+%26+-2+%26+6+%5Cend%7Bmatrix%7D+%5Cright%5D+&bg=ffffff&fg=000&s=2 "AX=B|A|=\left[ \begin{matrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{matrix} \right]")
Solve system of linear equations using matrix method in Questions 7 to 14:
Question 7.
5x + 2y = 4
7x + 3y = 5
Solution:
The given system of equations can be written as
Question 8.
2x – y = – 2
3x + 3y = 3
Solution:
The given system of equations can be written
Question 9.
4x – 3y = 3
3x – 5y = 7
Solution:
The given system of equations can be written as
where
![\begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 3 \\ 7 \end{matrix} \right] i.e,,AX=B](http://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D+4+%26+-3+%5C%5C+3+%26+-5+%5Cend%7Bbmatrix%7D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+x+%5C%5C+y+%5Cend%7Bmatrix%7D+%5Cright%5D+%3D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+3+%5C%5C+7+%5Cend%7Bmatrix%7D+%5Cright%5D+i.e%2C%2CAX%3DB&bg=ffffff&fg=000&s=2 "\begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 3 \\ 7 \end{matrix} \right] i.e,,AX=B")
Question 10.
5x + 2y = 3
3x + 2y = 5
Solution:
The given system of equations can be written as
where
![\begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 3 \\ 5 \end{matrix} \right] i.e,,AX=B](http://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D+5+%26+2+%5C%5C+3+%26+2+%5Cend%7Bbmatrix%7D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+x+%5C%5C+y+%5Cend%7Bmatrix%7D+%5Cright%5D+%3D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+3+%5C%5C+5+%5Cend%7Bmatrix%7D+%5Cright%5D+i.e%2C%2CAX%3DB&bg=ffffff&fg=000&s=2 "\begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 3 \\ 5 \end{matrix} \right] i.e,,AX=B")
Question 11.
2x + y + z = 1,
x – 2y – z = 3/2
3y – 5z = 9
Solution:
The given system of equations are
2x + y + z = 1,
x – 2y – z = 3/2,
3y – 5z = 9
We know AX = B => X = A-1B
Question 12.
x – y + z = 4
2x + y – 3z = 0
x + y + z = 2.
Solution:
The given system of equations can be written
![\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 0 \\ 2 \end{matrix} \right] i.e,,AX=B](http://s0.wp.com/latex.php?latex=%5Cleft%5B+%5Cbegin%7Bmatrix%7D+1+%26+-1+%26+1+%5C%5C+2+%26+1+%26+-3+%5C%5C+1+%26+1+%26+1+%5Cend%7Bmatrix%7D+%5Cright%5D+%5Cleft%5B+%5Cbegin%7Bmatrix%7D+x+%5C%5C+y+%5C%5C+z+%5Cend%7Bmatrix%7D+%5Cright%5D+%3D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+4+%5C%5C+0+%5C%5C+2+%5Cend%7Bmatrix%7D+%5Cright%5D+i.e%2C%2CAX%3DB&bg=ffffff&fg=000&s=2 "\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 0 \\ 2 \end{matrix} \right] i.e,,AX=B")
Question 13.
2x + 3y + 3z = 5
x – 2y + z = – 4
3x – y – 2z = 3
Solution:
The given system of equations can be written as:
![\left[ \begin{matrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 5 \\ -4 \\ 3 \end{matrix} \right] i.e,,AX=B](http://s0.wp.com/latex.php?latex=%5Cleft%5B+%5Cbegin%7Bmatrix%7D+2+%26+3+%26+3+%5C%5C+1+%26+-2+%26+1+%5C%5C+3+%26+-1+%26+-2+%5Cend%7Bmatrix%7D+%5Cright%5D+%5Cleft%5B+%5Cbegin%7Bmatrix%7D+x+%5C%5C+y+%5C%5C+z+%5Cend%7Bmatrix%7D+%5Cright%5D+%3D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+5+%5C%5C+-4+%5C%5C+3+%5Cend%7Bmatrix%7D+%5Cright%5D+i.e%2C%2CAX%3DB+&bg=ffffff&fg=000&s=2 "\left[ \begin{matrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 5 \\ -4 \\ 3 \end{matrix} \right] i.e,,AX=B")
Question 14.
x – y + 2z = 7
3x + 4y – 5z = – 5
2x – y + 3z = 12.
Solution:
The given system of equations can be written
![\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 7 \\ -5 \\ 12 \end{matrix} \right] i.e,,AX=B](http://s0.wp.com/latex.php?latex=%5Cleft%5B+%5Cbegin%7Bmatrix%7D+1+%26+-1+%26+2+%5C%5C+3+%26+4+%26+-5+%5C%5C+2+%26+-1+%26+3+%5Cend%7Bmatrix%7D+%5Cright%5D+%5Cleft%5B+%5Cbegin%7Bmatrix%7D+x+%5C%5C+y+%5C%5C+z+%5Cend%7Bmatrix%7D+%5Cright%5D+%3D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+7+%5C%5C+-5+%5C%5C+12+%5Cend%7Bmatrix%7D+%5Cright%5D+i.e%2C%2CAX%3DB+&bg=ffffff&fg=000&s=2 "\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 7 \\ -5 \\ 12 \end{matrix} \right] i.e,,AX=B")
Question 15.
If A = ![\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{matrix} \right]](http://s0.wp.com/latex.php?latex=%5Cleft%5B+%5Cbegin%7Bmatrix%7D+2+%26+-3+%26+5+%5C%5C+3+%26+2+%26+-4+%5C%5C+1+%26+1+%26+-2+%5Cend%7Bmatrix%7D+%5Cright%5D+&bg=ffffff&fg=000&s=2 "\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{matrix} \right]")
Solution:
We have AX = B
where
![A=\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right]](http://s0.wp.com/latex.php?latex=A%3D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+2+%26+-3+%26+5+%5C%5C+3+%26+2+%26+-4+%5C%5C+1+%26+1+%26+-2+%5Cend%7Bmatrix%7D+%5Cright%5D+%2CX%3D%5Cleft%5B+%5Cbegin%7Bmatrix%7D+x+%5C%5C+y+%5C%5C+z+%5Cend%7Bmatrix%7D+%5Cright%5D+&bg=ffffff&fg=000&s=2 "A=\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right]")
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
We hope the NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 help you. If you have any query regarding NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6, drop a comment below and we will get back to you at the earliest.
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