CBSE Sample Papers for Class 12 Maths Set 7 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Maths with Solutions Set 7 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Maths Set 7 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

• This Question paper contains five sections A, B, C, D, and E. Each Section is compulsory. However, there are internal choices in some questions.
• Section A has 18 MCQs and 2 Assertion-Reason-based questions of 1 mark each.
• Section B has 5 Very Short Answer (VSA) type questions of 2 marks each.
• Section C has 6 Short Answer (SA) type questions of 3 marks each.
• Section D has 4 Long Answer (LA) type questions of 5 marks each.
• Section E has 3 sources based/case based/passage based/integrated units of assessment (4 marks each) with subparts.

Section – A (20 Marks)
(Multiple Choice Questions Each question carries 1 mark)

Question 1.
If set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mapping from A to B is: [1]
(a) 600
(b) 56
(c) 65
(d) 0
Solution:
(d) 0
Explanation:
Since A and B are two non-empty finite sets containing m and n elements respectively, then the number of one-one and onto mapping from A to B is n, if m = n and 0, If m ≠ 0
there, m = 5 and n = 6
∴ m ≠ n
So, the number of mapping = 0

Question 2.
The principal value branch of cosec^-1x is: [1]
(a) \(\left\{-\frac{\pi}{2}, \frac{\pi}{2}\right\}\)
(b) \(\left\{-\frac{\pi}{2}, \frac{\pi}{2}\right\}\) – {0}
(c) {-∞, ∞}
(d) {-π, π} – {0}
Solution:
(b) \(\left\{-\frac{\pi}{2}, \frac{\pi}{2}\right\}\) – {0}
Explanation: The principal value branch of cosec^-1x is \(\left\{-\frac{\pi}{2}, \frac{\pi}{2}\right\}\) – {0}

Question 3.
What is the range of the function f(x) = \(\frac{|x-1|}{(x-1)}\). [1]
(a) {1, 3}
(b) {1, 2}
(c) {-1, 1}
(d) {-1, 3}
Solution:
(c) {-1, 1}
Explanation:

Question 4.
If A is a symmetric matrix, then A³ is: [1]
(a) symmetric matrix
(b) skew-symmetric matrix
(c) Identity matrix
(d) row matrix
Solution:
(a) Symmetric matrix
Explanation:
Since A is a symmetric matrix.
∴ A’ = A
Now (A³)’ = (A’)³ = A³ [∵ (A’)ⁿ = (Aⁿ)’]
Hence, A³ is a symmetric matrix.

Question 5.
The maximum value of ∆ = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+\sin \theta & 1 \\
1+\cos \theta & 1 & 1
\end{array}\right|\), where θ is a real number is: [1]
(a) 1
(b) \(\frac{1}{2}\)
(c) 3
(d) -1
Solution:
(b) \(\frac{1}{2}\)
Explanation:
We have, Δ = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+\sin \theta & 1 \\
1+\cos \theta & 1 & 1
\end{array}\right|\)
= 1(1 + sin θ – 1) – 1(1 – 1 – cos θ) + (1 – (1 + sin θ) (1 + cos θ))
= sin θ + cos θ + [1 – 1 – sin θ – cos θ – sin θ cos θ]
= sin θ + cos θ – sin θ – cos θ – sin θ cos θ
= -sin θ cos θ
= \(-\frac{1}{2}\) sin 2θ
Max. Δ = \(-\frac{1}{2}\)(-1) = \(\frac{1}{2}\) [∵ 1 ≤ sin θ ≤ 1]

Question 6.
If \(\left[\begin{array}{cc}
2 x+y & 4 x \\
5 x-7 & 4 x
\end{array}\right]=\left[\begin{array}{cc}
7 & 7 y-13 \\
y & x+6
\end{array}\right]\) then the value of x + y is: [1]
(a) 7
(b) 4
(c) 5
(d) 2
Solution:
(c) 5
Explanation:
On comparing the elements of the matrix, we get
4x = x + 6
⇒ 3x = 6
⇒ x = 2
∴ 2x + y = 7
⇒ y = 7 – 2x
⇒ y = 7 – 2 x 2 = 3
∴ x + y = 2 + 3 = 5

Question 7.
The difference between the order and the degree of the given differential equation is: [1]
\(\sqrt{x+\left(\frac{d y}{d x}\right)^2}=a\left(\frac{d^2 y}{d x^2}\right)^{1 / 3}\)
(a) 1
(b) 2
(c) -1
(d) 0
Solution:
(d) 0
Explanation:
Given \(\sqrt{x+\left(\frac{d y}{d x}\right)^2}=a\left(\frac{d^2 y}{d x^2}\right)^{1 / 3}\)
⇒ \(\left[x+\left(\frac{d y}{d x}\right)^2\right]^3=a^6\left(\frac{d^2 y}{d x^2}\right)^2\)
∴ The order of the differential equation is 2 and its degree is 2.

Question 8.
The value of \(\int\left(e^{x \log a}+e^{a \log x}+e^{a \log a}\right) d x\) is: [1]
(a) \(\frac{a^x}{\log a}+\frac{x^{a+1}}{a-1}+a x^a+C\)
(b) \(\frac{a^x}{\log a}+\frac{x^a}{a+1}+a x^a+C\)
(c) \(\frac{a^x}{\log a}+\frac{x^{a+1}}{a+1}+a^a x+C\)
(d) \(\frac{a^x}{\log a}+\frac{x^{a+1}}{a+1}+a^a x+C\)
Solution:
(d) \(\frac{a^x}{\log a}+\frac{x^{a+1}}{a+1}+a^a x+C\)
Explanation:

Question 9.
The function f(x) = \(\frac{x}{\log x}\) increases in the interval: [1]
(a) (-e, e)
(b) (e, ∞)
(c) (-∞, ∞)
(d) (-e, ∞)
Solution:
(b) (e, ∞)
Explanation:
Given: f(x) = \(\frac{x}{\log x}\) is defined for x > 0 and x ≠ 1
Also, \(f(x)=\frac{\log x \cdot 1-x \cdot \frac{1}{x}}{(\log x)^2}=\frac{\log x-1}{(\log x)^2}\)
∴ f(x) > 0
⇒ log x > 1
⇒ x > e
∴ x ∈ (e, ∞)

Question 10.
The integrating factor of \(\frac{d y}{d x}+y=\frac{1+y}{x}\) is: [1]
(a) \(\frac{e^x}{x}\)
(b) \(\frac{e^{-x}}{x}\)
(c) xe^x
(d) x²e^x
Solution:
(a) \(\frac{e^x}{x}\)
Explanation:

Question 11.
If 3x + 2y = sin y, then \(\frac{d y}{d x}\) is: [1]
(a) \(\frac{3}{\cos y-2}\)
(b) \(\frac{\sin y-1}{2}\)
(c) \(\frac{2-\sin y}{3}\)
(d) \(\frac{2-\cos y}{3}\)
Solution:
(a) \(\frac{3}{\cos y-2}\)
Explanation:
Given 3x + 2y = sin y
On differentiating w.r.t. x, we get

Question 12.
If a differential equation corresponding to a function y = Ae^x + Be^-x, A and B being arbitrary constants is formed, then the order of the differential equation is: [1]
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b) 2
Explanation:
As there are two arbitrary constants.
Hence, a differential equation is of order 2.

Question 13.
The area (in sq. units) bounded by the shaded figure: [1]

(a) \(\frac{1}{4}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{9}{4}\)
(d) \(\frac{5}{4}\)
Solution:
(c) \(\frac{9}{4}\)
Explanation:

Question 14.
What is the angle between vectors \(\vec{a} \vec{b}\) if \(|\vec{a}|\) = 1, \(|\vec{b}|\) = 2 and \(\vec{a} \times \vec{b}=\hat{i}+\hat{j}+\hat{k}\)? [1]
(a) \(\frac{\pi}{2}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{2\pi}{2}\)
(d) \(\frac{\pi}{6}\)
Solution:
(b) \(\frac{\pi}{3}\)
Explanation:

Question 15.
The area of a parallelogram whose one diagonal and one side are represented by \(2 \hat{i}\) and \(-3 \hat{j}\) is: [1]
(a) 6 sq. units
(b) 36 sq. units
(c) 3 sq. units
(d) \(\frac{3}{2}\) sq. units
Solution:
(a) 6 sq. units
Explanation:
Area = \(|2 \hat{i} \times(-3 \hat{j})|\)
= \(6|-\hat{k}|\)
= 6 × 1
= 6 sq. units

Question 16.
The probabilities that A and B will die within a year are p and q respectively, the probability that only one of them will be alive at the end of the year is: [1]
(a) p²
(b) pq + q
(c) p + q – 2pq
(d) (p + q)²
Solution:
(c) p + q – 2pq
Explanation:
P(A) = p, p(B) = q
P(\(\bar{A}\)) = 1 – p, P(\(\bar{B}\)) = 1 – q
P(only one is alive) = p(1 – q) + (1 – p)q
= p – pq + q – pq
= p + q – 2pq

Question 17.
The value of \(\int_0^{\pi / 2} e^x(\sin x+\cos x) \cdot d x\) is: [1]
(a) e
(b) \(e^{\pi / 2}\)
(c) \(e^{\pi / 2-1}\)
(d) e²
Solution:
(b) \(e^{\pi / 2}\)
Explanation:
Given

Question 18.
The area (in sq. units) enclosed by the curve shown in the given figure is: [1]

(a) \(\frac{8}{3}\)
(b) \(\frac{24}{7}\)
(c) \(\frac{32}{3}\)
(d) \(\frac{16}{3}\)
Solution:
(c) \(\frac{32}{3}\)
Explanation:

Assertion-Reason Based Questions
In the following questions, a statement of assertion (A) is followed by a statement of the reason (R).
Choose the correct answer out of the following choices.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 19.
Assertion (A): \(\int \frac{1}{\sqrt{x^2+2 x+10}} \cdot d x\) = \(\sin ^{-1}\left(\frac{x+1}{3}\right)+c\)
Reason (R): If a > 0, b² – 4ac < 0, then \(\int \frac{d x}{\sqrt{a x^2+b x+c}}\) = \(\frac{1}{\sqrt{a}} \sin ^{-1}\left(\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right)+c\). [1]
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Explanation:

So, reason explains assertion.

Question 20.
Assertion (A): \(\int_{-3}^3\left(x^3+5\right) d x=30\)
Reason (R): f(x) = x³ + 5 is an odd function. [1]
Solution:
(c) A is true but R is false.
Explanation:
Let f(x) = x³ + 5
f(-x) = (-x)³ + 5 = -x³ + 5
So, f(x) is neither odd nor even.
Hence, R is false.
\(\int_{-3}^3 x^3 d x=0\) [∵ x³ is odd]
\(\int_{-3}^3 5 . d x=5[x]_{-3}^3\) = 30
∴ \(\int_{-3}^3\left(x^3+5\right) \cdot d x\) = 0 + 30 = 30

Section – B (10 Marks)
This section comprises very short answer type-questions (VSA) of 2 marks

Question 21.
Let S be the set of points in a plane and R be a relation in S defined as R = {(A, B) : d(A, B) < 2} where d(A, B) represents the distance between the points A and B. Is R an equivalence relation?
OR
Find the value of \(\cot \left(\frac{\pi}{4}-2 \cot ^{-1} 3\right)\). [2]
Solution:
Relation R = {(A, B): d(A, B) < 2}
For reflexive: For every A ∈ S, d(A, A) = 0 < 2
Hence, (A, A) ∈ R, ∀ A ∈ S.
Hence, reflexive.
For symmetric: Let (A, B) ∈ R
⇒ d(A, B) < 2
⇒ d(B, A) < 2
⇒ (B, A) ∈ R, for all A, B ∈ S.
Hence, symmetric.
For transitive: Let (A, B) ∈ R and (B, C) ∈ R
Let d(A, B) = 1.5 and d(B, C) = 1.7 and A, B, C are collinear
Let d(A, C) = 1.5 + 1.7 = 3.2 > 2
Hence (A, B) ∈ R, (B, C) ∈ R
⇒ (A, C) ∈ R
Hence, not transitive
∴ Relation R is not an equivalence relation.
OR

= \(\frac{\frac{4}{3}+1}{\frac{4}{3}-1}=7\)

Question 22.
For the differential equation, find a particular solution satisfying the given condition (1 + sin²x) dy + (1 + y²) cos x dx = 0, given that when x = \(\frac{\pi}{2}\), y = 0 [2]
Solution:

Question 23.
For what of k is the function defined by f(x) = \(\left\{\begin{array}{cc}
\frac{\sin x+x \cos x}{x}, & \text { if } x \neq 0 \\
k, & \text { if } x=0
\end{array}\right\}\) continuous at x = 0?
OR
Evaluate: \(\int \frac{d x}{\sin ^2 x-5 \sin x \cos x}\). [2]
Solution:

Question 24.
If \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\), then show that \(\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}\). [2]
Solution:

Question 25.
If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are three mutually perpendicular vectors of equal magnitude, then show that \(\vec{a}+\vec{b}+\vec{c}\) is equally inclined to the vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\). [2]
Solution:

Section – C (18 Marks)
This section comprises short answer type questions (SA) of 3 marks

Question 26.
If y = \(\sin ^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)\) then show that \(\frac{d y}{d x}=\frac{-1}{2 \sqrt{1-x^2}}\). [3]
Solution:

Question 27.
A coin is biased so that the head is three times as likely to occur as the tail of the coin is tossed twice, find the probability distribution of a number of tails. Hence, find the mean of the number of tails.
OR
In a class, 5% of boys and 10% of girls have an IQ of more than 150. In the class 60% are boys and the rest are girls if a student is selected randomly and found to have an IQ of more than 150 then find the probability that the student is a boy. [3]
Solution:
Let X be the random variable, which denotes the no. of tails when a biased coin is tossed twice.
So, X may have values 0, 1, or 2.
Since the coin is biased which head is 3 times as likely to occur as the tail.
∴ P(H) = \(\frac{3}{4}\) and P(T) = \(\frac{1}{4}\)
P(X=0) = P(H, H)
= P(H) . P(H)
= \(\left(\frac{3}{4}\right)^2=\frac{9}{16}\)
P(X=1) = P(One tail and one head)
= P(HT, TH)
= P(HT) + P(TH)
= P(H) . P(T) + P(T) . P(H)
= \(\frac{3}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{3}{4}\)
= \(\frac{3}{16}+\frac{3}{16}=\frac{6}{16}=\frac{3}{8}\)
P(X=2) = P(two tails)
= P(TT)
= P(T) . P(T)
= \(\left(\frac{1}{4}\right)^2=\frac{1}{16}\)
The required probability distribution is as follows:

OR
Let E₁: Boy is selected
E₂: A girl is selected
A: Student has an IQ of more than 150 students
Now P(E₁) = 60% = \(\frac{60}{100}\)
P(E₂) = 40% = \(\frac{40}{100}\)
Now P(A/E₁) = Probability that boy has an IQ of more than 150 = 5% = \(\frac{5}{100}\)
and P(A/E₂) = Probability than girl has an IQ of more than 150 = 10% = \(\frac{10}{100}\)

Question 28.
Show that the relation S in the set R of real numbers, defined as S = {(a, b): a, b ∈ R and a ≤ b³} is neither reflexive nor symmetric nor transitive. [3]
Solution:
Given: S = {(a, b) ∈ R × R/a ≤ b³}
For reflexive: Let (-2, -2) ∈ S
⇒ -2 ≤ (-2)³
⇒ -2 ≤ -8, false.
Hence, not reflexive.
For symmetric: Let (-2, 2) ∈ S
⇒ -2 ≤ 2³
⇒ -2 ≤ 8, is true.
If symmetric then (2, -2) ∈ S
⇒ 2 ≤ (-2)³
⇒ 2 ≤ -8, false.
Hence, not symmetric
For transitive: Let (25, 3) ∈ S and (3, 2) ∈ S
⇒ 25 ≤ 3³ and 3 ≤ 2³
⇒ 25 ≤ 27 and 3 ≤ 8, true in both cases.
If transitive that (25, 2) ∈ S
⇒ 25 ≤ 2³
⇒ 25 ≤ 8, false.
Hence, not transitive.

Question 29.
If x¹⁶y⁹ = (x² + y)¹⁷, then prove that \(\frac{d y}{d x}=\frac{2 y}{x}\)
OR
If x = a(2θ – sin 2θ) and y = a(1 – cos 2θ), find \(\frac{d y}{d x}\) when θ = \(\frac{\pi}{3}\). [3]
Solution:
Consider x¹⁶y⁹ = (x² + y)¹⁷
Take log on both sides
16 log x + 9 log y = 17 log(x² + y)
Now differentiate both sides, w.r.t. x

Question 30.
If A = \(\left[\begin{array}{rr}
1 & -1 \\
2 & -1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
a & 1 \\
b & -1
\end{array}\right]\) and (A + B)² = A² + B², than find the values of a and b.
OR
Find x, y, z, w if \(\left[\begin{array}{cc}
x+y & x-y \\
y+z+w & 2 w-z
\end{array}\right]=\left[\begin{array}{cc}
2 & -1 \\
9 & 5
\end{array}\right]\). [3]
Solution:

Question 31.
Dot product of a vector with vectors \(\hat{i}-\hat{j}+\hat{k}\), \(2 \hat{i}+\hat{j}-3 \hat{k}\) and \(\hat{i}+\hat{j}+\hat{k}\) are respectively 4, 0 and 2. Find the vector. [3]
Solution:
Let the required vector is

⇒ a₁ – a₂ + a₃ = 4 ……(i)
Similarly 2a₁ + a₂ – 3a₃ = 0 …….(ii)
a₁ + a₂ + a₃ = 2 ……(iii)
Subtracting eq. (iii) from (i), we get
-2a₂ = 2
⇒ a₂ = -1
Put the value of a₂ in eq. (ii) & (iii)
2a₁ – 3a₃ = 1 …….(iv)
a₁ + a₃ = 3 …..(v)
On solving (iv) & (v), we get
a₁ = 2 and a₃ = 1
∴ Required vector is \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k}\)

Section – D (20 Marks)
This section comprises long answer-type questions (LA) of 5 marks

Question 32.
Solve the following problem graphically.
Minimise and maximise, Z = 3x + 9y
Subject to the constraints:
x + 3y ≤ 60
x + y ≥ 0
x ≤ y
x ≥ 0, y ≥ 0
OR
The feasible region of the system of linear constraints is given as:

Answer each of the following:
(A) Find the constraints for the LPP taking the x-along x-axis and the y-along y-axis.
(B) If Z = 450x + 300y is the objective function, then find the maximum value of x and the point of maximum. [5]
Solution:
Plotting the inequations
x + 3y ≤ 60, x + y ≥ 10, x ≤ y, x ≥ 0, y ≥ 0

The common shaded portion is the feasible solution.
Possible points for maximum and minimum z are A(5, 5), B(15, 5), C(0, 20) and D(0, 10)

Min. z is at A(5, 5) i.e. x = 5, y = 5, Min. z = 60.
Max. z is at B(15, 15) i.e. x = 15, y = 15
and (0, 20) i.e. x = 0, y = 20
Max. z = 180
OR
(A) BC passes through (0, 200) and (200, 0)
∴ Equation of BC is \(\frac{x}{200}+\frac{y}{200}=1\)
⇒ x + y = 200
The equation of AC is x = 20, as AC || y axis and x-coordinates of A is 20
Equation of AB is y = 9x as line passes through B(0, 0) and A(20, 80)
Then constraints are
x ≥ 20, y ≥, 0 x + y ≤ 200, y ≥ 4x
(B) The corner points are A(20, 80), B(40, 160) and C(20, 180)

Hence, the Max. value is 66000 at point B(40, 160)

Question 33.
Find the shortest distance between the following two lines.
\(\vec{r}=(1+\lambda) \hat{i}+(2-\lambda) \hat{j}+(\lambda+1) \hat{k}\)
\(\vec{r}=(2 \hat{i}-\hat{j}-\hat{k})+\mu(2 \hat{i}+\hat{j}+2 \hat{k})\) [5]
Solution:
Consider the lines

Question 34.
A metal box with a square base and vertical sides is to contain 1024 cm³. The material for the top and bottom costs ₹ 5/cm² and the material for the sides cost ₹ 2.50/cm². Then, find the least cost of the box.
OR
AB is a diameter of a circle and C is any point on the circle. Show that the area of ∆ABC is maximum when it is isosceles. [5]
Solution:

Question 35.
(A) Evaluate: \(\int_0^{2 \pi} \frac{x \sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} \cdot d x\)
(B) For x ≥ 0, let f(x) = \(\int_1^x \frac{\log _e t}{1+t} d t\). Find the function f(x) + f(\(\frac{1}{x}\)) and show that f(e) + f(\(\frac{1}{e}\)) = \(\frac{1}{2}\). [5]
Solution:

Section – E (12 Marks)

This section comprises of 3 case-study/passage-based questions of 4 marks each with two sub-part. First, two case-study questions have three sub-parts (A), (B), (C) of marks 1, 1, 2 respectively. The third case-study question has two subparts of 2 marks each.

Question 36.
Reshma was doing a project related to the average number of hours spent on a study by students selected at random. At the end of the survey, she prepared a report related to the data.

Let X denotes the average no. of hours spent on a study by students. The probability that X can take the values x, has the following form, where k is some unknown constant.
\(\mathrm{P}(\mathrm{X}=x)=\left\{\begin{array}{cl}
k & \text { if } x=0 \\
2 k, & \text { if } x=1 \\
3 k, & \text { if } x=2 \\
0, & \text { otherwise }
\end{array}\right.\)
Based on the above information, answer the following questions:
(A) What is the value of k? [1]
(B) What is the value P(X=2)? [1]
(C) What is the probability that the average study time of students is atleast 1 hour?
OR
Find the mean of the given data. [2]
Solution:
(A) Since, ΣP(x) = 1
∴ k + 2k + 3k + 0 = 1
⇒ 6k = 1
⇒ k = \(\frac{1}{6}\)
(B) From (i), k = \(\frac{1}{6}\)
∴ P(X=2) = 3k
= 3 × \(\frac{1}{6}\)
= \(\frac{1}{2}\)
(C) If a student has study time of atleast 1 hour, then either he/she had studied for 1 hour or 2 hours
∴ Required probability = P(X=1) + P(X=2)
= 2k + 3k
= 5k
= 5 × \(\frac{1}{6}\)
= \(\frac{5}{6}\)
Hence, required probability is \(\frac{5}{6}\)
OR
The probability distribution for the given data is:

Question 37.
Mohini purchased a rectangular parallelopiped shaped box and a spherical ball inside it as a showpiece. The sides of the box are x, 2x and \(\frac{x}{3}\), and the radius of the sphere is y.

The sum of the surface area of the parallelopiped and sphere is given to be constant.
Based on the above information, answer the following questions:
(A) Let the constant surface area given to be S, then what is the relation between x and y? [1]
(B) If the combined volume is denoted by V, then what is the value of V? [1]
(C) If volume V is minimum, then how are x and y is related to each other?
OR
If the shape has minimum volume when x = 2y, then what is the difference in the volume and surface area of the shape? [2]
Solution:

Question 38.
Minor of an element aij of a determinant is obtained by deleting its ith row and jth column in which element a_ij lies and is denoted by M_ij.
The cofactor of element a_ij denoted by A_ij is defined by A_ij = (-1)^i+j M_j, where M_ij is a minor of a_ij.
Based on the above information, answer the following questions:
(A) In the determinant \(\left|\begin{array}{ccc}
2 & -3 & 5 \\
4 & 0 & 6 \\
-7 & 5 & 1
\end{array}\right|\), what is the value of a₃₂.A₃₂. Find the sum of the cofactors of all the elements of \(\left|\begin{array}{cc}
-2 & 3 \\
1 & 4
\end{array}\right|\). [2]
(B) If A = \(\left|\begin{array}{ccc}
3 & 0 & -1 \\
2 & 3 & 0 \\
0 & 4 & 1
\end{array}\right|\), then find |adj(adj A)|. Show that all the positive integral power of a symmetric matrix is symmetric. [2]
Solution: