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Home/JEE MAINS/mathematics/determinants
Curated PYQ Collection

Top 50 Most Repeated DETERMINANTS PYQs | JEE MAINS

A curated collection of the most important questions from DETERMINANTS, fully solved with step-by-step concepts to prepare for JEE MAINS.

Question #1

Practice Question

Concept Applied

Compute $\det(A - \lambda I) = (3 - \lambda)(4 - \lambda) - (2)(1) = \lambda^2 - 7\lambda + 10$. The sum of roots of quadratic $a\lambda^2 + b\lambda ...

Read Full Step-by-Step Solution →

Question #2

Practice Question

A.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = 0$
B.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = 1$
C.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = -1$
D.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = 2$

Concept Applied

Step 1: To find the determinant of a 3x3 matrix, we can use the formula: det(A) = a(ei - fh) - b(di - fg) + c(dh - eg), where a, b, c, d, e, f, g, h, ...

Read Full Step-by-Step Solution →

Question #3

Practice Question

A.The characteristic polynomial of a matrix A is given by $det(A-\lambda I) = 0$
B.The characteristic polynomial of a matrix A is given by $det(A+\lambda I) = 0$
C.The characteristic polynomial of a matrix A is given by $det(A-\lambda I) = 1$
D.The characteristic polynomial of a matrix A is given by $det(A+\lambda I) = 1$

Concept Applied

Step 1: Recall the definition of the characteristic polynomial of a matrix. Step 2: Identify the correct expression for the characteristic polynomial ...

Read Full Step-by-Step Solution →

Question #4

Practice Question

Concept Applied

Determinant of product is product of determinants....

Read Full Step-by-Step Solution →

Question #5

Practice Question

Concept Applied

Expand along first row:

(0-24) - 2(0-20) + 3(0-5) = 1$....

Read Full Step-by-Step Solution →

Question #6

Practice Question

A.$M^{-1} = \frac{1}{\det M}\,\operatorname{adj}(M)$
B.$M^{-1} = \det M\,\operatorname{adj}(M)$
C.$M^{-1} = \operatorname{adj}(M)\,\det M$
D.$M^{-1} = \operatorname{adj}(M)^{-1}$

Concept Applied

For any invertible square matrix, $M^{-1} = \frac{1}{\det M}\,\operatorname{adj}(M)$. This follows from the definition of the adjugate and the identit...

Read Full Step-by-Step Solution →

Question #7

Practice Question

A.\(\lambda^{3} - 10\lambda^{2} + 27\lambda - 58 = 0\)
B.\(\lambda^{3} - 10\lambda^{2} + 27\lambda + 58 = 0\)
C.\(\lambda^{3} - 9\lambda^{2} + 27\lambda - 58 = 0\)
D.\(\lambda^{3} - 10\lambda^{2} + 26\lambda - 58 = 0\)

Concept Applied

Compute \(\det(B-\lambda I)\). Expanding the determinant gives \(-\lambda^{3}+10\lambda^{2}-27\lambda+58\), which after multiplying by \(-1\) yields t...

Read Full Step-by-Step Solution →

Question #8

Practice Question

Concept Applied

Element $a_{23} = -2$. Cofactor $C_{23} = (-1)^{2+3} \cdot \text{det}(M_{23})$, where $M_{23}$ is minor. Remove row 2, column 3: $M_{23} = \begin{bmat...

Read Full Step-by-Step Solution →

Question #9

Practice Question

Concept Applied

Expand along R1:

...

Read Full Step-by-Step Solution →

Question #10

Practice Question

Concept Applied

The area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is given by: $ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 -...

Read Full Step-by-Step Solution →

Question #11

Practice Question

A.The area is 4 square units
B.The area is 8 square units
C.The area is 2 square units
D.The area is -4 square units

Concept Applied

The determinant calculation gives 8 after substitutions. Taking absolute value and dividing by 2 yields 4. Common errors include omitting absolute val...

Read Full Step-by-Step Solution →

Question #12

Practice Question

A.The area is 10.5 square units
B.The area is 21 square units
C.The area is 42 square units
D.The area is 7 square units

Concept Applied

Using the determinant-based area formula, the correct value is half the absolute value of the determinant sum. Common errors include forgetting to div...

Read Full Step-by-Step Solution →

Question #13

Practice Question

Concept Applied

Compute $\det(A - \lambda I) = (3 - \lambda)(4 - \lambda) - (2)(1) = \lambda^2 - 7\lambda + 10$. The sum of roots of quadratic $a\lambda^2 + b\lambda ...

Read Full Step-by-Step Solution →

Question #14

Practice Question

A.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = 0$
B.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = 1$
C.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = -1$
D.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = 2$

Concept Applied

Step 1: To find the determinant of a 3x3 matrix, we can use the formula: det(A) = a(ei - fh) - b(di - fg) + c(dh - eg), where a, b, c, d, e, f, g, h, ...

Read Full Step-by-Step Solution →

Question #15

Practice Question

A.The characteristic polynomial of a matrix A is given by $det(A-\lambda I) = 0$
B.The characteristic polynomial of a matrix A is given by $det(A+\lambda I) = 0$
C.The characteristic polynomial of a matrix A is given by $det(A-\lambda I) = 1$
D.The characteristic polynomial of a matrix A is given by $det(A+\lambda I) = 1$

Concept Applied

Step 1: Recall the definition of the characteristic polynomial of a matrix. Step 2: Identify the correct expression for the characteristic polynomial ...

Read Full Step-by-Step Solution →

Question #16

Practice Question

Concept Applied

Determinant of product is product of determinants....

Read Full Step-by-Step Solution →

Question #17

Practice Question

Concept Applied

Expand along first row:

(0-24) - 2(0-20) + 3(0-5) = 1$....

Read Full Step-by-Step Solution →

Question #18

Practice Question

A.$M^{-1} = \frac{1}{\det M}\,\operatorname{adj}(M)$
B.$M^{-1} = \det M\,\operatorname{adj}(M)$
C.$M^{-1} = \operatorname{adj}(M)\,\det M$
D.$M^{-1} = \operatorname{adj}(M)^{-1}$

Concept Applied

For any invertible square matrix, $M^{-1} = \frac{1}{\det M}\,\operatorname{adj}(M)$. This follows from the definition of the adjugate and the identit...

Read Full Step-by-Step Solution →

Question #19

Practice Question

A.\(\lambda^{3} - 10\lambda^{2} + 27\lambda - 58 = 0\)
B.\(\lambda^{3} - 10\lambda^{2} + 27\lambda + 58 = 0\)
C.\(\lambda^{3} - 9\lambda^{2} + 27\lambda - 58 = 0\)
D.\(\lambda^{3} - 10\lambda^{2} + 26\lambda - 58 = 0\)

Concept Applied

Compute \(\det(B-\lambda I)\). Expanding the determinant gives \(-\lambda^{3}+10\lambda^{2}-27\lambda+58\), which after multiplying by \(-1\) yields t...

Read Full Step-by-Step Solution →

Question #20

Practice Question

Concept Applied

Element $a_{23} = -2$. Cofactor $C_{23} = (-1)^{2+3} \cdot \text{det}(M_{23})$, where $M_{23}$ is minor. Remove row 2, column 3: $M_{23} = \begin{bmat...

Read Full Step-by-Step Solution →

Question #21

Practice Question

Concept Applied

Expand along R1:

...

Read Full Step-by-Step Solution →

Question #22

Practice Question

Concept Applied

The area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is given by: $ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 -...

Read Full Step-by-Step Solution →

Question #23

Practice Question

A.The area is 4 square units
B.The area is 8 square units
C.The area is 2 square units
D.The area is -4 square units

Concept Applied

The determinant calculation gives 8 after substitutions. Taking absolute value and dividing by 2 yields 4. Common errors include omitting absolute val...

Read Full Step-by-Step Solution →

Question #24

Practice Question

A.The area is 10.5 square units
B.The area is 21 square units
C.The area is 42 square units
D.The area is 7 square units

Concept Applied

Using the determinant-based area formula, the correct value is half the absolute value of the determinant sum. Common errors include forgetting to div...

Read Full Step-by-Step Solution →

Question #25

Practice Question

Concept Applied

Compute $\det(A - \lambda I) = (3 - \lambda)(4 - \lambda) - (2)(1) = \lambda^2 - 7\lambda + 10$. The sum of roots of quadratic $a\lambda^2 + b\lambda ...

Read Full Step-by-Step Solution →

Question #26

Practice Question

A.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = 0$
B.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = 1$
C.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = -1$
D.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = 2$

Concept Applied

Step 1: To find the determinant of a 3x3 matrix, we can use the formula: det(A) = a(ei - fh) - b(di - fg) + c(dh - eg), where a, b, c, d, e, f, g, h, ...

Read Full Step-by-Step Solution →

Question #27

Practice Question

A.The characteristic polynomial of a matrix A is given by $det(A-\lambda I) = 0$
B.The characteristic polynomial of a matrix A is given by $det(A+\lambda I) = 0$
C.The characteristic polynomial of a matrix A is given by $det(A-\lambda I) = 1$
D.The characteristic polynomial of a matrix A is given by $det(A+\lambda I) = 1$

Concept Applied

Step 1: Recall the definition of the characteristic polynomial of a matrix. Step 2: Identify the correct expression for the characteristic polynomial ...

Read Full Step-by-Step Solution →

Question #28

Practice Question

Concept Applied

Determinant of product is product of determinants....

Read Full Step-by-Step Solution →

Question #29

Practice Question

Concept Applied

Expand along first row:

(0-24) - 2(0-20) + 3(0-5) = 1$....

Read Full Step-by-Step Solution →

Question #30

Practice Question

A.$M^{-1} = \frac{1}{\det M}\,\operatorname{adj}(M)$
B.$M^{-1} = \det M\,\operatorname{adj}(M)$
C.$M^{-1} = \operatorname{adj}(M)\,\det M$
D.$M^{-1} = \operatorname{adj}(M)^{-1}$

Concept Applied

For any invertible square matrix, $M^{-1} = \frac{1}{\det M}\,\operatorname{adj}(M)$. This follows from the definition of the adjugate and the identit...

Read Full Step-by-Step Solution →

Question #31

Practice Question

A.\(\lambda^{3} - 10\lambda^{2} + 27\lambda - 58 = 0\)
B.\(\lambda^{3} - 10\lambda^{2} + 27\lambda + 58 = 0\)
C.\(\lambda^{3} - 9\lambda^{2} + 27\lambda - 58 = 0\)
D.\(\lambda^{3} - 10\lambda^{2} + 26\lambda - 58 = 0\)

Concept Applied

Compute \(\det(B-\lambda I)\). Expanding the determinant gives \(-\lambda^{3}+10\lambda^{2}-27\lambda+58\), which after multiplying by \(-1\) yields t...

Read Full Step-by-Step Solution →

Question #32

Practice Question

Concept Applied

Element $a_{23} = -2$. Cofactor $C_{23} = (-1)^{2+3} \cdot \text{det}(M_{23})$, where $M_{23}$ is minor. Remove row 2, column 3: $M_{23} = \begin{bmat...

Read Full Step-by-Step Solution →

Question #33

Practice Question

Concept Applied

Expand along R1:

...

Read Full Step-by-Step Solution →

Question #34

Practice Question

Concept Applied

The area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is given by: $ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 -...

Read Full Step-by-Step Solution →

Question #35

Practice Question

A.The area is 4 square units
B.The area is 8 square units
C.The area is 2 square units
D.The area is -4 square units

Concept Applied

The determinant calculation gives 8 after substitutions. Taking absolute value and dividing by 2 yields 4. Common errors include omitting absolute val...

Read Full Step-by-Step Solution →

Question #36

Practice Question

A.The area is 10.5 square units
B.The area is 21 square units
C.The area is 42 square units
D.The area is 7 square units

Concept Applied

Using the determinant-based area formula, the correct value is half the absolute value of the determinant sum. Common errors include forgetting to div...

Read Full Step-by-Step Solution →

Question #37

Practice Question

Concept Applied

Compute $\det(A - \lambda I) = (3 - \lambda)(4 - \lambda) - (2)(1) = \lambda^2 - 7\lambda + 10$. The sum of roots of quadratic $a\lambda^2 + b\lambda ...

Read Full Step-by-Step Solution →

Question #38

Practice Question

A.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = 0$
B.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = 1$
C.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = -1$
D.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = 2$

Concept Applied

Step 1: To find the determinant of a 3x3 matrix, we can use the formula: det(A) = a(ei - fh) - b(di - fg) + c(dh - eg), where a, b, c, d, e, f, g, h, ...

Read Full Step-by-Step Solution →

Question #39

Practice Question

A.The characteristic polynomial of a matrix A is given by $det(A-\lambda I) = 0$
B.The characteristic polynomial of a matrix A is given by $det(A+\lambda I) = 0$
C.The characteristic polynomial of a matrix A is given by $det(A-\lambda I) = 1$
D.The characteristic polynomial of a matrix A is given by $det(A+\lambda I) = 1$

Concept Applied

Step 1: Recall the definition of the characteristic polynomial of a matrix. Step 2: Identify the correct expression for the characteristic polynomial ...

Read Full Step-by-Step Solution →

Question #40

Practice Question

Concept Applied

Determinant of product is product of determinants....

Read Full Step-by-Step Solution →

Question #41

Practice Question

Concept Applied

Expand along first row:

(0-24) - 2(0-20) + 3(0-5) = 1$....

Read Full Step-by-Step Solution →

Question #42

Practice Question

A.$M^{-1} = \frac{1}{\det M}\,\operatorname{adj}(M)$
B.$M^{-1} = \det M\,\operatorname{adj}(M)$
C.$M^{-1} = \operatorname{adj}(M)\,\det M$
D.$M^{-1} = \operatorname{adj}(M)^{-1}$

Concept Applied

For any invertible square matrix, $M^{-1} = \frac{1}{\det M}\,\operatorname{adj}(M)$. This follows from the definition of the adjugate and the identit...

Read Full Step-by-Step Solution →

Question #43

Practice Question

A.\(\lambda^{3} - 10\lambda^{2} + 27\lambda - 58 = 0\)
B.\(\lambda^{3} - 10\lambda^{2} + 27\lambda + 58 = 0\)
C.\(\lambda^{3} - 9\lambda^{2} + 27\lambda - 58 = 0\)
D.\(\lambda^{3} - 10\lambda^{2} + 26\lambda - 58 = 0\)

Concept Applied

Compute \(\det(B-\lambda I)\). Expanding the determinant gives \(-\lambda^{3}+10\lambda^{2}-27\lambda+58\), which after multiplying by \(-1\) yields t...

Read Full Step-by-Step Solution →

Question #44

Practice Question

Concept Applied

Element $a_{23} = -2$. Cofactor $C_{23} = (-1)^{2+3} \cdot \text{det}(M_{23})$, where $M_{23}$ is minor. Remove row 2, column 3: $M_{23} = \begin{bmat...

Read Full Step-by-Step Solution →

Question #45

Practice Question

Concept Applied

Expand along R1:

...

Read Full Step-by-Step Solution →

Question #46

Practice Question

Concept Applied

The area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is given by: $ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 -...

Read Full Step-by-Step Solution →

Question #47

Practice Question

A.The area is 4 square units
B.The area is 8 square units
C.The area is 2 square units
D.The area is -4 square units

Concept Applied

The determinant calculation gives 8 after substitutions. Taking absolute value and dividing by 2 yields 4. Common errors include omitting absolute val...

Read Full Step-by-Step Solution →

Question #48

Practice Question

A.The area is 10.5 square units
B.The area is 21 square units
C.The area is 42 square units
D.The area is 7 square units

Concept Applied

Using the determinant-based area formula, the correct value is half the absolute value of the determinant sum. Common errors include forgetting to div...

Read Full Step-by-Step Solution →

Question #49

Practice Question

Concept Applied

Compute $\det(A - \lambda I) = (3 - \lambda)(4 - \lambda) - (2)(1) = \lambda^2 - 7\lambda + 10$. The sum of roots of quadratic $a\lambda^2 + b\lambda ...

Read Full Step-by-Step Solution →

Question #50

Practice Question

A.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = 0$
B.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = 1$
C.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = -1$
D.$\begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{vmatrix} = 2$

Concept Applied

Step 1: To find the determinant of a 3x3 matrix, we can use the formula: det(A) = a(ei - fh) - b(di - fg) + c(dh - eg), where a, b, c, d, e, f, g, h, ...

Read Full Step-by-Step Solution →
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