CBSE Class 12 Mathematics Determinants Previous Year Questions Set 01

Here are the CBSE board Previous Year Questions on Chapter 4 Determinants for Class 12 Mathematics, from 2005 to 2020. Each question has a complete step-by-step solution with the marking split shown in brackets.

Very Short Answer Questions (1 Mark)

Question 1. A is a square matrix of order 3 and |A| = 7. Write the value of |adj A|.   [CBSE 2010, 1M]
Answer:
|A| = 7
|adj A| = |A|^{n − 1} = (7)² = 49   [1]

Question 2. If \( A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix} \), write A⁻¹ in terms of A.   [CBSE 2011, 1M]
Answer:
\( |A| = \begin{vmatrix} 2 & 3 \\ 5 & -2 \end{vmatrix} = -19 \neq 0 \)
\( \text{adj } A = \begin{bmatrix} -2 & -5 \\ -3 & 2 \end{bmatrix}^T = \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|}\,\text{adj } A = \frac{1}{19}\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix} \)
\( A^{-1} = \frac{1}{19}A \)   [1]

Question 3. Let A be a square matrix of order 3 × 3. Write the value of |2A|, where |A| = 4.   [CBSE 2012, 1M]
Answer:
Given A is a square matrix of order 3 × 3 and |A| = 4
∴ |2A| = 2³|A| = 8 × 4 = 32   [1]

Question 4. If A is a 3 × 3 matrix and |3A| = k|A|, then write the value of k.   [CBSE 2016, 1M]
Answer:
|3A| = 3³|A|
⇒ k = 27   [1]

Question 5. If for any 2 × 2 square matrix A, \( A(\text{adj } A) = \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} \), then write the value of |A|.   [CBSE 2017, 1M]
Answer:
Given \( A(\text{adj } A) = \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} \)
⇒ \( |A(\text{adj } A)| = \begin{vmatrix} 8 & 0 \\ 0 & 8 \end{vmatrix} \)
⇒ |A| |adj A| = 64
⇒ |A| |A| = 64   (∵ |adj A| = |A|²⁻¹ = |A|)
⇒ |A|² = 64
⇒ |A| = ±8   [1]

Question 6. If A is a square matrix satisfying A′A = I, write the value of |A|.   [CBSE 2019, 1M]
Answer:
|A′A| = |I|
|A′| |A| = |I|   [½]
⇒ |A|² = 1
⇒ |A| = 1 or |A| = −1   [½]

Question 7. If A is a non-singular square matrix of order 3 such that A² = 3A, then the value of |A| is   [CBSE 2020, 1M]
(A) −3    (B) 3    (C) 9    (D) 27
Answer:
Ans. (D)
Given A² = 3A, |A| ≠ 0, order of A is 3
∴ |A²| = |3A|
⇒ |A|² = 3³|A|   (∵ |A²| = |A|² and |kA| = kⁿ|A|)
⇒ |A| = 27   [1]

Short Answer Questions (2 Marks)

Question 8. If A is a skew-symmetric matrix of order 3, then prove that det A = 0.   [CBSE 2017, 2M]
Answer:
By definition of skew-symmetric matrix,
A′ = −A
⇒ |A′| = |−A|   [½]
⇒ |A| = (−1)³|A|   (∵ |A′| = |A|)   [½]
⇒ |A| = −|A|
⇒ 2|A| = 0   [½]
⇒ |A| = 0   [½]

Question 9. Given \( A = \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix} \), compute A⁻¹ and show that 2A⁻¹ = 9I − A.   [CBSE 2018, 2M]
Answer:
Given, \( A = \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix} \)
|A| = 14 − 12 = 2 ≠ 0, i.e., A is invertible.
\( \text{adj } A = \begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix} \)   [½]
∵ \( A^{-1} = \frac{\text{adj } A}{|A|} \)
\( A^{-1} = \frac{1}{2}\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix} \)   ......(i)
\( = \begin{bmatrix} 7/2 & 3/2 \\ 2 & 1 \end{bmatrix} \)   [½]
Now RHS = 9I − A
\( = 9\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix} = \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} - \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix} = \begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix} \)
= 2A⁻¹ = LHS   [using (i)]   [1]

Long Answer Questions-I (4 Marks)

Question 10. A typist charges ₹ 145 for typing 10 English and 3 Hindi pages, while charges for typing 3 English and 10 Hindi pages are ₹ 180. Using matrices, find the charges of typing one English and one Hindi page separately. However, the typist charged only ₹ 2 per page from a poor student Shyam for 5 Hindi pages. How much less was charged from this poor boy?   [CBSE 2016, 4M]
Answer:
Let ₹ x and ₹ y be the charges of typing one English and one Hindi page respectively.
10x + 3y = 145
3x + 10y = 180
The matrix form is: \( \begin{bmatrix} 10 & 3 \\ 3 & 10 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 145 \\ 180 \end{bmatrix} \)   [1½]
AX = B   ......(i)
|A| = 91 and \( \text{adj } A = \begin{bmatrix} 10 & -3 \\ -3 & 10 \end{bmatrix} \)
∵ \( A^{-1} = \frac{\text{adj } A}{|A|} = \frac{1}{91}\begin{bmatrix} 10 & -3 \\ -3 & 10 \end{bmatrix} \)
By (i), X = A⁻¹B
\( X = \frac{1}{91}\begin{bmatrix} 10 & -3 \\ -3 & 10 \end{bmatrix}\begin{bmatrix} 145 \\ 180 \end{bmatrix} = \frac{1}{91}\begin{bmatrix} 910 \\ 1365 \end{bmatrix} = \begin{bmatrix} 10 \\ 15 \end{bmatrix} \)
\( \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 15 \end{bmatrix} \)   [1½]
x = ₹ 10, y = ₹ 15
Typist charged from poor student = 2 × 5 = ₹ 10
Actual price = 15 × 5 = ₹ 75
The poor boy was charged 75 − 10 = ₹ 65 less.   [1]

Long Answer Questions-II (6 Marks)

Question 11. Using matrix method, solve the following system of linear equations:   [CBSE 2005, 07, 11, 12, 6M]
x + 2y + z = 7
x + 3z = 11
2x − 3y = 1
Answer:
We can write the given equations as
AX = B   .....(1)
where \( A = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{bmatrix};\; X = \begin{bmatrix} x \\ y \\ z \end{bmatrix};\; B = \begin{bmatrix} 7 \\ 11 \\ 1 \end{bmatrix} \)
Now \( |A| = \begin{vmatrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{vmatrix} \)
= 1(0 + 9) − 2(0 − 6) + 1(−3 − 0)
= 9 + 12 − 3 = 18 ≠ 0
⇒ A⁻¹ exists and it is given by   [1]
\[ A^{-1} = \frac{1}{|A|}\,\text{adj } A = \frac{1}{18}\begin{bmatrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{bmatrix} \][2]
From eq. (1), X = A⁻¹B
\[ \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{18}\begin{bmatrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{bmatrix}\begin{bmatrix} 7 \\ 11 \\ 1 \end{bmatrix} = \frac{1}{18}\begin{bmatrix} 36 \\ 18 \\ 54 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} \][2]
⇒ x = 2, y = 1, z = 3.   [1]

Question 12. The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category.   [CBSE 2013, 6M]
Answer:
x = awarded members for honesty
y = awarded members for helping (cooperation)
z = awarded members for supervision
Sum of all the awardees is 12,
so x + y + z = 12   .........(i)
Three times the sum of awardees for y and z added to two times x is 33:
3(y + z) + 2x = 33
2x + 3y + 3z = 33   .........(ii)
The sum of the numbers for x and z is twice y:
x + z = 2y
x − 2y + z = 0   ........(iii)   (for any two correct)   [2]
All the above three equations can be written in matrix form as
\( \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 12 \\ 33 \\ 0 \end{bmatrix} \)
Let AX = B
⇒ X = A⁻¹B   ......(i)   [1]
|A| = 1(3 + 6) − 1(2 − 3) + 1(−4 − 3)
= 9 + 1 − 7 = 3
adj A = [C_ij]^T
\[ = \begin{bmatrix} 9 & 1 & -7 \\ -3 & 0 & 3 \\ 0 & -1 & 1 \end{bmatrix}^T = \begin{bmatrix} 9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{bmatrix} \][1]
By (i), \( X = \frac{\text{adj } A}{|A|}\cdot B \)
\[ X = \frac{1}{3}\begin{bmatrix} 9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{bmatrix}\begin{bmatrix} 12 \\ 33 \\ 0 \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 108 - 99 + 0 \\ 12 + 0 + 0 \\ -84 + 99 + 0 \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 9 \\ 12 \\ 15 \end{bmatrix} \][1]
\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix} \) ⇒ x = 3, y = 4, z = 5   [1]

Question 13. If \( A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} \), find A⁻¹. Use it to solve the system of equations   [CBSE 2017, 2018, 6M]
2x − 3y + 5z = 11
3x + 2y − 4z = −5
x + y − 2z = −3.
Answer:
\( A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} \)
∴ |A| = 2(−4 + 4) + 3(−6 + 4) + 5(3 − 2) = 0 − 6 + 5 = −1 ≠ 0
∵ |A| ≠ 0, so A is invertible.
Now, A₁₁ = 0, A₁₂ = 2, A₁₃ = 1
A₂₁ = −1, A₂₂ = −9, A₂₃ = −5
A₃₁ = 2, A₃₂ = 23, A₃₃ = 13
∴ \( \text{adj } A = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} \)   [2]
\[ \therefore\; A^{-1} = \frac{1}{|A|}(\text{adj } A) = -\begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \]......(i)   [1]
Given equations,
2x − 3y + 5z = 11
3x + 2y − 4z = −5
x + y − 2z = −3
The given system of equations can be written as AX = B,
where \( A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix},\; X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\text{ and } B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \)   [1]
⇒ X = A⁻¹B
\[ \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}\begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \][Using (i)]   [1]
\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 - 5 + 6 \\ -22 - 45 + 69 \\ -11 - 25 + 39 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \]
Hence, x = 1, y = 2 and z = 3.   [1]

Question 14. Show that for the matrix \( A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} \), A³ − 6A² + 5A + 11I = O. Hence, find A⁻¹.   [CBSE 2019, 6M]
OR
Using matrix method, solve the following system of equations:
3x − 2y + 3z = 8
2x + y − z = 1
4x − 3y + 2z = 4
Answer:
Given, \( A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} \)
To prove: A³ − 6A² + 5A + 11I = O
\[ \Rightarrow A^2 = A\cdot A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} \]
\[ = \begin{bmatrix} 1+1+2 & 1+2-1 & 1-3+3 \\ 1+2-6 & 1+4+3 & 1-6-9 \\ 2-1+6 & 2-2-3 & 2+3+9 \end{bmatrix} \]
\[ \Rightarrow A^2 = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} \][1½]
and A³ = A²·A
\[ = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} \]
\[ = \begin{bmatrix} 4+2+2 & 4+4-1 & 4-6+3 \\ -3+8-28 & -3+16+14 & -3-24-42 \\ 7-3+28 & 7-6-14 & 7+9+42 \end{bmatrix} \]
\[ \Rightarrow A^3 = \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix} \][1½]
L.H.S. = A³ − 6A² + 5A + 11I
\[ = \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix} - 6\begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} + 5\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} + 11\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
\[ = \begin{bmatrix} 8-24+5+11 & 7-12+5+0 & 1-6+5+0 \\ -23+18+5+0 & 27-48+10+11 & -69+84-15+0 \\ 32-42+10+0 & -13+18-5+0 & 58-84+15+11 \end{bmatrix} \]
\[ = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O \][1]
∵ A³ − 6A² + 5A + 11I = O
⇒ A²(AA⁻¹) − 6A(AA⁻¹) + 5(AA⁻¹) + 11IA⁻¹ = O
⇒ A² − 6A + 5I + 11A⁻¹ = O   [∵ AA⁻¹ = I]
⇒ \( A^{-1} = \frac{-1}{11}\left(A^2 - 6A + 5I\right) \)   [1]
\[ = \frac{-1}{11}\left\{\begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} - 6\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} + 5\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\right\} \]
\[ \text{Hence } A^{-1} = \begin{bmatrix} -3/11 & 4/11 & 5/11 \\ 9/11 & -1/11 & -4/11 \\ 5/11 & -3/11 & -1/11 \end{bmatrix} \][1]
OR
Given system of equations is:
3x − 2y + 3z = 8
2x + y − z = 1
and 4x − 3y + 2z = 4
By using matrix method, the given system of equations can be written as:
AX = B   ……(i)
where \( A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix},\; X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\text{ and } B = \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix} \)   [1]
Now, \( |A| = \begin{vmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{vmatrix} \)
= −3 + 16 − 30 = −17 ≠ 0
∵ |A| ≠ 0, so A⁻¹ exists.
Now, \( A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix} \)
Cofactors are given as:
C₁₁ = −1, C₁₂ = −8, C₁₃ = −10
C₂₁ = −5, C₂₂ = −6, C₂₃ = 1
C₃₁ = −1, C₃₂ = 9, C₃₃ = 7
\( \text{adj } A = \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix} \)   [2]
∵ \( A^{-1} = \frac{\text{adj } A}{|A|} = \frac{-1}{17}\begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix} \)   [1]
X = A⁻¹B   [By using (i)]
\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = -\frac{1}{17}\begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix}\begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix} \]
\[ = -\frac{1}{17}\begin{bmatrix} -8-5-4 \\ -64-6+36 \\ -80+1+28 \end{bmatrix} = -\frac{1}{17}\begin{bmatrix} -17 \\ -34 \\ -51 \end{bmatrix} \]
\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \)
⇒ x = 1, y = 2, z = 3   [2]

Question 15. If \( A = \begin{bmatrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{bmatrix} \), find A⁻¹ and use it to solve the following system of equations:   [CBSE 2020, 6M]
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
Answer:
Given \( A = \begin{bmatrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{bmatrix} \)
∴ \( |A| = \begin{vmatrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{vmatrix} \) = 5(18 + 10) + 1(12 − 25) + 4(−4 − 15) = 51 ≠ 0.
Since |A| ≠ 0, A is a non-singular matrix, i.e. A⁻¹ exists.   [1]
Now, cofactors of elements of |A| are
A₁₁ = 28, A₁₂ = 13, A₁₃ = −19
A₂₁ = −2, A₂₂ = 10, A₂₃ = 5
A₃₁ = −17, A₃₂ = −17, A₃₃ = 17
\[ \therefore\; \text{adj } A = \begin{bmatrix} 28 & 13 & -19 \\ -2 & 10 & 5 \\ -17 & -17 & 17 \end{bmatrix}^T = \begin{bmatrix} 28 & -2 & -17 \\ 13 & 10 & -17 \\ -19 & 5 & 17 \end{bmatrix} \][1]
∵ \( A^{-1} = \frac{1}{|A|}\,\text{adj } A \)
\[ A^{-1} = \frac{1}{51}\begin{bmatrix} 28 & -2 & -17 \\ 13 & 10 & -17 \\ -19 & 5 & 17 \end{bmatrix} \][1]
Given system of equations is
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
This system is written in matrix equation form as
\( \begin{bmatrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ 2 \\ -1 \end{bmatrix} \)   [1]
AX = B ⇒ X = A⁻¹B
\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{51}\begin{bmatrix} 28 & -2 & -17 \\ 13 & 10 & -17 \\ -19 & 5 & 17 \end{bmatrix}\begin{bmatrix} 5 \\ 2 \\ -1 \end{bmatrix} \]
\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{51}\begin{bmatrix} 140 - 4 + 17 \\ 65 + 20 + 17 \\ -95 + 10 - 17 \end{bmatrix} \][1]
\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{51}\begin{bmatrix} 153 \\ 102 \\ -102 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \\ -2 \end{bmatrix} \]
So, x = 3, y = 2, z = −2   [1]