Determinants — Class 12 Maths practice questions

Question 1

A school canteen manager records the daily purchase of three food items — Sandwich (S), Juice (J) and Salad (L) — over three days as follows:

• Day 1: 2 Sandwiches, 1 Juice and 1 Salad cost ₹ 170.
• Day 2: 1 Sandwich, 2 Juices and 1 Salad cost ₹ 160.
• Day 3: 1 Sandwich, 1 Juice and 2 Salads cost ₹ 150.

Let the price (in ₹) of one Sandwich, one Juice and one Salad be x, y and z respectively.

(i) Express the above situation as a matrix equation of the form AX = B.
(ii) Find |A|.
(iii) Using the matrix method (A⁻¹), find the price of each food item. OR (iii) If the canteen offers a 'Combo Deal' comprising 1 Sandwich, 1 Juice and 1 Salad, find the total cost of the Combo Deal.

Accepted Answer

(i) The matrix equation AX = B is:

A = [2 1 1 / 1 2 1 / 1 1 2], X = [x / y / z], B = [170 / 160 / 150]

So the system is:

[2 1 1]   [x]   [170]
[1 2 1] × [y] = [160]
[1 1 2]   [z]   [150]

∴ AX = B.                                                              (1 mark)

(ii) Expanding |A| along R₁:

|A| = 2[(2)(2) − (1)(1)] − 1[(1)(2) − (1)(1)] + 1[(1)(1) − (2)(1)]
    = 2[4 − 1] − 1[2 − 1] + 1[1 − 2]
    = 2(3) − 1(1) + 1(−1)
    = 6 − 1 − 1

∴ |A| = 4 ≠ 0, so A⁻¹ exists.                                         (1 mark)

(iii) Finding A⁻¹:

Cofactors of A:
C₁₁ = +[(2)(2)−(1)(1)] = +3
C₁₂ = −[(1)(2)−(1)(1)] = −1
C₁₃ = +[(1)(1)−(2)(1)] = −1
C₂₁ = −[(1)(2)−(1)(1)] = −1
C₂₂ = +[(2)(2)−(1)(1)] = +3
C₂₃ = −[(2)(1)−(1)(1)] = −1
C₃₁ = +[(1)(1)−(2)(1)] = −1
C₃₂ = −[(2)(1)−(1)(1)] = −1
C₃₃ = +[(2)(2)−(1)(1)] = +3

adj A = (Cofactor matrix)ᵀ =
[ 3  −1  −1]
[−1   3  −1]
[−1  −1   3]

A⁻¹ = (1/|A|) adj A = (1/4) [ 3  −1  −1 / −1  3  −1 / −1  −1  3]

Now, X = A⁻¹B:

x = (1/4)[(3)(170) + (−1)(160) + (−1)(150)]
  = (1/4)[510 − 160 − 150]
  = (1/4)(200) = 50

y = (1/4)[(−1)(170) + (3)(160) + (−1)(150)]
  = (1/4)[−170 + 480 − 150]
  = (1/4)(160) = 40

z = (1/4)[(−1)(170) + (−1)(160) + (3)(150)]
  = (1/4)[−170 − 160 + 450]
  = (1/4)(120) = 30

∴ Price of one Sandwich = ₹ 50, one Juice = ₹ 40, one Salad = ₹ 30.   (2 marks)

OR

(iii) From the values obtained above (x = 50, y = 40, z = 30):

Total cost of Combo Deal = x + y + z
= 50 + 40 + 30

∴ Total cost of the Combo Deal = ₹ 120.                                (2 marks)

Question 2

A school canteen serves two types of meal combos — Type P (a nutritious meal with roti, dal, and salad) and Type Q (a special meal with rice, paneer, and dessert). The canteen manager recorded the following information for two consecutive days:

• On Day 1: 30 Type P combos and 20 Type Q combos were sold, collecting a total of ₹4,900.
• On Day 2: 20 Type P combos and 40 Type Q combos were sold, collecting a total of ₹5,800.

Let ₹x be the price of one Type P combo and ₹y be the price of one Type Q combo.

Based on the above information, answer the following questions:

(i) Represent the given situation as a matrix equation of the form AX = B, where A is the coefficient matrix, X = [x, y]ᵀ and B is the constant matrix. Write the matrix A.

(ii) Find |A|, the determinant of matrix A.

(iii) Find the prices ₹x and ₹y of each combo using the matrix (inverse) method. Hence find the total cost of 2 Type P combos and 3 Type Q combos.

Accepted Answer

(i) The two equations formed from the given data are:
30x + 20y = 4900   ...(i)
20x + 40y = 5800   ...(ii)

In matrix form AX = B:

A = \begin{bmatrix} 30 & 20 \ 20 & 40 \end{bmatrix},   X = \begin{bmatrix} x \ y \end{bmatrix},   B = \begin{bmatrix} 4900 \ 5800 \end{bmatrix}

∴ A = \begin{bmatrix} 30 & 20 \ 20 & 40 \end{bmatrix}                                                [1 mark]

(ii) |A| = (30)(40) − (20)(20)
       = 1200 − 400
∴ |A| = 800                                                                            [1 mark]

(iii) Since |A| = 800 ≠ 0, A⁻¹ exists.

The cofactor matrix of A is:
C₁₁ = 40,   C₁₂ = −20,   C₂₁ = −20,   C₂₂ = 30

adj A = \begin{bmatrix} 40 & -20 \ -20 & 30 \end{bmatrix}

A⁻¹ = (1/|A|) · adj A = (1/800) \begin{bmatrix} 40 & -20 \ -20 & 30 \end{bmatrix}

Now, X = A⁻¹B:

\begin{bmatrix} x \ y \end{bmatrix} = (1/800) \begin{bmatrix} 40 & -20 \ -20 & 30 \end{bmatrix} \begin{bmatrix} 4900 \ 5800 \end{bmatrix}

= (1/800) \begin{bmatrix} (40)(4900) + (−20)(5800) \ (−20)(4900) + (30)(5800) \end{bmatrix}

= (1/800) \begin{bmatrix} 196000 − 116000 \ −98000 + 174000 \end{bmatrix}

= (1/800) \begin{bmatrix} 80000 \ 76000 \end{bmatrix}

= \begin{bmatrix} 100 \ 95 \end{bmatrix}

∴ x = ₹100  and  y = ₹95

∴ Price of 1 Type P combo = ₹100 and price of 1 Type Q combo = ₹95.

Total cost of 2 Type P combos and 3 Type Q combos
= 2x + 3y
= 2(100) + 3(95)
= 200 + 285

∴ Required total cost = ₹485                                                        [2 marks]

Question 3

A school treasurer is managing funds using a system of two equations. The amounts (in ₹ thousands) allocated to sports and library are represented by variables x and y. The system is encoded as a matrix equation AX = B, where

A = [[3, 1], [5, 2]], X = [[x], [y]], B = [[11], [19]]

Based on the above information, answer the following questions:

(i) Find |A|. [1 mark]
(ii) Find A⁻¹. [1 mark]
(iii) Using A⁻¹, find the amounts allocated to sports (x) and library (y). [2 marks]

Accepted Answer

(i) Finding |A|:

|A| = (3)(2) − (1)(5) = 6 − 5 = 1

∴ |A| = 1

(ii) Finding A⁻¹:

Since |A| = 1 ≠ 0, A⁻¹ exists.

The cofactor matrix of A is:

C₁₁ = 2,  C₁₂ = −5,  C₂₁ = −1,  C₂₂ = 3

adj A = [[2, −1], [−5, 3]]

A⁻¹ = (1/|A|) · adj A = (1/1) · [[2, −1], [−5, 3]]

∴ A⁻¹ = [[2, −1], [−5, 3]]

(iii) Finding x and y:

Since AX = B, we have X = A⁻¹B.

[[x], [y]] = [[2, −1], [−5, 3]] · [[11], [19]]

x = (2)(11) + (−1)(19) = 22 − 19 = 3

y = (−5)(11) + (3)(19) = −55 + 57 = 2

∴ Amount allocated to sports = ₹3 thousand and amount allocated to library = ₹2 thousand.

Question 4

A city planner is designing a triangular park with vertices at three locations on a coordinate grid: P(k, 2), Q(4, 6), and R(1, 4). The planner wants the area of the park to be exactly 4 square units.

(i) Using determinants, write the expression for the area of triangle PQR in terms of k.   [1 mark]
(ii) Set up the equation obtained when the area equals 4 square units and simplify it.   [1 mark]
(iii) Find all possible values of k. Hence determine the two possible positions of vertex P.   [2 marks]

Accepted Answer

(i) Area of △PQR using determinants:

Area = (1/2)|det| = (1/2) |k(6 − 4) + 4(4 − 2) + 1(2 − 6)|

Expanding the determinant along the formula:

Area = (1/2) |k·(d − f) + x₂·(f − y₁) + x₃·(y₁ − d)|

∴ Area = (1/2) |k(6 − 4) + 4(4 − 2) + 1(2 − 6)|

= (1/2) |2k + 8 − 4|

∴ Area = (1/2)|2k + 4|   ...(i)

(ii) Setting area = 4 square units:

From (i), (1/2)|2k + 4| = 4

⟹ |2k + 4| = 8

∴ 2k + 4 = 8   or   2k + 4 = −8   ...(ii)

(iii) Solving both cases from (ii):

Case 1: 2k + 4 = 8 ⟹ 2k = 4 ⟹ k = 2

Case 2: 2k + 4 = −8 ⟹ 2k = −12 ⟹ k = −6

∴ The two possible values of k are 2 and −6.

∴ The two possible positions of vertex P are (2, 2) and (−6, 2).

Question 5

A structural engineering firm uses a system of three linear equations to model the load distribution (in kN) across three support beams of a bridge:

2x + 3y − z = 10
    x − y + 2z = 4
    3x + 2y + kz = 16

where x, y, z represent the loads on the three beams respectively, and k is a design parameter.

(i) Write the coefficient matrix A of the system. Find |A| in terms of k.   [1 mark]
(ii) For what value of k does the system fail to have a unique solution? Justify using the determinant condition.   [1 mark]
(iii) For k = 3, find |adj A| and hence determine whether A⁻¹ exists. Also find the value of |3A⁻¹| for this case.   [2 marks]

Accepted Answer

(i) The coefficient matrix of the system is:

A = ⎡ 2   3  −1 ⎤
    ⎢ 1  −1   2 ⎥
    ⎣ 3   2   k ⎦

Expanding |A| along R₁:

|A| = 2[(−1)(k) − (2)(2)] − 3[(1)(k) − (2)(3)] + (−1)[(1)(2) − (−1)(3)]

⟹ |A| = 2[−k − 4] − 3[k − 6] − 1[2 + 3]

⟹ |A| = −2k − 8 − 3k + 18 − 5

∴ |A| = −5k + 5  =  5(1 − k)

(ii) The system fails to have a unique solution when the coefficient matrix A is singular, i.e., when |A| = 0.

∵ |A| = 5(1 − k)

|A| = 0  ⟹  5(1 − k) = 0  ⟹  k = 1

∴ For k = 1, the determinant |A| = 0, so A is singular and the system does not have a unique solution.

(iii) For k = 3:

|A| = 5(1 − 3) = 5 × (−2) = −10

∵ |A| = −10 ≠ 0, A is non-singular.
∴ A⁻¹ exists.

|adj A| = |A|³⁻¹ = |A|² = (−10)² = 100

∴ |adj A| = 100

Finding |3A⁻¹|:
Using the property |kA| = kⁿ|A| for an n × n matrix:

|3A⁻¹| = 3³ · |A⁻¹| = 27 × (1/|A|) = 27 × (1/(−10))

∴ |3A⁻¹| = −27/10

Question 6

A school treasurer is managing three school funds — Sports Fund (x), Cultural Fund (y), and Science Fund (z) — in thousands of rupees. Based on the school budget records, the following three conditions are noted:

(i) The total of all three funds is ₹6000.
(ii) The Science Fund exceeds the Cultural Fund by ₹1000.
(iii) Twice the Sports Fund, added to the Cultural Fund, equals the Science Fund by ₹5000 in total.

This translates to the system of equations:
  x + y + z = 6
  y - z = -1
  2x + y - z = 5

(where amounts are in thousands of rupees)

(a) Write the above system in the matrix form AX = B, identifying matrices A, X, and B.  [1]
(b) Find |A| (the determinant of the coefficient matrix A).  [1]
(c) Find A⁻¹ using the adjoint method, and hence solve for x, y, and z.  [2]

Accepted Answer

(a) The system x + y + z = 6, y − z = −1, 2x + y − z = 5 is written as AX = B, where

A = [1  1  1 / 0  1  −1 / 2  1  −1],   X = [x / y / z],   B = [6 / −1 / 5]    …(1 mark)

(b) Expanding |A| along R₁:

|A| = 1·|(1  −1 / 1  −1)| − 1·|(0  −1 / 2  −1)| + 1·|(0  1 / 2  1)|

= 1·[(1)(−1) − (−1)(1)] − 1·[(0)(−1) − (−1)(2)] + 1·[(0)(1) − (1)(2)]

= 1·[−1 + 1] − 1·[0 + 2] + 1·[0 − 2]

= 1·(0) − 1·(2) + 1·(−2)

= 0 − 2 − 2

∴ |A| = −4    …(1 mark)

∵ |A| = −4 ≠ 0, so A⁻¹ exists.

(c) Computing the cofactors of A:

C₁₁ = +(1·(−1) − (−1)·1) = +(−1 + 1) = 0
C₁₂ = −(0·(−1) − (−1)·2) = −(0 + 2) = −2
C₁₃ = +(0·1 − 1·2) = +(0 − 2) = −2

C₂₁ = −(1·(−1) − 1·1) = −(−1 − 1) = 2
C₂₂ = +(1·(−1) − 1·2) = +(−1 − 2) = −3
C₂₃ = +(1·1 − 1·2) = −(1 − 2) = 1

[Note: C₂₃ = −(1·1 − 1·2) = −(1−2) = 1]

C₃₁ = +(1·(−1) − 1·1) = +(−1 − 1) = −2
C₃₂ = −(1·(−1) − 1·0) = −(−1 − 0) = 1
C₃₃ = +(1·1 − 1·0) = +(1 − 0) = 1

adj A = (Cofactor matrix)ᵀ =
[ C₁₁  C₂₁  C₃₁ /  C₁₂  C₂₂  C₃₂ /  C₁₃  C₂₃  C₃₃ ]
= [0   2   −2 / −2  −3   1 / −2   1   1]

∴ A⁻¹ = (1/|A|) · adj A = (1/−4) · [0  2  −2 / −2  −3  1 / −2  1  1]

Since X = A⁻¹B:

X = (1/−4) · [0  2  −2 / −2  −3  1 / −2  1  1] · [6 / −1 / 5]

Row 1: 0(6) + 2(−1) + (−2)(5) = 0 − 2 − 10 = −12
Row 2: (−2)(6) + (−3)(−1) + 1(5) = −12 + 3 + 5 = −4
Row 3: (−2)(6) + 1(−1) + 1(5) = −12 − 1 + 5 = −8

∴ X = (1/−4) · [−12 / −4 / −8] = [3 / 1 / 2]

∴ x = 3, y = 1, z = 2

Hence, the Sports Fund = ₹3000, Cultural Fund = ₹1000, and Science Fund = ₹2000.    …(2 marks)

Question 7

A small business owner models her monthly profit-and-loss data using a 3×3 matrix A, where the entries represent coefficients of a system of three linear equations in three unknowns (inventory, labour, overhead). She knows the following facts about matrix A:

(i) A(adj A) = kI₃, where k is a real constant and I₃ is the 3×3 identity matrix.

(ii) |A| = 5.

(iii) The matrix B = 3A.

Based on this information, answer the following:

(a) Find the value of k. [1 mark]

(b) Find |adj A|. [1 mark]

(c) Find |adj B|. [2 marks]

Accepted Answer

(a) Finding the value of k:

Using the standard result for any square matrix A of order n:

A(adj A) = |A| · Iₙ

Here n = 3, so A(adj A) = |A| · I₃.

Comparing with the given A(adj A) = k I₃,

∴ k = |A| = 5

(b) Finding |adj A|:

Using the standard result for a square matrix of order n:

|adj A| = |A|ⁿ⁻¹

Here n = 3 and |A| = 5,

∴ |adj A| = |A|³⁻¹ = |A|² = 5² = 25

(c) Finding |adj B|:

Since B = 3A and A is a 3×3 matrix, using the scalar-multiplication property of determinants:

|B| = |3A| = 3³ · |A| = 27 × 5 = 135

Now applying the standard result |adj B| = |B|ⁿ⁻¹ with n = 3:

|adj B| = |B|² = (135)² = 18225

∴ k = 5, |adj A| = 25, and |adj B| = 18225.

Question 8

A city planner is designing a triangular park with vertices at coordinates A(1, 2), B(4, k) and C(7, 8) on a grid map, where each unit represents 10 metres. The municipal authority requires that:
(i) The area of the park must be exactly 900 square metres (i.e., 9 square units on the grid).
(ii) If the area condition gives two possible values of k, the park boundary must not be a straight line (i.e., the three vertices must not be collinear).

Using the determinant method:
(a) Set up a determinant expression for the area of triangle ABC in terms of k. [1 mark]
(b) Find all possible values of k satisfying the area requirement. [2 marks]
(c) Determine which value(s) of k are valid for the park design, giving a reason using the concept of collinearity. [1 mark]

Accepted Answer

(a) Setting up the determinant expression:

The area of triangle with vertices A(1, 2), B(4, k) and C(7, 8) is given by:

Area = (1/2) |det| where det = |1  2  1|
                                   |4  k  1|
                                   |7  8  1|

∴ Area = (1/2) |1(k − 8) − 2(4 − 7) + 1(32 − 7k)|

[1 mark — for correct determinant setup with three vertices in matrix form]

(b) Finding the values of k:

Expanding the determinant:

= 1·(k·1 − 8·1) − 2·(4·1 − 7·1) + 1·(4·8 − 7·k)

= (k − 8) − 2(4 − 7) + (32 − 7k)

= (k − 8) − 2(−3) + (32 − 7k)

= k − 8 + 6 + 32 − 7k

= −6k + 30

Area = (1/2)|−6k + 30|

Setting Area = 9:

(1/2)|−6k + 30| = 9

|−6k + 30| = 18

Case 1: −6k + 30 = 18
⟹ −6k = −12
⟹ k = 2

Case 2: −6k + 30 = −18
⟹ −6k = −48
⟹ k = 8

∴ The two possible values of k are 2 and 8.  [2 marks — 1 for correct expansion to |−6k + 30| = 18; 1 for both values k = 2 and k = 8]

(c) Validity check using collinearity:

Three points are collinear if and only if the area of the triangle formed by them is zero, i.e., the determinant = 0.

Check k = 2: The vertices become A(1, 2), B(4, 2) and C(7, 8).
The determinant value = −6(2) + 30 = −12 + 30 = 18 ≠ 0 ∴ A, B, C are not collinear. ✓

Check k = 8: The vertices become A(1, 2), B(4, 8) and C(7, 8).
The determinant value = −6(8) + 30 = −48 + 30 = −18 ≠ 0 ∴ A, B, C are not collinear. ✓

∵ both values k = 2 and k = 8 give non-collinear vertices, both satisfy the municipal design requirement.

∴ Both k = 2 and k = 8 are valid for the park design, as in each case the three vertices form a proper (non-degenerate) triangle of area exactly 9 square units, i.e., 900 square metres.  [1 mark — for correct collinearity reasoning and valid conclusion for both values]

Question 9

If A is a square matrix of order 3 and |A| = 5, then |adj A| is:

Accepted Answer

Option (B) is correct.

Explanation: For a square matrix A of order n, the standard result is |adj A| = |A|^(n−1).

Here n = 3 and |A| = 5, so |adj A| = 5^(3−1) = 5² = 25.

Question 10

If A = $\begin{bmatrix} 3 & x \ 2 & 5 \end{bmatrix}$ is a singular matrix, then the value of x is:
(a) 15/2  (b) 6  (c) 5/3  (d) 15

Accepted Answer

Option (a) is correct.

Explanation: A matrix is singular if and only if its determinant equals zero.

|A| = (3)(5) − (x)(2) = 15 − 2x

Setting |A| = 0: 15 − 2x = 0 ⟹ x = 15/2

Question 11

If A is a square matrix of order 3 and |A| = 5, then |2A| is equal to:
(a) 10   (b) 25   (c) 40   (d) 15

Accepted Answer

Option (c) is correct. Explanation: For a square matrix A of order n, |kA| = kn|A|. Here n = 3 and k = 2, so |2A| = 23|A| = 8 × 5 = 40.

Question 12

If A is a square matrix of order 3 and |A| = 5, then |adj A| is equal to:
(A) 5
(B) 25
(C) 125
(D) 15

Accepted Answer

Option (B) is correct. Explanation: For a square matrix A of order n, the standard result is |adj A| = |A|n−1. Here n = 3 and |A| = 5, so |adj A| = 53−1 = 52 = 25.

Question 13

If A is a square matrix of order 3 and |A| = −5, then |adj A| is:

Accepted Answer

Option (C) is correct.

Explanation: For a square matrix A of order n, the standard result is |adj A| = |A|^(n−1).

Here n = 3 and |A| = −5, so |adj A| = (−5)^(3−1) = (−5)² = 25.

Question 14

If A is a square matrix of order 3 and |A| = 5, then |2A| is

Accepted Answer

Option (C) is correct.

Explanation: For a square matrix A of order n, the property |kA| = kⁿ|A| holds for any scalar k.

Here n = 3 and k = 2, so |2A| = 2³|A| = 8 × 5 = 40.

Question 15

Let A and B be two square matrices of order 3 such that |A| = −2 and |B| = 5. Find the value of |3A²B|.

Accepted Answer

Using the property |kM| = k³|M| for a 3×3 matrix M,

|3A²B| = 3³ · |A²| · |B|

= 27 · |A|² · |B|

= 27 × (−2)² × 5

= 27 × 4 × 5

∴ |3A²B| = 540

Question 16

Let A and B be two square matrices of order 3 such that |A| = −2 and |B| = 5. Find the value of |3A²B|.

Accepted Answer

Using the property |kM| = k³|M| for a square matrix M of order 3,

|3A²B| = 3³ · |A²| · |B|

⟹ |3A²B| = 27 · |A|² · |B|

⟹ |3A²B| = 27 × (−2)² × 5

⟹ |3A²B| = 27 × 4 × 5

∴ |3A²B| = 540

Question 17

Let A and B be two square matrices of order 3 such that |A| = 5 and |B| = −2. Find the value of |3AB|.

Accepted Answer

Using the property |kA| = k³|A| for a square matrix of order 3,

|3AB| = 3³ · |AB|

⟹ |3AB| = 27 · |A| · |B|          [∵ |AB| = |A||B|]

⟹ |3AB| = 27 × 5 × (−2)

∴ |3AB| = −270

Question 18

Let A and B be two square matrices of order 3 such that |A| = 5 and |B| = −2. Find the value of |−3AB|.

Accepted Answer

Using the property |kA| = k³|A| for a square matrix A of order 3,

|−3AB| = (−3)³ |AB|

⟹ |−3AB| = −27 · |A| · |B|          [∵ |AB| = |A||B|]

⟹ |−3AB| = −27 × 5 × (−2)

∴ |−3AB| = 270

Question 19

If A = \begin{bmatrix} 2 & -1 & 1 \ -1 & 2 & -1 \ 1 & -1 & 2 \end{bmatrix}, find A⁻¹ and use it to solve the system of equations:  2x − y + z = 3,  −x + 2y − z = −4,  x − y + 2z = 5.

Accepted Answer

Given: A = \begin{bmatrix} 2 & -1 & 1 \ -1 & 2 & -1 \ 1 & -1 & 2 \end{bmatrix}

Step 1: Find |A|.

Expanding along R₁,

|A| = 2[(2)(2) − (−1)(−1)] − (−1)[(−1)(2) − (−1)(1)] + 1[(−1)(−1) − (2)(1)]

⟹ |A| = 2[4 − 1] + 1[−2 + 1] + 1[1 − 2]

⟹ |A| = 2(3) + 1(−1) + 1(−1)

⟹ |A| = 6 − 1 − 1 = 4

∵ |A| = 4 ≠ 0, ∴ A⁻¹ exists.

Step 2: Find the cofactor matrix of A.

C₁₁ = +[(2)(2) − (−1)(−1)] = 4 − 1 = 3
C₁₂ = −[(−1)(2) − (−1)(1)] = −[−2 + 1] = 1
C₁₃ = +[(−1)(−1) − (2)(1)] = 1 − 2 = −1

C₂₁ = −[(−1)(2) − (1)(−1)] = −[−2 + 1] = 1
C₂₂ = +[(2)(2) − (1)(1)] = 4 − 1 = 3
C₂₃ = −[(2)(−1) − (−1)(1)] = −[−2 + 1] = 1

C₃₁ = +[(−1)(−1) − (1)(2)] = 1 − 2 = −1
C₃₂ = −[(2)(−1) − (1)(−1)] = −[−2 + 1] = 1
C₃₃ = +[(2)(2) − (−1)(−1)] = 4 − 1 = 3

Cofactor matrix = \begin{bmatrix} 3 & 1 & -1 \ 1 & 3 & 1 \ -1 & 1 & 3 \end{bmatrix}

adj A = (Cofactor matrix)' = \begin{bmatrix} 3 & 1 & -1 \ 1 & 3 & 1 \ -1 & 1 & 3 \end{bmatrix}

(Note: the cofactor matrix is symmetric here, so adj A equals the cofactor matrix.)

∴ A⁻¹ = (1/|A|) adj A = (1/4) \begin{bmatrix} 3 & 1 & -1 \ 1 & 3 & 1 \ -1 & 1 & 3 \end{bmatrix}

Step 3: Write the system in matrix form AX = B.

The given system 2x − y + z = 3, −x + 2y − z = −4, x − y + 2z = 5 is AX = B, where

X = \begin{bmatrix} x \ y \ z \end{bmatrix}, B = \begin{bmatrix} 3 \ -4 \ 5 \end{bmatrix}

Step 4: Solve X = A⁻¹B.

X = (1/4) \begin{bmatrix} 3 & 1 & -1 \ 1 & 3 & 1 \ -1 & 1 & 3 \end{bmatrix} \begin{bmatrix} 3 \ -4 \ 5 \end{bmatrix}

⟹ X = (1/4) \begin{bmatrix} 3(3) + 1(−4) + (−1)(5) \ 1(3) + 3(−4) + 1(5) \ (−1)(3) + 1(−4) + 3(5) \end{bmatrix}

⟹ X = (1/4) \begin{bmatrix} 9 − 4 − 5 \ 3 − 12 + 5 \ −3 − 4 + 15 \end{bmatrix}

⟹ X = (1/4) \begin{bmatrix} 0 \ -4 \ 8 \end{bmatrix}

⟹ X = \begin{bmatrix} 0 \ -1 \ 2 \end{bmatrix}

∴ x = 0, y = −1, z = 2.

Question 20

If A is a square matrix of order 3 such that det(A) = −5, find the value of det(3A) + det(−A²).

Accepted Answer

We use the property: for a square matrix A of order n, det(kA) = kⁿ det(A).

Here, A is of order 3, so n = 3.

**Finding det(3A):**

det(3A) = 3³ · det(A)

⟹ det(3A) = 27 × (−5)

⟹ det(3A) = −135  ...(i)

**Finding det(−A²):**

Using det(kA) = k³ det(A) with k = −1 and matrix A²:

det(−A²) = (−1)³ · det(A²)

⟹ det(−A²) = −det(A²)

Now, det(A²) = det(A · A) = det(A) · det(A) = (−5)² = 25

⟹ det(−A²) = −25  ...(ii)

**Finding det(3A) + det(−A²):**

From (i) and (ii),

det(3A) + det(−A²) = −135 + (−25)

∴ det(3A) + det(−A²) = −160

Determinants: Class 12 Maths Practice Questions

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