These NCERT Solutions for Class 12 Maths Chapter 4 Determinants Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Miscellaneous Exercise
Question 1.
Prove that the determinant \(\left|\begin{array}{ccc}
x & \sin \theta & \cos \theta \\
-\sin \theta & -x & 1 \\
\cos \theta & 1 & x
\end{array}\right|\) is independent of θ.
Solution:
Question 2.
Without expanding the determinant, prove that \(\left|\begin{array}{lll}
a & a^{2} & b c \\
b & b^{2} & c a \\
c & c^{2} & a b
\end{array}\right|=\left|\begin{array}{lll}
1 & a^{2} & a^{3} \\
1 & b^{2} & b^{3} \\
1 & c^{2} & c^{3}
\end{array}\right|\)
Solution:
Question 3.
Evaluate \(\left|\begin{array}{ccc}
\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
-\sin \beta & \cos \beta & 0 \\
\sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha
\end{array}\right|\)
Solution:
Question 4.
If a, b and c are real numbers, and ∆ = \(\left|\begin{array}{lll}
b+c & c+a & a+b \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|\) = 0 Show that either a + b + c = 0 or a = b = c.
Solution:
Expanding along C,
= 2(a + b + c) ((c – b)(b – c) – (a- b)(a – c))
= 2(a + b + c) (bc – c² – b² + bc – a² + ac + ab – bc)
= – (a + b + c) (2a² + 2b² + 2c² – 2ab – 2bc – 2ac)
= – (a + b + c) (a² + b² – 2ab + b² + c² – 2bc + c² + a² – 2ac)
= – (a + b + c) ((a – b)² + (b – c)² + (c – a)²)
∴ \(\left|\begin{array}{lll}
b+c & c+a & a+b \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|\) ⇒ – (a + b + c)((a – b)² + (b – c)² + (c – a)²) = 0
⇒ (a + b + c) = 0 or (a – b)² + (b – c)² + (c – a)² = 0
⇒ a + b + c = 0 or a – b = 0, b – c = 0 and c – a = 0
⇒ a + b + c = 0 or a = b = c
Question 5.
Solve the equation \(\left|\begin{array}{ccc}
x+a & x & x \\
x & x+a & x \\
x & x & x+a
\end{array}\right|\) = 0, a ≠ 0
Solution:
Question 6.
Prove that \(\left|\begin{array}{ccc}
a^{2} & b c & a c+c^{2} \\
a^{2}+a b & b^{2} & a c \\
a b & b^{2}+b c & c^{2}
\end{array}\right|\) = 4a²b²c²
Solution:
Question 7.
If A-1 \(=\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]\), find (AB)-1
Solution:
Question 8.
Let A = \(\left[\begin{array}{ccc}
1 & -2 & 1 \\
-2 & 3 & 1 \\
1 & 1 & 5
\end{array}\right]\). Verify that i. [adj A]-1 = adj(A-1) ii. (A-1)-1 = A
Solution:
From (1) and (2) we get (adj A)-1 = adj(A-1).
Question 9.
Evaluate \(\left|\begin{array}{ccc}
x & y & x+y \\
y & x+y & x \\
x+y & x & y
\end{array}\right|\)
Solution:
Question 10.
Evaluate \(\left|\begin{array}{ccc}
1 & x & y \\
1 & x+y & y \\
1 & x & x+y
\end{array}\right|\)
Solution:
Question 11.
\(\left|\begin{array}{lll}
\alpha & \alpha^{2} & \beta+\gamma \\
\beta & \beta^{2} & \gamma+\alpha \\
\gamma & \gamma^{2} & \alpha+\beta
\end{array}\right|\) = (α – ß) (ß – γ) (γ – α) (α + ß + γ)
Solution:
Taking out (ß – α) from R2 and (γ – α) from
R3
Expanding along C3,
= (α + ß + γ)(γ – α)(γ – α)(γ + α – ß – α)
= (α + ß + γ )(ß – α)(γ – α)(γ – ß)
= (ß – γ)(γ – α)(α – ß)(α + ß + γ)
Question 12.
\(\left|\begin{array}{lll}
x & x^{2} & 1+p x^{3} \\
y & y^{2} & 1+p y^{3} \\
z_{1} & z^{2} & 1+p z^{3}
\end{array}\right|\) = (1 + pxyz) (x – y) (y – z) (z – x)
Solution:
Expanding along C1,
= (1 + pxyz) (y – x) (z – x) (z + x – y – x)
= (1 + pxyz) (y – x) (z – x) (z – y)
= (1 + pxyz) (x – y) (y – z) (z – x)
Question 13.
\(\left|\begin{array}{ccc}
3 a & -a+b & -a+c \\
-b+a & 3 b & -b+c \\
-c+a & -c+b & 3 c
\end{array}\right|\) = 3(a + b + c) (ab + bc + ca)
Solution:
Expanding along C1,
(a + b + c)\(\left|\begin{array}{ll}
2 b+a & -b+a \\
-c+a & 2 c+a
\end{array}\right|\)
= (a + b + c) [(2b + a) (2c + a) – (a – c) (a – b)]
= (a + b + c) [4bc + 2ab + 2ca + a² – (a² – ab – ac + bc)]
= (a + b + c) [3bc + 3ab + 3ac] = 3 (a + b + c) (ab + bc + ca)
Question 14.
\(\left|\begin{array}{ccc}
1 & 1+p & 1+p+q \\
2 & 3+2 p & 4+3 p+2 q \\
3 & 6+3 p & 10+6 p+3 q
\end{array}\right|\) = 1.
Solution:
Question 15.
\(\left|\begin{array}{lll}
\sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\
\sin \beta & \cos \beta & \cos (\beta+\delta) \\
\sin \gamma & \cos \gamma & \cos (\gamma+\delta)
\end{array}\right|\) = 0
Solution:
Question 16.
Solve the system of the following 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 \(\frac { 1 }{ x }\) = a, \(\frac { 1 }{ y }\) = b and \(\frac { 1 }{ z }\) = c
∴ The given equations become
2a + 3b + 10c = 4
4a – 6b + 5c = 1
6a + 9b – 20c = 2
The equations can be written in the form AX = B
Where A = \(\left[\begin{array}{ccc}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right]/latex]; X = [latex]\left[\begin{array}{l}
a \\
b \\
c
\end{array}\right]\); B = \(\left[\begin{array}{l}
4 \\
1 \\
2
\end{array}\right]\)
∴ a = \(\frac { 1 }{ 2 }\); \(\frac { 1 }{ 3 }\) = b and \(\frac { 1 }{ 15 }\) = c
Hence x = \(\frac { 1 }{ a }\) = 2; y = \(\frac { 1 }{ b }\) = 3; z = \(\frac { 1 }{ c }\) = 5
Question 17.
If a, b, c are in A.P. then determinant
\(\left|\begin{array}{lll}
x+2 & x+3 & x+2 a \\
x+3 & x+4 & x+2 b \\
x+4 & x+5 & x+2 c
\end{array}\right|\) is
a. 0
b. 1
c. x
d. 2x
Solution:
a. 0
Question 18.
If x, y, z are nonzero real numbers, then the inverse of matrix A = \(\left[\begin{array}{lll}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z
\end{array}\right]\) is
a. \(\left[\begin{array}{ccc}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]\)
b. \(x y z\left[\begin{array}{ccc}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]\)
c. \(\frac{1}{x y z}\left[\begin{array}{lll}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z
\end{array}\right]\)
d. \(\frac{1}{x y z}\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Solution:
Question 19.
Let A = \(\left[\begin{array}{ccc}
1 & \sin \theta & 1 \\
-\sin \theta & 1 & \sin \theta \\
-1 & -\sin \theta & 1
\end{array}\right]\), where 0 ≤ θ ≤ 2π. Then
a. Det (A) = 0
b. Det (A) ∈ (2, ∞)
c. Det (A) ∈ (2, 4)
d. Det (A) ∈ [2, 4]
Solution:
d. Det (A) ∈ [2, 4]
