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

6.  Answer:  (c)  \(\frac{\sqrt3}{2}\)

Explanation:  cos(A + B) = cos A cos B - sin A sin B

We have \(\begin{vmatrix} cos 15^\circ & sin 15^\circ\\[0.3em] sin 15^\circ & cos 15^\circ \\[0.3em] \end{vmatrix}\)

On expanding the above, 

⇒ {cos 15°} {cos 15°} – {sin 15°} {sin 15°}

On applying formula cos(A + B) = cos A cos B - sin A sin B

= cos (15 + 15) 

= cos (30°)

= \(\frac{\sqrt3}{2}\) 

7. Answer: (b) singular matrix

Explanation: A matrix can be singular, only if it has a determinant of zero.

8. Answer: (d) multiplied to column

9. Answer: (a) Cramer's rule

10. Answer: (d) (A + B)^-1 = B^-1 + A^-1 

11. Answer: (c) k³|A|

Explanation: If A is a square matrix of order n , then we can take k common from each of the n rows of kA
Hence, ∣kA∣=kⁿ∣A∣

So, for a square matrix of of order 3, 
∣kA∣=k³∣A∣

12. Answer: (a) y = 2x

Explanation: We have to find the equation of line joining (1,2) and (3,6)

Let the third point on the line be (x,y)

The area of a triangle with vertices (x,y),(1,2),(3,6) is \(\frac{1}{2}\begin{vmatrix} x & y&1 \\[0.3em] 1& 2&1\\[0.3em] 3& 6&1\end{vmatrix}\)

Since the three points are collinear, the area formed will be zero

\(\Rightarrow\begin{vmatrix} x & y&1 \\[0.3em] 1& 2&1\\[0.3em] 3& 6&1\end{vmatrix}=0\)

⟹x(2−6)−y(1−3)+1(6−6)=0

⟹−4x+2y=0

⟹2x−y=0.

Hence, the equation of line joining (1,2) and (3,6) is 2x−y=0.

13.  Answer:  (b) \(\frac{1}{2}\)

Explanation:  sin(A + B) = sin A cos B + cos A sin B

We have, \(\begin{vmatrix} sin 23^ \circ & -sin7^ \circ \\[0.3em] cos23^ \circ & cos7^ \circ \end{vmatrix}\)

on expanding the above,

On expanding the above, 

⇒ (sin 23°) (cos 7°) – (cos 23°) (-sin 7°) 

⇒ (sin 23°) (cos 7°) + (cos 23°) (sin 7°) 

On applying formula sin(A + B) = sin A cos B + cos A sin B

= sin (23 + 7) = sin (30°)

= \(\frac{1}{2}\)

14. Answer: (c) (a² + b² + c² + d²) 

Explanation: We have \(\begin{vmatrix} a+ ib & c + id\\[0.3em] -c + id & a - id \end{vmatrix}\)

On expanding the above, 

⇒ (a + ib) (a – ib) – (-c + id) (c + id)

⇒ (a² – iab + iba – i²b²) - (-c² – icd + icd + i²d²) 

⇒ {a² – iab + iba – (-1)b²} – {-c² – icd + icd + (-1)d²} 

⇒ {a² – iab + iba + 1b²} – {-c² – icd + icd - 1d²} 

⇒ a² + b² + c² + d²  

15. Answer: (d) 12, -2.

Explanation: Area of triangle having vertices as (x1,y1),(x2,y2),(x3,y3) is given as

\(=\frac{1}{2}\begin{vmatrix} x_1 & y_1&1 \\[0.3em] x_2& y_2&1\\[0.3em] x_3& y_3&1\end{vmatrix}\)

\(\Rightarrow35=\frac{1}{2}\begin{vmatrix} 2 & -6&1 \\[0.3em] 5& 4&1\\[0.3em] k& 4&1\end{vmatrix}\)

⇒35= 1/2 ∣2(0)+6(5−k)+1(20−4k)∣

⇒∣50−10k∣=70

⇒50−10k=±70

⇒50−10k=70 and 50−10k=−70

⇒−10k=20 and −10k=−120

⇒k=−2 and k=12 

16. Answer: (b) |A|²  

17. Answer: (c) xy 

Explanation: We have, \(\begin{vmatrix} 1 & 1 & 1 \\[0.3em] 1 & 1 + x & 1 \\[0.3em] 1 & 1& 1 + y \end{vmatrix}\)

Applying R₁ → R₂ - R₁

\(=\begin{vmatrix} 0 & -x & 0 \\[0.3em] 1 & 1 + x & 1 \\[0.3em] 1 & 1& 1 + y \end{vmatrix}\)

Expanding along R₁

⇒ [x{(1)(1+y)-(1)(1)}] 

⇒ [x{1+y-1}] 

⇒ xy

18. Answer: (c) -6

Explanation: 

We have, \(\begin{vmatrix} 5 & 3 & -1 \\[0.3em] -7 & \text{x} & 2 \\[0.3em] 9 & 6& 2 \end{vmatrix}\) = 0

Applying R₁ → 2R₁ 

\(\Rightarrow\begin{vmatrix} 10 & 6 & -2 \\[0.3em] -7 & \text{x} & 2 \\[0.3em] 9 & 6& -2 \end{vmatrix}\)

Applying R₁ → R₁ -R₃

 \(\Rightarrow\begin{vmatrix} 1 & 0 & 0 \\[0.3em] -7 & \text{x} & 2 \\[0.3em] 9 & 6& -2 \end{vmatrix}\)

Expanding along R₁ 

⇒ [1{(x)(-2) – (6)(2)}] = 0 

⇒ [1{-2x – 12}] = 0 

⇒ -2x-12 = 0 

⇒ -2x = 12 

⇒ x = -6

19. Answer: (b) 1/det(A)

20. Answer: (d) If det A = ± 1, then A^-1 exists and all its entries are integers.

Explanation: Since each entry of A is an integer,
∴ co-factor of each entry is also an integer.
Hence, each entry of the adjoint is an integer.
Also det A = ± 1 and A^-1 = 1det(A) (adj A).
Hence, all entries of A-1 are integers.

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