Author name: Prasanna

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

These NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-4-ex-4-2/

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.2

Class 12 Maths Ex 4.2 Solutions NCERT Solutions Question 1.
\(\left| \begin{matrix} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{matrix} \right|\) = 0
Solution:
L.H.S = \(\left| \begin{matrix} x & a & x \\ y & b & y \\ z & c & z \end{matrix} \right| +\left| \begin{matrix} x & a & a \\ y & b & b \\ z & c & c \end{matrix} \right| \)
(C₁ = C₃ and C₂ = C₃)
= 0 + 0
= 0
= R.H.S

Ncert Solutions Class 12 Maths Chapter 4 Exercise 4.2 Question 2.
\(\left| \begin{matrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{matrix} \right| =0\)
Solution:

Exercise 4.2 Class 12 NCERT Solutions Question 3.
\(\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| =0\)
Solution:
\(\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| =\left| \begin{matrix} 2 & 7 & 0 \\ 3 & 8 & 0 \\ 5 & 9 & 0 \end{matrix} \right| \)
\({ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }-{ 9C }_{ 2 }=0\)

Ex 4.2 Class 12 Maths NCERT Solutions  Question 4.
\(\left| \begin{matrix} 1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b) \end{matrix} \right| =0\)
Solution:

Ex 4.2 Class 12 NCERT Solutions Question 5.
\(\left| \begin{matrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{matrix} \right| =2\left| \begin{matrix} a & p & x \\ b & q & y \\ c & r & z \end{matrix} \right| \)
Solution:

By using properties of determinants in Exercise 6 to 14, show that:

Question 6.
\(\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right| =0\)
Solution:

Question 7.
\(\left| \begin{matrix} { -a }^{ 2 } & ab & ac \\ ba & { -b }^{ 2 } & bc \\ ac & cb & { -c }^{ 2 } \end{matrix} \right| ={ 4a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }\)
Solution:

Question 8.
(a) \(\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { b }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| =(a-b)(b-c)(c-a)\)
(b) \(\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ { a }^{ 3 } & { b }^{ 3 } & { c }^{ 3 } \end{matrix} \right| =(a-b)(b-c)(c-a)(a+b+c)\)
Solution:
(a) \(\left|\begin{array}{lll}
1 & a & a^{2} \\
1 & b & b^{2} \\
1 & c & c^{2}
\end{array}\right|\)

Taking out (b – a) from R₂ and (c – a) from R₃.
Expanding along C₁
= (b – a)(c – a)\(\left|\begin{array}{ll}
1 & b+a \\
1 & c+a
\end{array}\right|\)
= (b – a) (c – a) (c + a – b – a) = (b – a) (c – a) (c – b)
= (-1 )(a – b)(c – a)(-1)(b – c)
= (a- b) (b – c) (c – a)

(b) \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^{3} & b^{3} & c^{3}
\end{array}\right|\)

Question 9.
\(\left| \begin{matrix} x & x^{ 2 } & yx \\ y & { y }^{ 2 } & zx \\ z & { z }^{ 2 } & xy \end{matrix} \right| =(x-y)(y-z)(z-x)(xy+yz+zx)\)
Solution:
Let ∆ = \(\left| \begin{matrix} x & x^{ 2 } & yx \\ y & { y }^{ 2 } & zx \\ z & { z }^{ 2 } & xy \end{matrix} \right|\)
Applying R₁ → R₁ – R₂, R₂ → R₂ – R₃

Question 10.
(a) \(\left| \begin{matrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{matrix} \right| =(5x+4){ (4-x) }^{ 2 }\)
(b) \(\left| \begin{matrix} y+x & y & y \\ y & y+k & y \\ y & y & y+k \end{matrix} \right| ={ k }^{ 2 }(3y+k) \)
Solution:

Question 11.
(a) \(\left| \begin{matrix} a-b-c & \quad 2a & \quad 2a \\ 2b & \quad b-c-a & \quad 2b \\ 2c & 2c & \quad c-a-b \end{matrix} \right| ={ (a+b+c) }^{ 3 } \)
(b) \(\left| \begin{matrix} x+y+2z & \quad z & \quad z \\ x & \quad y+z+2x & \quad x \\ y & y & \quad z+x+2y \end{matrix} \right| ={ 2(x+y+z) }^{ 3 } \)
Solution:

Question 12.
\(\left| \begin{matrix} 1 & \quad x & { \quad x }^{ 2 } \\ { x }^{ 2 } & \quad 1 & x \\ x & { \quad x }^{ 2 } & 1 \end{matrix} \right| ={ { (1-x }^{ 3 }) }^{ 2 } \)
Solution:

Question 13.
\(\left| \begin{matrix} 1+{ a }^{ 2 }-{ b }^{ 2 } & \quad 2ab & \quad -2b \\ 2ab & \quad 1-{ a }^{ 2 }+{ b }^{ 2 } & \quad 2a \\ 2b & \quad -2a & \quad 1-{ a }^{ 2 }+{ b }^{ 2 } \end{matrix} \right| ={ (1+{ a }^{ 2 }+{ b }^{ 2 }) }^{ 3 } \)
Solution:

Question 14.
\(\left| \begin{matrix} { a }^{ 2 }+1 & \quad ab & \quad ac \\ ab\quad & \quad b^{ 2 }+1 & \quad bc \\ ca\quad & \quad cb & \quad { c }^{ 2 }+1 \end{matrix} \right| =1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \)
Solution:
Let
∆ = \(\left| \begin{matrix} { a }^{ 2 }+1 & \quad ab & \quad ac \\ ab\quad & \quad b^{ 2 }+1 & \quad bc \\ ca\quad & \quad cb & \quad { c }^{ 2 }+1 \end{matrix} \right| \)
\(\left| \begin{matrix} { a }^{ 2 }+1 & \quad ab+0 & \quad ac+0 \\ ab+0\quad & \quad b^{ 2 }+1 & \quad bc+0 \\ ca+0\quad & \quad cb+0 & \quad { c }^{ 2 }+1 \end{matrix} \right| \)
This may be expressed as the sum of 8 determinants

Question 15.
If A be a square matrix of order 3×3, then | kA | is equal to
(a) k|A|
(b) k² |A|
(c) k³ |A|
(d) 3k|A|
Solution:
Option (c) is correct.

Question 16.
Which of the following is correct:
(a) Determinant is a square matrix
(b) Determinant is a number associated to a matrix
(c) Determinant is a number associated to a square matrix
(d) None of these
Solution:
Option (c) is correct.

These NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-4-ex-4-5/

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.5

Ex 4.5 Class 12 NCERT Solutions Question 1.
\(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Adj A = \(\begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}\)

Exercise 4.5 Class 12 NCERT Solutions Question 2.
\(\left[ \begin{matrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{matrix} \right]\)
Solution:

Exercise 4.5 Class 12 Maths Question 3.
\(\begin{bmatrix} 2 & 3 \\ -4 & 6 \end{bmatrix}\)
Solution:

4.5 Class 12 NCERT Solutions Question 4.
\(\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right]\)
Solution:
|A| = \(\left|\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right|\)
A₁₁ = 0, A₁₂ = – 11, A₁₃ = 0,
A₂₁ = – 3, A₂₂ = 1, A₂₃ = 1,
A₃₁ = – 2, A₃₂ = 8, A₃₃ = 3

Ex 4.5 Class 12 Maths Ncert Solutions Question 5.
\(\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}\)
Solution:

Exercise 4.5 Maths Class 12 Question 6.
\(\begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}\)
Solution:

Exercise 4.5 Class 12 Maths Solutions Question 7.
\(\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{matrix} \right]\)
Solution:

Class 12 Maths Ch 4 Ex 4.5 Question 8.
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{matrix} \right]\)
Solution:

Ex4.5 Class 12 NCERT Solutions Question 9.
\(\left[ \begin{matrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{matrix} \right]\)
Solution:

Ex 4.5 Class 12 Maths NCERT Solutions Question 10.
\(\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} \right]\)
Solution:

Class 12 Maths Chapter 4 Exercise 4.5 Solutions Question 11.
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & cos\alpha & sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{matrix} \right] \)
Solution:

Ncert Solutions For Class 12 Maths Chapter 4 Exercise 4.5 Question 12.
Let \(A=\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix},B=\begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}\), verify that (AB)^-1 = B^-1A^-1
Solution:

Class 12 Maths Chapter 4 Exercise 4.5 Question 13.
If \(A=\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \) show that A² – 5A + 7I = 0, hence find A^-1.
Solution:

Class 12 Maths Chapter 4 Exercise 4.5 Solution Question 14.
For the matrix A = \(\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \) find file numbers a and b such that A² + aA + bI² = 0. Hence, find A^-1.
Solution:

Class 12 Maths Ex 4.5 Solutions Question 15.
For the matrix \(A=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{matrix} \right] \) Show that A³ – 6A² + 5A + 11I₃=0. Hence find A^-1
Solution:

i.e., A³ – 6A² + 5A + 11I₃ = 0 … (1)
|A| = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right|\)
= 1(3) – 1(9) + 1(- 5) = – 11 ≠ 0
∴ A^-1 exists
⇒ (1) = A.A.A – 6A.A. + 5A = – 11 I
⇒ AA(AA^-1 – 6A(AA^-1) + 5AA^-1 = – 11 IA^-1 (Multiplying by A^-1)
⇒ A²I – 6AI + 5I = – 11A^-1
⇒ A² – 6A + 5I = – 11A^-1
⇒ A^-1 = \(\frac { -1 }{ 11 }\)[A² – 6A + 5I]

Class 12 Ex 4.5 NCERT Solutions Question 16.
If \(A=\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right] \) show that A³ – 6A² + 9A – 4I = 0 and hence, find A^-1
Solution:

Exercise 4.5 Class 12 Maths Ncert Solutions Question 17.
Let A be a non-singular square matrix of order 3 x 3. Then | Adj A | is equal to:
(a) | A |
(b) | A |²
(c) | A |³
(d) 3 | A |
Solution:

Hence option (b) is correct.

Class 12 Maths Ex 4.5 NCERT Solutions Question 18.
If A is an invertible matrix of order 2, then det. (A^-1) is equal to:
(a) det. (A)
(b) \(\\ \frac { 1 }{ det.(A) } \)
(c) 1
(d) 0
Solution:
|A|≠0
⇒ A^-1 exists ⇒ A^-1 = I
|AA^-1| = |I| = I
⇒ |A||A^-1| = I
\(|{ A }^{ -1 }|=\frac { 1 }{ |A| } \)
Hence option (b) is correct.

These NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-4-ex-4-3/

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.3

Ex 4.3 Class 12 NCERT Solutions Question 1.
Find the area of the triangle with vertices at the point given in each of the following:
(i) (1, 0), (6, 0) (4, 3)
(ii) (2, 7), (1, 1), (10, 8)
(iii) (- 2, – 3), (3, 2), (- 1, – 8)
Solution:

Exercise 4.3 Class 12 NCERT Solutions Question 2.
Show that the points A (a, b + c), B (b, c + a) C (c, a+b) are collinear.
Solution:

∴ The given points are collinear.

4.3 Class 12 NCERT Solutions Question 3.
Find the value of k if area of triangle is 4 square units and vertices are
(i) (k, 0), (4, 0), (0, 2)
(ii) (-2, 0), (0, 4), (0, k).
Solution:
(i) Area of ∆ = 4 (Given)
\(\frac { 1 }{ 2 } \left| \begin{matrix} k\quad & 0 & \quad 1 \\ 4\quad & 0 & \quad 1 \\ 0\quad & 2 & \quad 1 \end{matrix} \right| \)
= \(\\ \frac { 1 }{ 2 } \) [-2k+8]
= -k+4
Case (a): -k + 4 = 4 ⇒ k = 0
Case (b): -k + 4 = -4 ⇒ k = 8
Hence, k = 0, 8

(ii) The area of the triangle whose vertices are (-2,0), (0,4), (0, k)
\(\frac{1}{2}\left|\begin{array}{ccc} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{array}\right|\) =
= ± 4 ⇒ \(\frac{1}{2}\)[- 2(4 – k) – 0 + (0)] = ± 4
⇒ 8 + 2k = ± 8 ⇒ 2k = 16 or 2k = 0 or k = 8

Exercise 4.3 Class 12 Maths NCERT Solutions Question 4.
(i) Find the equation of line joining (1, 2) and (3,6) using determinants.
(ii) Find the equation of line joining (3,1), (9,3) using determinants.
Solution:
(i) Let (x, y) be a point on the line, then the points (x, y), (1, 2) and (3, 6) are collinear.
∴ \(\left|\begin{array}{lll}
x & y & 1 \\
1 & 2 & 1 \\
3 & 6 & 1
\end{array}\right|\) = 0
⇒ x(2 – 6) – y(1 – 3) + 1(6 – 6) = 0
⇒ – 4x + 2y = 0 ⇒ 2x – y = 0
⇒ y = 2x

(ii) Let (x, y) be a point on the line. Then (r, y), (3, 1) and (9, 3) are collinear.
∴ \(\left|\begin{array}{lll}
x & y & 1 \\
3 & 1 & 1 \\
9 & 3 & 1
\end{array}\right|\) = 0
⇒ x(1 – 3) – y(3 – 9) + 1(9 – 9) = 0
⇒ 2x + 6y = 0 ⇒ x – 3y = 0

Ex 4.3 Class 12 Maths Ncert Solutions Question 5.
If area of triangle is 35 sq. units with vertices (2, – 6), (5,4) and (k, 4). Then k is
(a) 12
(b) – 2
(c) -12,-2
(d) 12,-2
Solution:
(d) Area of ∆ = \(\frac { 1 }{ 2 } \left| \begin{matrix} 2\quad & -6 & \quad 1 \\ 5\quad & 4 & \quad 1 \\ k\quad & 4 & \quad 1 \end{matrix} \right| \)
= \(\\ \frac { 1 }{ 2 } \) [50 – 10k] = 25 – 5k
∴ 25 – 5k = 35 or 25 – 5k = – 35
– 5k = 10 or 5k = 60
⇒ k = -2 or k = 12

These NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.4 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-4-ex-4-4/

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.4

Ex 4.4 Class 12 NCERT Solutions  Question 1.
Write the minors and cofactors of the elements of following determinants:
(i) \(\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}\)
(ii) \(\begin{vmatrix} a & c \\ b & d \end{vmatrix}\)
Solution:

Exercise 4.4 Class 12 NCERT Solutions Question 2.
Write Minors and Cofactor of elements of following determinant
(i) \(\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right| \)
(ii) \(\left| \begin{matrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{matrix} \right| \)
Solution:

4.4 Class 12 NCERT Solutions Question 3.
Using cofactors of elements of second row, evaluate \(\Delta =\left| \begin{matrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{matrix} \right| \)
Solution:

Exercise 4.4 Class 12 Maths NCERT Solutions Question 4.
Using Cofactors of elements of third column, evaluate \(\Delta =\left| \begin{matrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{matrix} \right| \)
Solution:

= yz² – y²z + zx² – xz² + xy² – x²y
= – xz² – zy² + z²y – x²y + x²z + xy²
= xyz – xz²- zy² + z²y – x²y + x²z + xy² – xyz (adding and substracting xyz)
= xz(y – z) – zy(y – z) – x²(y – z) + xy(y – z)
= (xz – zy) (y – z) – (x² – xy)(y – z)
= z(x – y)(y – z) – x(x – y)(y – z)
= (x – y)(y – z)(z – x)

Exercise 4.4 Class 12 Maths Solutions Question 5.
If \(\Delta =\left| \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right| \) and Aij is the cofactors of aij? then value of ∆ is given by
(a) a₁₁A₃₁ + a₁₂A₃₂ + a₁₃A₃₃
(b) a₁₁A₁₁ + a₁₂A₂₁ + a₁₃A₃₁
(c) a₂₁A₁₁ + a₂₂A₁₂ + a₂₃A₁₃
(d) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁
Solution:
Option (d) is correct.

These NCERT Solutions for Class 12 Maths Chapter 3 Matrices Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-3-miscellaneous-exercise/

NCERT Solutions for Class 12 Maths Chapter 3 Matrices Miscellaneous Exercise

Question 1.
Let A = \(\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right]\) show that (aI + bA)ⁿ = aⁿI + na^n-1 bA, where I is the identity matrix of order 2 and n ∈ N .
Solution:
Let P(n) : (aI + bA)ⁿ = aⁿI + na^n-1 bA
P(1) : (aI + bA)¹ = a¹I + 1a°bA
⇒ P(1) : aI + bA = aI + bA
∴ P(1) is true
Let us assume that P(k) is true.
i.e., P(k) : (aI + bA)^k = a^kI + ka^k-1bA
P(k+ 1) : (aI + bA)^k+1
= (aI + bA)^k.(aI + bA)
= (a^kI + ka^k-1A) (aI + bA)
= a^k.a.I.l + a^k.b. I.A + ka^k-1.b.a. AI + ka^k-1b.bA.A
= a^k+1I + a^kbA + ka^kbA + ka^k-1b²A²
= a^k+1I + (k + 1)a^kbA + ka^k-1b²A² … (1)
A² = \(\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\) = \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\) = 0
∴ (1) ⇒ P(k + 1) = a^k+1 I + (k + 1)a^kbA + kak^k-1b² x 0
= a^k+1 I + (k + 1)a^(k+1)-1Ab
which is true whenever P(L) is true.
Hence by the Principle of Mathematical Induction, the result is true for all n ∈ N .

Question 2.
If A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\), prove that Aⁿ = \(\left[\begin{array}{ccc}
3^{n-1} & 3^{n-1} & 3^{n-1} \\
3^{n-1} & 3^{n-1} & 3^{n-1} \\
3^{n-1} & 3^{n-1} & 3^{n-1}
\end{array}\right]\), n ∈ N
Solution:

which is true
whenever P(k) is true.
Hence by the Principle of Mathematical Induction P(n) is true for all n ∈ N.

Question 3.
If A = \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\), then prove that Aⁿ = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\), where n is any positive integer.
Solution:

∴ P(k + 1) is true, whenever P(k) is true.
Hence by the Principle of Mathematical Induction P(n) is true for all n ∈ N.

Question 4.
If A and Bare symmetric matrices, prove that AB – BA is a skew symmetric matrix.
Solution:
A and B are symmetric matrices
∴ A’ = A and B’ = B

(i) (AB – BA)’ = (AB)’ – (BA)’
= B’A’ – A’B’ since (AB)’ = B’A’
= BA – AB since A’ = A and
B’ = B
= – (AB – BA)
∴ AB – BA is a skew-symmetric matrix.

(ii) Let AB = BA
Taking transpose on both sides,
(AB)’ = (BA)’
i.e.,(AB)’ = A’B’ since (BA)’ = A’B’
i.e., (AB)’ = AB, since A’ A and B’ = B
∴ AB ¡s a symmetric matrix.
Let AB be a symmetric matrix.
Then (AB)’ = AB
⇒ B’A’ = AB
⇒ BA = AB since B’ = B and A’ = A
∴ AB = BA

Question 5.
Show that the matrix B’AB is symmetric or skew symmetric according as A is sym¬metric or skew symmetric.
Solution:
Let A be a symmetric matrix.
Then A’ = A
∴ (B’AB)’ = (B’(AB))’
= (AB)’(B’)’
= B’A’B
= B’AB since A’ = A
∴ B’AB is a symmetric matrix
Let A be a skew-symmetric matrix.
Then A’ = – A
∴ (B’AB)’ (B’(AB))’
= (AB)’(B’)’
= B’A’B
= B’(-A)B since A’= – A
= – B’AB
∴ B’AB is a skew-symmetric matrix.

Question 6.
Find the values of x, y, z if the matrix A = \(\left[\begin{array}{ccc}
0 & 2 y & z \\
x & y & -z \\
x & -y & z
\end{array}\right]\) satisfied the equation A’A = I.
Solution:

Question 7.
For what values of x if
\(\left[\begin{array}{lll}
1 & 2 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 0 \\
2 & 0 & 1 \\
1 & 0 & 2
\end{array}\right]\left[\begin{array}{l}
0 \\
2 \\
x
\end{array}\right]\) = 0
Solution:

Question 8.
If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\), show that A² – 5A + 7I = 0.
Solution:

Question 9.
Find x, if
\(\left[\begin{array}{lll}
x & -5 & -1
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
4 \\
1
\end{array}\right]\) = 0
Solution:

Question 10.
A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:

(i) If unit sale prices of x, y and z are ₹ 2.50, ₹ 1.50 and ₹ 1.00 respectively, find the total revenue in each market with the help of matrix algebra.
(ii) If the unit costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively, find the gross profit.
Solution:
(i) Let

represents annual sale of products in two markets and the column matrix
B = \(\left[\begin{array}{l|l}
2.50 \\
1.50 \\
1.00
\end{array}\right] \begin{aligned}
&x \\
&y \\
&z
\end{aligned}\) represents the unit sale price of the commodities x, y, z.
∴ The revenue collected by each market is given by AB.

The revenue in market I = ₹ 46,000 and
the revenue in market II = ₹ 53,000
Hence gross revenue in two markets
= ₹ 46,000 + 53,000 = ₹ 99,000

(ii) Let the column matrix C = \(\left[\begin{array}{c}
2.00 \\
1.00 \\
0.50
\end{array}\right]\)
Repre-sent the unit cost price of the commodities x, y and z.
∴ The cost price of the articles in the two markets is AC.

The cost price of articles in market I = ₹ 31,000 and the cost price of articles in market II = ₹ 36,000.
Hence the gross cost price
= ₹ 31,000 + 36,000 = ₹ 67,000
Gross profit = Gross revenue – Gross cost price
= ₹ 99,000 – 67, 000 = ₹ 32, 000.

Question 11.
Find the matrix X so that
X\(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6
\end{array}\right]\) = \(\left[\begin{array}{ccc}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right]\)
Solution:
X\(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6
\end{array}\right]\) = \(\left[\begin{array}{ccc}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right]\)
It is clear that the order of X is 2 x 2
since the product exists. Let X = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\).

Equating the corresponding elements
∴ a + 4b = – 7 … (1)
2a + 56 = – 8 … (2)
3a + 66 = – 9 … (3)
c + 4d = 2 … (4)
2c + 5d = 4 … (5)
3c + 6d = 6 … (6)
(1) + (2) ⇒ 3a + 96 = – 15
(3) ⇒ 3a + 6b = – 9
Subtracting we get 3b = – 6
⇒ 6 = – 2
∴ (1) ⇒ a = -7 + 8 = 1
(4) + (5) ⇒ 3c + 9d = 6
(6) ⇒ 3c + 6d = 6
Subtracting 3d = 0 ⇒ d = 0
(4) ⇒ c = 2
∴ X = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\) = \(\left[\begin{array}{cc}
1 & -2 \\
2 & 0
\end{array}\right]\)

Question 12.
If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABⁿ = BⁿA. Further, prove that (AB)ⁿ = AⁿBⁿ for all n ∈ N.
Solution:
Let P(n) : ABⁿ = BⁿA be the given statement.
P(1): AB’ = B’A which is true.
Assume that P(&) is true, i.e., AB^k = B^kA
AB^k+1 = (AB)B^k = (BA)B^k
= B(AB^k)
= B(B^kA) since P(k) is true
= B^k+1A
Hence by the Principle of Mathematical In-duction, P(n) is true for all values of n ∈ N .
Let P(n) : (AB)ⁿ = AⁿBⁿ
P(1) : (AB)’ = A’B’ ⇒ AB = AB which is true.
Assume that P(k) is true i.e., (AB)^k = A^kB^k
(AB)^k+1 = (AB)^k.(AB) = A^k.B^k(AB)
= A^k(B^kA)B
= A^k(AB^k)B (since B^k A = AB^k)
= (A^k.A)(B^kB)
= A^k+1 B^k+1
∴ By the Principle of Mathematical Induction P(n) is true for all n ∈ N.

Choose the correct answer in the following questions.

Question 13.
If A = \(\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]\) is such that A² = I, then
a. 1 + α² + βγ = 0
b. 1 – α² + βγ = 0
c. 1 – α² – βγ = 0
d. 1 + α² – βγ = 0
Solution:
c. 1 – α² – βγ = 0

Question 14.
If the matrix A is both symmetric and skew symmetric, then
a. A is a diagonal matrix
b. A is a zero matrix
c. A is a square matrix
d. None of these
Solution:
b. A is a zero matrix
For symmetric matrix a_ij = a_ji
For skew symmetric matrix a_ij = – a_ji
These conditions are satisfied only if a_ij = 0 for all i and j.
Hence the matrix is a zero matrix.

Question 15.
If A is square matrix such that A² = A, then (I + A)³ – 7A is equal to
a. A
b. I – A
c. I
d. 3A
Solution:
c. I
(I + A)³ – 7A = I³ + 3I²A + 3IA² + A³ – 7A
= I + 3A + 3A² + A²A – 7A
= I + 3A + 3A + A.A – 7A
= I + 6A + A – 7A
= I + 7A – 7A = I