You will find a list of lass 12 Maths MCQ Questions of Determinants with Answers here to start your preparation. Understand the concepts and practice Objective Type Questions for 12 Class Maths the maximum amount as you’ll be able to get well. MCQ Questions for class 12 Maths needs lots of practice. Prepare effectively with Class 12 Maths Chapter Wise Multiple-Choice Questions provided and score well in your board or competitive exams.
Go through MCQ questions for class 12 determinants are given here with detailed solutions. The solutions are given in a step-by-step procedure so that students can understand in an easy and better way and score good marks in class 12 final exams.
Practice MCQ Question for Class 12 Maths
1. 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
2. The area of a triangle with vertices (-3, 0), (3, 0) and (0, k) is 9 sq. units. The value of k will be
(a) 9
(b) 3
(c) -9
(d) 6
3. If the points (3, -2), (k, 2), (8, 8) are collinear, then find the value of k.
(a) 2
(b) 3
(c) 4
(d) 5
4. The value of \(\begin{vmatrix} sin20^\circ & -cos^\circ \\[0.3em] sin70^\circ& cos70^\circ\end{vmatrix}\) is......
(a) 5
(b) 1
(c) 2
(d) 3
5. \(\begin{vmatrix} cos 70^\circ & sin 20^\circ\\[0.3em] sin 70^\circ & cos 20^\circ \\[0.3em] \end{vmatrix}\) = ?
(a) 1
(b) 0
(c) cos 50°
(d) sin 50°
6. \(\begin{bmatrix} cos 15^\circ & sin 15^\circ\\[0.3em] sin 15^\circ & cos 15^\circ \\[0.3em] \end{bmatrix}\) = ?
(a) 1
(b) 1/2
(c) \(\frac{\sqrt3}{2}\)
(d) none of these
7. If |A| = 0, then A is
(a) zero matrix
(b) singular matrix
(c) non-singular matrix
(d) 0
8. According to determinant properties, the determinant equals to zero if column is
(a) divided to row
(b) divided to column
(c) multiplied to row
(d) multiplied to column
9. The rule which provides method of solving the determinants is classified as
(a) Cramer's rule
(b) determinant rule
(c) solving rule
(d) thumb rule
10. If A and B are invertible matrices, then which of the following is not correct?
(a) adj A = |A|.A-1
(b) det (a)-1 = [det (a)]-1
(c) (AB)-1 = B-1A-1
(d) (A + B)-1 = B-1 + A-1
11. Let 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|
12. Using determinants, find the equation of the line joining the points (1, 2) and (3, 6).
(a) y = 2x
(b) x = 3y
(c) y = x
(d) 4x – y = 5
13. \(\begin{vmatrix} sin 23^ \circ & -sin7^ \circ \\[0.3em] cos23^ \circ & cos7^ \circ \end{vmatrix}\) = ?
(a) \(\frac{\sqrt{3}}{2}\)
(b) \(\frac{1}{2}\)
(c) sin 16°
(d) cos 16°
14. \(\begin{vmatrix} a+ ib & c + id\\[0.3em] -c + id & a - id \end{vmatrix}\) = ?
(a) (a2 + b2 – c2 – d2)
(b) (a2 – b2 + c2 – d2)
(c) (a2 + b2 + c2 + d2)
(d) none of these
15. 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.
16. Let A be a non-singular matrix of order 3 × 3. Then |adj. A| is equal to
(a) |A|
(b) |A|2
(c) |A|3
(d) 3|A|
17. \(\begin{vmatrix} 1 & 1 & 1 \\[0.3em] 1 & 1 + x & 1 \\[0.3em] 1 & 1& 1 + y \end{vmatrix}\)= ?
(a) (x + y)
(b) (x – y)
(c) xy
(d) none of these
18. If \(\begin{vmatrix} 5 & 3 & -1 \\[0.3em] -7 & \text{x} & 2 \\[0.3em] 9 & 6& 2 \end{vmatrix}\)= 0 then x = ?
(a) 0
(b) 6
(c) -6
(d) 9
19. If A is an invertible matrix of order 2, then det (A-1) is equal to
(a) det (A)
(b) 1/det(A)
(c) 1
(d) 0
20. Let A be a square matrix all of whose entries are integers. Then which of the following is true?
(a) If det A = ± 1, then A-1 need not exist
(b) If det A = ± 1, then A-1 exists but all entries are not necessarily integers.
(c) If det A ≠ ± 1, then A-1 exists and all its entries are non-integers
(d) If det A = ± 1, then A-1 exists and all its entries are integers.
Answer:
1. Answer: (c) Determinant is a number associated to a square matrix
Explanation: Determinant is defined only for a square matrix. And its denotes the value of that square matrix.
2. Answer: (b) 3
Explanation: We know that , area of a triangle with vertices (a1,y1),(x2,y2) and (x3,y3) is given by
\(\Delta=\frac{1}{2}\)\(\begin{vmatrix} x_1 & y_1&1 \\[0.3em] x_2& y_2&1\\[0.3em] x_3& y_3&1\end{vmatrix}\)
\(\therefore\Delta=\frac{1}{2}\)\(\begin{vmatrix} -3 & 0&1 \\[0.3em] 3& 0&1\\[0.3em] 0& k&1\end{vmatrix}\)
Expanding along R1
9= 1/2[−3(−k)−0+1(3k)]
⇒18=3k+3k=6k
∴K= 18/6 = 3
3. Answer: (d) 5
Explanation: If the points are collinear than the value determinant is zero.
\(\therefore \Delta\) \(=\begin{vmatrix} 3 & -2&1 \\[0.3em] k& 2&1\\[0.3em] 8& 8&1\end{vmatrix}\)
Apply R1→(R1 −R3)
\(\Delta=\begin{vmatrix} -5 & -10&0 \\[0.3em] k& 2&1\\[0.3em] 8& 8&1\end{vmatrix}\)
Apply R2→(R2−R3)
\(\Delta=\begin{vmatrix} -5 & -10&0 \\[0.3em] k-8& -6&0\\[0.3em] 8& 8&1\end{vmatrix}\)
Expand by c3
1[30+10(k−8)]=0
∴k−8=−3
∴
k=5
4. Answer: (b) 1
Explanation:
\(\begin{vmatrix} sin\,20° & -cos\,20° \\[0.3em] sin\,70°& cos\,70° \\[0.3em] \end{vmatrix}\)
= sin 20° . cos 70° + cos 20° . sin 70°
= sin (20° + 70°)
= sin 90°
= 1.
5. Answer: (b) 0
Explanation: cos\(\theta\) = sin(90 - \(\theta\))
We have \(\begin{vmatrix} cos 70^\circ & sin 20^\circ\\[0.3em] sin 70^\circ & cos 20^\circ \\[0.3em] \end{vmatrix}\)
On expanding the above,
⇒ {cos 70°} {cos 20°} – {sin 70°} {sin 20°}
On applying formula cos\(\theta\) = sin(90 - \(\theta\) )
⇒ {sin (90 – 70)} {sin (90 – 20)} - {sin 70°} {sin 20°}
⇒ {sin 20°} {sin 70°} - {sin 70°} {sin 20°}
= 0