Determinants MCQ AP Inter 2nd Year Maths Chapter 4

Practice AP Inter 2nd Year Maths Study Material Chapter 4 Determinants MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Determinants MCQ

Question 1.
If \(\left|\begin{array}{cc}
x & 2 \\
18 & x
\end{array}\right|=\left|\begin{array}{cc}
6 & 2 \\
18 & 6
\end{array}\right|\), then x is equal to
1) 6
2) ±6
3) -6
4) 0
Solution:
2) ±6
G.E = x² – 36 = 36 – 36 ⇒ x² – 36 = 0 ⇒ x² = 36 ⇒ x = ±6

Question 2.
If A is 3 × 3 matrix and det (3A) = k (deta A), then k =
1) 9
2) 6
3) 1
4) 27
Solution:
4) 27
Given A_{3 × 3} ∴ |3A| = 3³|A| = 27 (det A) = k (det A) ⇒ k = 27

Question 3.
Value of k, which \(\) is singular
1) 4
2) -4
3) ±4
4) 0
Solution:
3) ±4
If A is singular |A| = 0 ⇒ \(\left|\begin{array}{ll}
\mathrm{k} & 2 \\
8 & \mathrm{k}
\end{array}\right|\) = 0 ⇒ k² – 16 = 0 ⇒ k = ±4

Question 4.
The area of a triangle with vertices (-3, 0), (0, 3) and (0, k) is 9 sq. units the value of k will be
1) 9
2) 3
3) -9
4) 6
Solution:
1) 9
Area of the triangle formed by (x₁, y₁) (x₂, y₂) (x₃, y₃) is
= \(\frac{1}{2}\left|\begin{array}{lll}
1 & x_1 & y_1 \\
1 & x_2 & y_2 \\
1 & x_3 & y_3
\end{array}\right|=9 \Rightarrow \frac{1}{2}\left|\begin{array}{ccc}
1 & -3 & 0 \\
1 & 0 & 3 \\
1 & 0 & k
\end{array}\right|=9 \Rightarrow\left|\begin{array}{ccc}
1 & -3 & 0 \\
1 & 0 & 3 \\
1 & 0 & k
\end{array}\right|\) = 18 ⇒ 3|k – 3| = 18 ⇒ |k – 3| = 6
k – 3 = 6 ⇒ k = 9; k – 3 = -6 ⇒ k = -3

Question 5.
If A is square matrix of order 3 and |A| = -4, then |adj A| is equal to
1) -4
2) 4
3) -16
4) 16
Solution:
4) 16
If A_n×n, then |AdjA| = |A|^n-1
We have A_3×3 ∴ |AdjA| = (|A|)² = (-4)² = 16

Question 6.
If area of triangle is 35 sq units with vertices (2, -6), (5, 4) and (k, 4). Then k is
1) 12
2) -2
3) -12, -2
4) 12, -2
Solution:
4) 12, -2
Area of triangle = 35
⇒ \(\frac{1}{2}\left|\begin{array}{ccc}
1 & 2 & -6 \\
1 & 5 & 4 \\
1 & \mathrm{k} & 4
\end{array}\right|\) = 35 ⇒ |1(20 – 4k) – 2(4 – 4) – 6(k – 5)| = 70
⇒ 20 – 4k + 0 – 6k + 30 = ±70 ⇒ 50 – 10k = ±70 ⇒ 5 – k = ±7
5 – k = 7 ⇒ k = -2; 5 – k = -7 ⇒ k = 12

Question 7.
If A = \(\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|\) and A_ij is Cofactors of a_ij then value of ∆ is given by
1) a₁₁A₃₁ + a₁₁A₃₂ + a₁₃A₃₃
2) a₁₁A₁₁ + a₁₂A₂₁ + a₁₃A₃₁
3) a₂₁A₁₁ + a₂₂A₁₂ + a₂₃A₁₃
4) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁
Solution:
4) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁
∆ = Determinant of a mathix
= sum of the products of the elements of a row (or) column with the corresponding co-factors = (a₁₁)A₁₁ + (a₁₂)A₁₂ + (a₃₁)A₃₁ [using 1^st row]

Question 8.
Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to
1) |A|
2) |A|²
3) |A|³
4) 3|A|
Solution:
2) |A|²
If A_n×n, then |AdjA| = |A|^n-1
We have A_3×3 ∴ |AdjA| = |A|²

Question 9.
If A is an invertible matrix of order 2, then det (A^-1) is equal to
1) det(A)
2) \(\frac{1}{{det}(\mathrm{~A})}\)
3) 1
4) 0
Solution:
2) \(\frac{1}{{det}(\mathrm{~A})}\)
A_2×2, and A, exists ∵ AA^-1 = A^-1A = I
Consider AA^-1 = I
|AA^-1| = |I| ⇒ |A||A^-1| = I ⇒ |A^-1| = \(\frac{1}{|\mathrm{~A}|}\) ⇒ det(A^-1| = \(\frac{1}{{det} A}\)

Question 10.
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
1) \(\left[\begin{array}{ccc}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]\)
2) \(x y z\left[\begin{array}{ccc}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]\)
3) \(\frac{1}{\mathrm{xyz}}\left[\begin{array}{ccc}
\mathrm{x} & 0 & 0 \\
0 & \mathrm{y} & 0 \\
0 & 0 & \mathrm{z}
\end{array}\right]\)
4) \(\frac{1}{\mathrm{xyz}}\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Solution:
1) \(\left[\begin{array}{ccc}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]\)
If A = diag[a b c] then A^-1 = \(\frac{1}{|\mathrm{~A}|}\) AdjA = \(\left[\begin{array}{lll}
a^{-1} & b^{-1} & c^{-1}
\end{array}\right]=\left[\begin{array}{ccc}
1 / a & 0 & 0 \\
0 & 1 / b & 0 \\
0 & 0 & 1 / c
\end{array}\right]\)
∴ A = \(\left[\begin{array}{lll}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z
\end{array}\right]\) = diag [x y z]
⇒ A^-1 = diag\(\left[\begin{array}{lll}
x^{-1} & y^{-1} & z^{-1}
\end{array}\right]=\left[\begin{array}{ccc}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]\)

Question 11.
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
1) Det(A) = 0
2) Det(A) ∈ (2, ∞)
3) Det(A) ∈ (2, 4)
4) Det(A) ∈ [2, 4]
Solution:
4) Det(A) ∈ [2, 4]
|A| = 1(1 + sin²θ) – sinθ[-sin θ + sin θ] + 1[sin² θ + 1]
= 1 + sin² θ – 0 + sin² θ + 1 = 2 + 2sin² θ
⇒ det A = 2 + 2sin² θ
∵ 0 ≤ sin² θ ≤ 1 ⇒ 0 ≤ 2sin² θ ≤ 2 ⇒ 0 + 2 ≤ (2 + 2sin² θ) ≤ 2 + 2 ⇒ 2 ≤ |A| ≤ 4
⇒ |A| ∈ [2, 4]