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NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 FAQs
What if the determinant is 0?
The matrix of the determinant is non-singular and not invertible. The matrix of the determinant may be a zero matrix. The system of equations associated with the matrix is linearly dependent.
Can determinant be negative?
Generally, a determinant is a real number, and it is not a matrix. But a determinant can be a negative number. Most importantly, it is not linked with absolute value at all except that they both use vertical lines.
Do all matrices have determinants?
The determinant of a matrix is the product of its eigenvalues. Non-square matrices don't have eigenvalues, so you can't define determinants for them.
Is determinant function linear?
As a function of row vectors, the determinant has certain properties. In particular, it is multilinear. This means that it is linear in each input.
What is the difference between a matrix and a determinant?
In a matrix, the set of numbers are arranged by two rectangular brackets whereas, in a determinant, the set of numbers are covered by two bars or by modulus sign. A matrix is a group of numbers, but a determinant is a unique number related to that matrix.
NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 Determinants
NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6: This page consist of detail and accurate NCERT Solutions For Class 12 Maths Chapter 4-Determinants Exercise 4.6 prepared as per the requirements of CBSE board.
Krati Saraswat16 Jan, 2024
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NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 Determinants
NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 Determinants is prepared by academic team of Physics Wallah. We have prepared
NCERT Solutions
for all exercise of chapter 4. Given below is step by step solutions of all questions given in NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6.
NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 Overview
NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 Determinants is prepared by our experts to help students understand the concepts of the chapter better.
Students can solve these questions before their examinations these questions will help them to understand the concepts better and by doing these questions students can easily ace their examinations.
NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 PDF
To assist students in comprehending and applying the concepts covered in the chapter, the Physics Wallah team has created comprehensive solutions for NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6. These questions are meant to aid students in understanding explanations.
By following this link, you can download the NCERT Solutions for Class 12 Maths, Chapter 4 PDF:
NCERT Solutions Class 12 Maths Chapter 4 PDF Download Link
NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6
Solve The Following Questions NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6
Examine the consistency of the system of equations in Exercises 1 to 3.
Question
1
.
x
+ 2
y
= 2
2
x
+ 3
y
= 3
Solution :
Matrix form of given equations is AX = B
∴
A
is non-singular.
Therefore,
A
−1
exists.
Hence, the given system of equations is consistent.
Solution :
Matrix form of given equations is AX = B
Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.
Question
6.
5
x
−
y
+ 4
z
= 5
2
x
+ 3
y
+ 5
z
= 2
5
x
− 2
y
+ 6
z
= −1
Solution :
Matrix form of given equations is AX = B
∴
A
is non-singular.
Therefore,
A
−1
exists.
Hence, the given system of equations is consistent.
Solve the system of linear equations, using matrix method, in Exercise 7 to 10.
Question
7.
5x + 2y =4
7x + 3y = 5
Solution :
Matrix form of given equations is AX = B
Question
8.
2x - y = - 2
3x + 4y = 3
Solution :
Matrix form of given equations is AX = B
Question
9.
4x - 3y = 3
3x - 5y = 7
Solution :
Matrix form of given equations is AX = B
Question
10.
5
x
+ 2
y
= 3
3
x
+ 2
y
= 5
Solution :
Matrix form of given equations is AX = B
Thus,
A
is non-singular. Therefore, its inverse exists.
Solve the system of linear equations, using matrix method, in Exercise 11 to 14.
Question
11.
Solution :
Matrix form of given equations is AX = B
Question
12.
x
−
y
+
z
= 4
2
x
+
y
− 3
z
= 0
x
+
y
+
z
= 2
Solution :
Matrix form of given equations is AX = B
Question
13.
2
x
+ 3
y
+ 3
z
= 5
x
− 2
y
+
z
= −4
3
x
−
y
− 2
z
= 3
Solution :
Matrix form of given equations is AX = B
Question
14.
x
−
y
+ 2
z
= 7
3
x
+ 4
y
− 5
z
= −5
2
x
−
y
+ 3
z
= 12
Solution :
Matrix form of given equations is AX = B
Question
15. If A =
find
A
−1
. Using A
−1
solve the system of equations
Solution :
Question
16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ` 60. The cost of 2 kg onion, 4 kg wheat and 2 kg rice is ` 90. The cost of 6 kg onion, 2 k wheat and 3 kg rice is ` 70. Find cost of each item per kg by matrix method.
Solution :
Let the cost of onions, wheat, and rice per kg be Rs
x
, Rs
y
,and Rs
z
respectively.
Then, the given situation can be represented by a system of equations as:
4x + 3y + 2z = 60
2x + 4y + 6z = 90
6x + 2y + 3z = 70
This system of equations can be written in the form of
AX
=
B
, where
Hence, the cost of onions is Rs 5 per kg, the cost of wheat is Rs 8 per kg, and the cost of rice is Rs 8 per kg.
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find
A
−1
. Using A
−1
solve the system of equations
