NCERT Solutions for Class 12 Maths Determinants Chapter 4

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Exercise 4.1

Evaluate the determinants in Exercises 1 and 2.

Question 1:
\(\begin{vmatrix}
2 & 4 \\-5 & -1
\end{vmatrix}\)
Solution
\(\begin{vmatrix}
2 & 4 \\-5 & -1
\end{vmatrix}\) = 2 \(\times\) (-1) – 4 \(\times\) (-5) = -2 + 20 = 18

Question 2:
(i) \(\begin{vmatrix}
\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta
\end{vmatrix}\) \(\quad\) (ii) \(\begin{vmatrix}
x^2-x+1 & x -1 \\ x +1 & x + 1
\end{vmatrix}\)
Solution
(i) \(\begin{vmatrix}
\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta
\end{vmatrix}\) = \(\cos \theta \times \cos \theta\) – \((-sin \theta) \times \sin \theta\) = \(\cos^{2}\theta + \sin^{2}\theta\) = 1

(ii) \(\begin{vmatrix}
x^2-x+1 & x -1 \\ x +1 & x + 1
\end{vmatrix}\) = (x + 1) (x² – x + 1) – (x – 1) (x + 1) = (x³ + 1 )- (x² – 1) = x³ + 1 – x² + 1 = x³ – x² + 2

Question 3:
If A =\(\begin{bmatrix}
1&2 \\ 4& 2
\end{bmatrix}\), then show that | 2A | = 4 | A |
Solution
A =\(\begin{bmatrix}
1&2 \\ 4& 2
\end{bmatrix}\) and 2A = 2\(\begin{bmatrix}
1&2 \\ 4& 2
\end{bmatrix}\) = \(\begin{bmatrix}
2&4 \\ 8& 4
\end{bmatrix}\)
|A| = 1\(\times\)2 – 2\(\times\)4 = 2 – 8= -6
|2A| = 2 \(\times\)4 – 4\(\times\)8 = 8 – 32 = -24
LHS
\(\quad\)= |2A| = -24
RHS
\(\quad\)= 4|A| = 4 \(\times\) (-6) = -24

Question 4:
If A =\(\begin{bmatrix}
1&0&1 \\0&1&2\\0&0&4
\end{bmatrix}\), then show that | 3A | = 27 | A |
Solution
A =\(\begin{bmatrix}
1&0&1 \\0&1&2\\0&0&4
\end{bmatrix}\) and 3A =\(\begin{bmatrix}
3&0&3 \\0&3&6\\0&0&12
\end{bmatrix}\)

|A| = \(\begin{vmatrix}
1&0&1 \\0&1&2\\0&0&4
\end{vmatrix}\) = 1\(\begin{vmatrix}
1&2 \\0&4
\end{vmatrix}\) – 0 \(\begin{vmatrix}
0&2 \\0&4
\end{vmatrix}\) + 1\(\begin{vmatrix}
0&1 \\0&0
\end{vmatrix}\) = 1 (4 – 0) – 0(0 – 0) + 1 (0 – 0) = 4

|3A| = \(\begin{vmatrix}
3&0&3 \\0&3&6\\0&0&12
\end{vmatrix}\) =3\(\begin{vmatrix}
3&6 \\0&12
\end{vmatrix}\) – 0 \(\begin{vmatrix}
0&6 \\0&12
\end{vmatrix}\) + 3\(\begin{vmatrix}
0&3 \\0&0
\end{vmatrix}\) = 3 (36 – 0) – 0(0 – 0) + 3 (0 – 0) = 108

LHS = |3A| = 108
RHS = 27|A| = 27(4) = 108

Question 5:
Evaluate the determinants:
(i) \(\begin{vmatrix}
3&-1&-2 \\0&0&-1\\3&-5&0
\end{vmatrix}\) \(\quad\) (ii) \(\begin{vmatrix}
3&-4&5 \\1&1&-2\\2&3&1
\end{vmatrix}\)\(\quad\) (iii) \(\begin{vmatrix}
0&1&2 \\-1&0&-3\\-2&3&0
\end{vmatrix}\)\(\quad\) (iv) \(\begin{vmatrix}
2&-1&-2 \\0&2&-1\\3&-5&0
\end{vmatrix}\)
Solution
(i) \(\begin{vmatrix}
3&-1&-2 \\0&0&-1\\3&-5&0
\end{vmatrix}\) = 3\(\begin{vmatrix}
0&-1 \\-5&0
\end{vmatrix}\) – (-1) \(\begin{vmatrix}
0&-1 \\3&0
\end{vmatrix}\) + (-2)\(\begin{vmatrix}
0&0 \\3&-5
\end{vmatrix}\) = 3 (0 – 5) + 1(0 + 3) – 2 (0 – 0) = -15 + 3 = -12

(ii) \(\begin{vmatrix}
3&-4&5 \\1&1&-2\\2&3&1
\end{vmatrix}\) = 3\(\begin{vmatrix}
1&-2 \\3&1
\end{vmatrix}\) – (-4) \(\begin{vmatrix}
1&-2 \\2&1
\end{vmatrix}\) + 5\(\begin{vmatrix}
1&1 \\2&3
\end{vmatrix}\) = 3 (1 + 6) + 4(1 + 4) + 5 (3 – 2) = 21 + 20 + 5 = 46

(iii) \(\begin{vmatrix}
0&1&2 \\-1&0&-3\\-2&3&0
\end{vmatrix}\) = 0\(\begin{vmatrix}
0&-3 \\3&0
\end{vmatrix}\) – 1 \(\begin{vmatrix}
-1&-3 \\-2&0
\end{vmatrix}\) + 2\(\begin{vmatrix}
-1&0 \\-2&3
\end{vmatrix}\) = 0 (0 + 9) – 1(0 – 6) + 2 (-3 – 0) = 0 + 6 – 6 = 0

(iv) \(\begin{vmatrix}
2&-1&-2 \\0&2&-1\\3&-5&0
\end{vmatrix}\)= 2\(\begin{vmatrix}
2&-1 \\-5&0
\end{vmatrix}\) – (-1) \(\begin{vmatrix}
0&-1 \\3&0
\end{vmatrix}\) + (-2)\(\begin{vmatrix}
0&2 \\3&-5
\end{vmatrix}\) = 2 (0 – 5) + 1(0 + 3) – 2 (0 – 6) = -10 + 3 + 12= 5

Question 6:
If A = \(\begin{bmatrix}
1&1&-2 \\2&1&-3\\5&4&-9
\end{bmatrix}\), find |A|.
Solution
|A| = \(\begin{vmatrix}
1&1&-2 \\2&1&-3\\5&4&-9
\end{vmatrix}\) = 1\(\begin{vmatrix}
1&-3 \\4&-9
\end{vmatrix}\) – 1 \(\begin{vmatrix}
2&-3 \\5&-9
\end{vmatrix}\) + (-2)\(\begin{vmatrix}
2&1 \\5&4
\end{vmatrix}\) = 1 (-9 + 12) – 1(-18 + 15) – 2 (8 – 5) = 3 + 3 – 6= 0

Question 7:
Find the values of x, if
(i) \(\begin{vmatrix}
2&4\\5&1
\end{vmatrix}\) = \(\begin{vmatrix}
2x&4\\6&x
\end{vmatrix}\) \(\quad\)(ii) \(\begin{vmatrix}
2&3\\4&5
\end{vmatrix}\) = \(\begin{vmatrix}
x&3\\2x&5
\end{vmatrix}\)
Solution
(i) \(\begin{vmatrix}
2&4\\5&1
\end{vmatrix}\) = \(\begin{vmatrix}
2x&4\\6&x
\end{vmatrix}\)
\(\Rightarrow\) 2 \(\times\) 1- 4 \(\times\) 5 = 2x\(\times\)x – 4\(\times\)6
\(\Rightarrow\) 2 – 20 = 2x² – 24 \(\Rightarrow\) -18 = 2x² – 24 \(\Rightarrow\) – 18 + 24 = 2x²
\(\Rightarrow\) 2x² = 6 \(\Rightarrow\) x² = 3 \(\Rightarrow\) x = \(\pm \sqrt{3}\)

(ii) \(\begin{vmatrix}
2&3\\4&5
\end{vmatrix}\) = \(\begin{vmatrix}
x&3\\2x&5
\end{vmatrix}\)
\(\Rightarrow\) 2 \(\times\) 5- 3 \(\times\) 4 = x\(\times\)5 – 3\(\times\)2x
\(\Rightarrow\) 10 – 12 = 5x – 6x \(\Rightarrow\) -2 = -x \(\Rightarrow\) x = 2

Question 8:
If \(\begin{vmatrix}
x&2\\18&x
\end{vmatrix}\) = \(\begin{vmatrix}
6&2\\18&6
\end{vmatrix}\), then x is equal to
(A) 6 \(\quad\) (B) \(\pm\)6 \(\quad\) (C) – 6 \(\quad\) (D) 0
Solution
\(\begin{vmatrix}
x&2\\18&x
\end{vmatrix}\) = \(\begin{vmatrix}
6&2\\18&6
\end{vmatrix}\)
\(\Rightarrow\) x \(\times\) x- 2 \(\times\) 18 = 6 \(\times\)6 – 2 \(\times\) 18
\(\Rightarrow\) x² – 36 = 36 -36 \(\Rightarrow\) x² = 36 \(\Rightarrow\) x = \(\pm\)6 (B)

NCERT Solutions for Class 12 Maths Chapter 4 Determinants

Exercise 4.2

Question 1:
Find 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
(i) (1, 0), (6, 0), (4, 3)
\(\quad\) \(\triangle\) = \(\frac{1}{2}\begin{vmatrix} 1&0&1\\6&0&1\\4&3&1\end{vmatrix}\)

(ii) (2, 7), (1, 1), (10, 8)
\(\quad\) \(\triangle\) = \(\frac{1}{2}\begin{vmatrix} 2&7&1\\1&1&1\\10&8&1\end{vmatrix}\)

(iii) (–2, –3), (3, 2), (–1, –8)
\(\quad\) \(\triangle\) = \(\frac{1}{2}\begin{vmatrix} -2&-3&1\\3&2&1\\-1&-8&1\end{vmatrix}\)

Question 2:
Show that points A (a, b + c), B (b, c + a), C (c, a + b) are collinear.

Question 3:
Find values of k if area of triangle is 4 sq. units and vertices are
(i) (k, 0), (4, 0), (0, 2) (ii) (–2, 0), (0, 4), (0, k)

Question 4:
(i) Find equation of line joining (1, 2) and (3, 6) using determinants.
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

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

NCERT Solutions for Class 12 Maths Chapter 4 Determinants

Exercise 4.3

Write Minors and Cofactors of the elements of following determinants:

Question 1:
(i) \(\begin{vmatrix}
2&-4\\0&3
\end{vmatrix}\)\(\quad\quad\) (ii) \(\begin{vmatrix}
a&c\\b&d
\end{vmatrix}\)

Question 2:
(i) \(\begin{vmatrix}
1&0&0\\0&1&0\\0&0&1
\end{vmatrix}\) \(\quad\quad\)(ii) \(\begin{vmatrix}
1&0&4\\3&5&-1\\0&1&2
\end{vmatrix}\)

Question 3:
Using Cofactors of elements of second row, evaluate \(\triangle\) = \(\begin{vmatrix}
5&3&8\\2&0&1\\1&2&3
\end{vmatrix}\)

Question 4:
Using Cofactors of elements of third column, evaluate \(\triangle\) = \(\begin{vmatrix}
1&x&yz\\1&y&zx\\1&z&xy
\end{vmatrix}\)

Question 5:
If \(\triangle\) = \(\begin{vmatrix}
a_11&a_12&a_13\\a_21&a_22&a_23\\a_31&a_32&a_33
\end{vmatrix}\) and A_ij is is Cofactors of a_ij, then value of ∆ is given by
(A) a₁₁ A₃₁+ a₁₂ A₃₂ + a₁₃ A₃₃ \(\quad\) (B) a₁₁ A₁₁+ a₁₂ A₂₁ + a₁₃ A₃₁ \(\quad\)
(C) a₂₁ A₁₁ + a₂₂ A₁₂ + a₂₃ A₁₃ \(\quad\)(D) a₁₁ A₁₁ + a₂₁ A₂₁ + a₃₁ A₃₁

NCERT Solutions for Class 12 Maths Chapter 4 Determinants

Exercise 4.4

Find adjoint of each of the matrices in Exercises 1 and 2.

Question 1:
\(\begin{bmatrix}
1&2\\3&4
\end{bmatrix}\)

Question 2:
\(\begin{bmatrix}
1&-1&2\\2&3&4\\-2&0&1
\end{bmatrix}\)

Verify A (adj A) = (adj A) A = |A| I in Exercises 3 and 4

Question 3:
\(\begin{bmatrix}
2&3\\-4&-6
\end{bmatrix}\)

Question 4:
\(\begin{bmatrix}
1&-1&2\\3&0&-2\\1&0&3
\end{bmatrix}\)

Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.

Question 5:
\(\begin{bmatrix}
2&-2\\4&3
\end{bmatrix}\)

Question 6:
\(\begin{bmatrix}
-1&5\\-3&2
\end{bmatrix}\)

Question 7:
\(\begin{bmatrix}
1&2&3\\0&2&4\\0&0&5
\end{bmatrix}\)

Question 8:
\(\begin{bmatrix}
1&0&0\\3&3&0\\5&2&-1
\end{bmatrix}\)

Question 9:
\(\begin{bmatrix}
2&1&3\\4&-1&0\\-7&2&1
\end{bmatrix}\)

Question 10:
\(\begin{bmatrix}
1&-1&2\\0&2&-3\\3&-2&4
\end{bmatrix}\)

Question 11:
\(\begin{bmatrix}
1&0&0\\0&\cos \alpha&\sin \alpha\\0&\sin \alpha&-\cos \alpha
\end{bmatrix}\)

Question 12:
Let A = \(\begin{bmatrix}
3&7\\2&5
\end{bmatrix}\) and B = \(\begin{bmatrix}
6&8\\7&9
\end{bmatrix}\). Verify that (AB)^-1 = B^-1A^-1.

Question 13:
If A = \(\begin{bmatrix}
3&1\\-1&2
\end{bmatrix}\), show that A² – 5A + 7I = 0. Hence find A^-1.

Question 14:
For the matrix A = \(\begin{bmatrix}
3&2\\1&1
\end{bmatrix}\), find the numbers a and b such that A² + aA + bI = O.

Question 15:
For the matrix A = \(\begin{bmatrix}
1&1&1\\1&2&-3\\2&-1&3
\end{bmatrix}\)
Show that A3 – 6A2 + 5A + 11 I = 0. Hence, find A^-1.

Question 16:
If A = \(\begin{bmatrix}
2&-1&1\\-1&2&-1\\1&-1&2
\end{bmatrix}\)
Verify that A³ – 6A² + 9A – 4 I = 0 and hence find A^-1.

Question 18:
If A is an invertible matrix of order 2, then det (A^–1) is equal to
(A) det (A) \(\quad\) (B) \(\frac{1}{det (A)}\) \(\quad\) (C) 1 \(\quad\) (D) 0

NCERT Solutions for Class 12 Maths Chapter 4 Determinants

Exercise 4.5

Examine the consistency of the system of equations in Exercises 1 to 6.

Question 1:
x + 2y = 2;
2x + 3y = 3
Solution
x + 2y = 2
2x + 3y = 3
Matrix form of given equations is AX = B
A = \(\begin{bmatrix}
1&2\\2&3
\end{bmatrix}\) and B = \(\begin{bmatrix}
2\\3\end{bmatrix}\) and X = \(\begin{bmatrix}
x\\y\end{bmatrix}\).
Now |A| = \(\begin{vmatrix}
1&2\\2&3
\end{vmatrix}\) = 1 \(\times\) 3 – 2 \(\times\) 2 = 3 – 4 = -1 \(\neq\) 0
\(\therefore\) A is non-singular.
Therefore, A^-1 exists.
Hence, the given system of equations is consistent.

Question 2:
2x – y = 5
x + y = 4
Solution
2x – y = 5
x + y = 4
Matrix form of given equations is AX = B
A = \(\begin{bmatrix}
2&-1\\1&1
\end{bmatrix}\) and B = \(\begin{bmatrix}
5\\4\end{bmatrix}\) and X = \(\begin{bmatrix}
x\\y\end{bmatrix}\).
Now |A| = \(\begin{vmatrix}
2&-1\\1&1
\end{vmatrix}\) = 2 \(\times\) 1 – (-1) \(\times\) 1 = 2 + 1 = 3 \(\neq\) 0
\(\therefore\) A is non-singular.
Therefore, A^-1 exists.
Hence, the given system of equations is consistent.

Question 3:
x + 3y = 5
2x + 6y = 8
Solution
x + 3y = 5
2x + 6y = 8
Matrix form of given equations is AX = B
A = \(\begin{bmatrix}
1&3\\2&6
\end{bmatrix}\) and B = \(\begin{bmatrix}
5\\8\end{bmatrix}\) and X = \(\begin{bmatrix}
x\\y\end{bmatrix}\).
Now |A| = \(\begin{vmatrix}
1&3\\2&6
\end{vmatrix}\) = 1 \(\times\) 6 – 3 \(\times\) 2 = 6 – 6 = 0
\(\therefore\) A is singular.
adj A = \(\begin{bmatrix}
6&-3\\-2&1
\end{bmatrix}\)
(adj A) B = \(\begin{bmatrix}
1&3\\2&6
\end{bmatrix} \begin{bmatrix}
5\\8
\end{bmatrix}\) = \(\begin{bmatrix}
30-24\\-10+8
\end{bmatrix}\) = \(\begin{bmatrix}
6\\-2
\end{bmatrix}\) \(\neq\) 0
Thus, the solution of the given system of equations does not exist.
Hence, the system of equations is inconsistent.

Question 4:
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Solution
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Matrix form of given equations is AX = B
A = \(\begin{bmatrix}
1&1&1\\2&3&2\\a&a&2a
\end{bmatrix}\) and B = \(\begin{bmatrix}
1\\2\\4\end{bmatrix}\) and X = \(\begin{bmatrix}
x\\y\\z\end{bmatrix}\).
Now |A| = \(\begin{vmatrix}
1&1&1\\2&3&2\\a&a&2a
\end{vmatrix}\) = 1 (6a – 2a)- 1 (4a – 2a) + 1 (2a – 3a) = 4a – 2a – a = a \(\neq\) 0
\(\therefore\) A is non-singular.
Therefore, A^-1 exists.
Hence, the given system of equations is consistent.

Question 5:
3x–y – 2z = 2
2y – z = –1
3x – 5y = 3
Solution
3x–y – 2z = 2
2y – z = –1
3x – 5y = 3
Matrix form of given equations is AX = B
A = \(\begin{bmatrix}
3&-1&-2\\0&2&-1\\3&-5&0
\end{bmatrix}\) and B = \(\begin{bmatrix}
2\\-1\\3\end{bmatrix}\) and X = \(\begin{bmatrix}
x\\y\\z\end{bmatrix}\).
Now |A| = \(\begin{vmatrix}
3&-1&-2\\0&2&-1\\3&-5&0
\end{vmatrix}\) = 3 (0 – 5)- (-1) (0 + 3) + (-2) (0 – 6) = -15 + 3 + 12 = 0
\(\therefore\) A is singular.
Now,
adj A = \(\begin{bmatrix}
-5&10&5\\-3&6&3\\-6&12&6
\end{bmatrix}\)
(adj A) B = \(\begin{bmatrix}
-5&10&5\\-3&6&3\\-6&12&6
\end{bmatrix} \begin{bmatrix}
2\\-1\\3
\end{bmatrix}\) = \(\begin{bmatrix}
-10-10+15\\-6-6+9\\-12-12+18
\end{bmatrix}\) = \(\begin{bmatrix}
-5\\-3\\-6
\end{bmatrix}\) \(\neq\) 0
Thus, the solution of the given system of equations does not exist.
Hence, the system of equations is inconsistent.

Question 6:
5x – y + 4z = 5; 2x + 3y + 5z = 2; 5x – 2y + 6z = –1

Solve system of linear equations, using matrix method, in Exercises 7 to 14.

Question 7:
5x + 2y = 4 ; 7x + 3y = 5

Question 8:
2x – y = –2 ; 3x + 4y = 3

Question 9:
4x – 3y = 3; 3x – 5y = 7

Question 10:
5x + 2y = 3; 3x + 2y = 5

Question 11:
2x + y + z = 1; x – 2y – z = \(\frac{3}{2}\) ; 3y – 5z = 9

Question 12:
x – y + z = 4; 2x + y – 3z = 0; x + y + z = 2

Question 13:
2x + 3y +3 z = 5; x – 2y + z = – 4; 3x – y – 2z = 3

Question 14:
x – y + 2z = 7; 3x + 4y – 5z = – 5; 2x – y + 3z = 12

Question 15:
If A = \(\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2 \end{bmatrix}\), find A^-1. Using A^-1 solve the system of equations
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3

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 6 kg rice is ₹ 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹ 70. Find cost of each item per kg by matrix method.

NCERT Solutions for Class 12 Maths Chapter 4 Determinants

Miscellaneous Exercises on Chapter 4

Question 1:
Prove that the determinant \(\begin{vmatrix}
x & \sin \theta &\cos \theta \\ -\sin \theta & -x & 1\\ \cos \theta &1 &x
\end{vmatrix}\) is independent of θ.

Question 2:
Evaluate \(\begin{vmatrix}
\cos \alpha \; \cos \beta & \cos \alpha \; \sin \beta &-\sin \alpha\\ -\sin \beta & \cos \beta & 0\\ \sin \alpha \;\cos \beta &\sin \alpha \;\sin \beta &\cos \alpha
\end{vmatrix}\)

Question 3:
If A^-1 = \(\begin{vmatrix} 3&-1&1\\-15&6&-5\\5&-2&2\end{vmatrix}\) and B^-1 = \(\begin{vmatrix} 1&2&-2\\-1&3&0\\0&-2&1\end{vmatrix}\), find (AB)^-1.

Question 4:
Let A = \(\begin{vmatrix} 1&2&1\\2&3&1\\1&1&5\end{vmatrix}\). Verfify that
\(\quad\) (i) [adj A]^-1 = adj (A^-1) \(\quad \quad\) (ii) (A^-1)^-1 = A

Question 5:
Evaluate \(\begin{vmatrix} x&y&x + y\\y&x + y&x\\x + y&x&y\end{vmatrix}\)

Question 6:
Evaluate \(\begin{vmatrix} 1&x&y\\1&x + y&y\\1&x&x + y\end{vmatrix}\)

Question 7:
Solve the system of equations
\(\frac{2}{x}\) + \(\frac{3}{y}\) + \(\frac{10}{z}\) = 4
\(\frac{4}{x}\) – \(\frac{6}{y}\) + \(\frac{5}{z}\) = 1
\(\frac{6}{x}\) + \(\frac{9}{y}\) – \(\frac{20}{z}\) = 2
Solution
Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) =q, \(\frac{1}{z}\) = r
2p + 3q + 10r = 4
4p – 6q + 5r = 1
6p + 9q – 20r = 2
This system can be written in the form of AX = B, where
A = \(\begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20 \end{bmatrix}\), X = \(\begin{bmatrix}p\\q\\r\end{bmatrix}\) and B = \(\begin{bmatrix}4\\1\\2\end{bmatrix}\)

Question 8:
If x, y, z are nonzero real numbers, then the inverse of matrix A = \(\begin{bmatrix} x&0&0 \\0&y&0\\0&0&z \end{bmatrix}\) is
(A) \(\begin{bmatrix} x^{-1}&0&0 \\0&y^{-1}&0\\0&0&z^{-1} \end{bmatrix} \quad\) (B) xyz \(\begin{bmatrix} x^{-1}&0&0 \\0&y^{-1}&0\\0&0&z^{-1} \end{bmatrix}\)
(C) \(\frac{1}{xyz} \begin{bmatrix} x&0&0 \\0&y&0\\0&0&z \end{bmatrix} \quad \quad\) (D) \(\frac{1}{xyz} \begin{bmatrix} 1&0&0 \\0&1&0\\0&0&1 \end{bmatrix}\)
Solution
A = \(\begin{bmatrix} x&0&0 \\0&y&0\\0&0&z \end{bmatrix}\)
|A| = x (yz – 0 ) – 0(0 – 0) + 0 (0 – 0) = xyz \(\ne\) 0.
Now,
adj A = \(\begin{bmatrix} yz&0&0 \\0&xz&0\\0&0&xy \end{bmatrix}\)
\(\therefore\) A^-1 = \(\frac{1}{|A|}\) adj A
= \(\frac{1}{xyz} \begin{bmatrix} yz&0&0 \\0&xz&0\\0&0&xy \end{bmatrix}\) = \(\begin{bmatrix} \frac{1}{x}&0&0 \\0&\frac{1}{y}&0\\0&0&\frac{1}{z} \end{bmatrix}\) = \(\begin{bmatrix} x^{-1}&0&0 \\0&y^{-1}&0\\0&0&z^{-1} \end{bmatrix}\)
Answer is (A).

Question 9:
Let A = \(\begin{bmatrix} 1&\sin \theta &1 \\- \sin \theta &1&\sin \theta\\-1&-\sin \theta &1 \end{bmatrix}\), where\(\le \theta \le 2 \pi\).
(A) Det(A) = 0 \(\quad\) (B) Det(A) ∈ (2, \(\infty\)) \(\quad\) (C) Det(A) ∈ (2, 4) \(\quad\) (D) Det(A) ∈ [2, 4]
Solution
A = \(\begin{bmatrix} 1&\sin \theta &1 \\- \sin \theta &1&\sin \theta\\-1&-\sin \theta &1 \end{bmatrix}\)
|A| = 1 (1 + \(\sin^2 \theta\)) – \(\sin \theta\)(-\(\sin \theta + \sin \theta\)) + 1 (\(\sin^2 \theta\) + 1) = 2 + 2\(\sin^2 \theta\) \(\ne\) 0.
Now, 0 \(\le \theta \le 2\pi\)
\(\Rightarrow\) 0 \(\le \sin \theta \le 1\)
\(\Rightarrow\) 0 \(\le \sin^2 \theta \le 1\)
\(\Rightarrow\) 1 \(\le + 1 + \sin \theta \le 2\)
\(\Rightarrow\) 2 \(\le 2(1 + \sin^2 \theta) \le 4\)
\(\therefore\) Det (A) \(\in\)[2, 4]
Answer is D

NCERT Solutions for Class 12 Maths Chapter 4 Determinants

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NCERT Solutions for Class 12 Maths Determinants