Ch 1Applications of Matrices and Determinants
5-Mark Questions
For the given $3\times3$ adjoint matrix in the textbook, compute $|\operatorname{adj}(A)|$.
Answer: The preserved working gives $|\operatorname{adj}(A)|=2(24-0)+4(-6-14)+2(0+24)=48-80+48=16$. The source matrix itself is missing from the text extract, so this card is not marked validated. <div
For the given $3\times3$ adjoint matrix in the textbook, compute $|\operatorname{adj} A|$.
Answer: The preserved working gives $|\operatorname{adj} A|=0+2(36-18)+0=36$. The source matrix itself is missing from the text extract, so this card is not marked validated. <div
2-Mark Questions
Find the adjoint of the matrices: (i) $-\begin{bmatrix}3&4\\6&2\end{bmatrix}$ (ii) $\begin{bmatrix}2&3&1\\3&4&1\\3&7&2\end{bmatrix}$. The third matrix in this textbook item needs source-text restoration before validation.
Answer: (i) $\operatorname{adj}(A)=\begin{bmatrix}-2&4\\6&-3\end{bmatrix}$. (ii) $\operatorname{adj}(A)=\begin{bmatrix}1&1&-1\\-3&1&1\\9&-5&-1\end{bmatrix}$. The missing third matrix is intentionally left unverified here; no answer is marked validated until the exact textbook entries are restored. <div
Find the inverse (if it exists) of the following: (i) −[[2 4][1 3]] (ii) [[5 1 1][1 5 1][1 1 5]] (iii) [[2 3 1][3 4 1][3 7 2]]
(i) A^{-1} = [[-3/2, 2], [1/2, -1]]. (ii) A^{-1} = [[3/14, -1/28, -1/28], [-1/28, 3/14, -1/28], [-1/28, -1/28, 3/14]]. (iii) Using det(A)=2 and adj(A) from earlier, A^{-1} = (1/2) · [[1,1,-1],[-3,1,1],[9,-5,-1]] = [[1/2,1/2,-1/2],[-3/2,1/2,1/2],[9/2,-5/2,-1/2]].
If F(α) = [[cosα, -sinα, 0], [sinα, cosα, 0], [0, 0, 1]], show that F(α)F(-α) = I.
F(α)F(-α) = I. Reason: Multiply the matrices using cos(-α)=cosα and sin(-α)=-sinα. The (1,1) entry becomes cos^2α+sin^2α=1, off-diagonal terms cancel, and the third row/column give the identity entry for the 3rd coordinate. Hence the product is the identity matrix.
1-Mark Questions (MCQ)
By using Gaussian elimination method, balance the chemical reaction equation: C6H12O6 + O2 → CO2 + H2O.
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O
Ch 2Complex Numbers
5-Mark Questions
Simplify: (1) $i^{1947}+i^{1950}$ (2) $i^{1948}-i^{-1869}$ (3) $\sum_{n=1}^{12}i^n$.
(1) $-1-i$ (2) $1+i$ (3) $0$.
Evaluate the following if $z=5-2i$ and $w=-1+3i$: (i) $z+w$ (ii) $z-iw$ (iii) $2z+3w$ (iv) $zw$ (v) $z^2+2zw+w^2$ (vi) $(z+w)^2$.
(i) $4+i$ (ii) $8-i$ (iii) $7+5i$ (iv) $1+17i$ (v) $15+8i$ (vi) $15+8i$.
2-Mark Questions
Simplify: (4) $i^{59}+\dfrac{1}{i^{59}}$ (5) $i\,i^2i^3\cdots i^{2000}$ (6) $\sum_{n=1}^{10}i^{n+50}$.
(4) $0$ (5) $1$ (6) $1-i$.
Find the real numbers $x$ and $y$ if $(3-i)x-(2-i)y+2i+5$ and $2x+(-1+2i)y+3+2i$ are equal.
$x=-1$, $y=1$.
If $z_1=1-3i$, $z_2=-4i$, and $z_3=5$, show that (i) $(z_1+z_2)+z_3=z_1+(z_2+z_3)$ (ii) $(z_1z_2)z_3=z_1(z_2z_3)$.
Both identities are verified: the common values are (i) $6-7i$ and (ii) $-60-20i$.
1-Mark Questions (MCQ)
If $z=2-2i$, find its rotation through $\theta$ radians counterclockwise about the origin when (i) $\theta=\pi/3$ (ii) $\theta=2\pi/3$ (iii) $\theta=3\pi/2$.
(i) $(1+\sqrt3)+i(\sqrt3-1)$ (ii) $(\sqrt3-1)+i(\sqrt3+1)$ (iii) $-2-2i$.
Ch 3Theory of Equations
5-Mark Questions
If α, β and γ are the roots of the cubic equation x³ + 2x² + 3x + 4 = 0, form a cubic equation whose roots are (i) 2α, 2β, 2γ, (ii) $\frac{1}{α}$, $\frac{1}{β}$, $\frac{1}{γ}$ (iii) – α, – β, – γ
⇒ x 3 + x 2 (α + β + γ) + x (αβ + βγ + γα) + (αβγ) = 0 ⇒ x 3 + x 2 (-2) + x (3) – 4 = 0 ⇒ x 3 – 2x 2 + 3x – 4 = 0
Solve the equation 3x^3-16x^2+23x-6=0 if the product of two roots is 1.
Roots: 3, 2, 1/3
2-Mark Questions
If the sides of a cubic box are increased by 1, 2, 3 units respectively to form a cuboid, then the volume is increased by 52 cubic units. Find the volume of the cuboid.
60
Construct a cubic equation with roots (i) 1, 2, and 3 (ii) 1, 1, and −2 (iii) 2, 1/2, and 1.
(i) x^3-6x^2+11x-6=0 (ii) x^3-3x+2=0 (iii) 2x^3-7x^2+7x-2=0
Find the sum of squares of roots of the equation 2x 4 – 8x³ + 6x² – 3 = 0.
= (4) 2 – 2(3) = 16 – 6 = 10
1-Mark Questions (MCQ)
Find the exact number of real roots and imaginary of the equation x 9 + 9x 7 + 7x 5 + 5x³ + 3x.
P(x) = x 9 + 9x 7 + 7x 5 + 5x 3 + 3x. There is no change in the sign of P(x) and P(-x), P(x) has no positive and no negative real roots, but 0 is the root of the polynomial equation P(x).
Ch 4Inverse Trigonometric Functions
5-Mark Questions
Find all values of x such that (i) −10π ≤ x ≤ 10π and sin x = 0. (ii) −3π ≤ x ≤ 3π and sin x = −1.
(i) x = kπ, k = −10, −9, …, 0, …, 9, 10. (ii) x = 3π/2 + 2kπ; within the interval these are x = −5π/2, −π/2, 3π/2.
Find the domain of (i) f(x) = sin^{-1}(x+1/2) (ii) g(x) = sin^{-1}(2x−1/4).
(i) −3/2 ≤ x ≤ 1/2. (ii) (1/4 −1)/2 ≤ x ≤ (1/4 +1)/2 i.e. −3/4 ≤ x ≤ 5/4.
2-Mark Questions
Find the period and amplitude of (i) y = sin 7x (ii) y = − sin(x/3) (iii) y = −4 sin 2x.
(i) Amplitude 1, period 2π/7. (ii) Amplitude 1, period 6π. (iii) Amplitude 4, period π.
Sketch the graph of y = sin(x/3) for 0 ≤ x < 6π. (Describe key points.)
One full sine wave from x=0 to x=6π: zeros at x=0, 3π, 6π; maximum y=1 at x=3π/2; minimum y=−1 at x=9π/2.
Evaluate (i) sin(sin^{-1}(1/2)) (ii) sin^{-1}(sin(π/5)).
(i) 1/2. (ii) π/5.
1-Mark Questions (MCQ)
Find the number of the solutions of the equations tan -1 (x – 1) + tan -1 x + tan -1 (x + 1) = tan -1 3x
x = 0, x² = 1 x = ±1 Number of solutions are three (0, 1 -1)
Ch 5Two Dimensional Analytical Geometry-II
5-Mark Questions
Find the equation of circles that touch both the axes and pass through (-4,2) in general form.
Two circles: (x+10)^2+(y-10)^2=100 and (x+2)^2+(y-2)^2=4. In general form: x^2+y^2+20x-20y=0 and x^2+y^2+4x-4y=0.
If the equation 3x² + (3 – p) xy + qy² – 2px = 8pq represents a circle, find p and q. Also determine the centre and radius of the circle.
Centre (-g, -f) = (1, 0) Radius = $\sqrt{g^2+f^2-c}$ = $\sqrt{1+0+24}$ = $\sqrt{25}$ = 5
2-Mark Questions
Obtain the equation of the circles with radius 5 cm and touching x-axis at the origin in general form.
Two circles: x^2+(y-5)^2=25 and x^2+(y+5)^2=25. In general form: x^2+y^2-10y=0 and x^2+y^2+10y=0.
Find the equation of the circle with centre (2,1) and passing through the point (3,6) in standard form.
(x-2)^2+(y-1)^2=26.
Find the equation of the circle with centre (2,3) and passing through the intersection of the lines 3x-2y-1=0 and 4x+2y-7=0. (Assumed second line is 4x+2y-7=0.)
(x-2)^2+(y-3)^2 = 769/196.
1-Mark Questions (MCQ)
At a water fountain, water attains a maximum height of 4 m at horizontal distance of 0.5 m from its origin. If the path of water is a parabola, find the height of water at a horizontal distance of 0.75 m from the point of origin.
3 m.
Ch 6Applications of Vector Algebra
5-Mark Questions
Prove by vector method that if a line is drawn from the centre of a circle to the midpoint of a chord, then the line is perpendicular to the chord.
The line from centre to midpoint is perpendicular to the chord.
Prove by vector method that the median to the base of an isosceles triangle is perpendicular to the base.
The median from apex to base is perpendicular to base.
2-Mark Questions
Prove by vector method that an angle in a semi-circle is a right angle.
Angle subtended by a semicircle is 90°.
Using vector method, prove that if the diagonals of a parallelogram are equal, then it is a rectangle.
If diagonals equal in length then adjacent sides are perpendicular, so parallelogram is rectangle.
If G is the centroid of a triangle ABC, prove that area(GAB)=area(GBC)=area(GCA)=1/3 area(ABC).
Each triangle formed with centroid has area one-third of ABC.
1-Mark Questions (MCQ)
If $\vec a\cdot\vec b=\vec b\cdot\vec c=\vec c\cdot\vec a=0$, then the value of $[\vec a\ \vec b\ \vec c]$ is (A) $|\vec a||\vec b||\vec c|$ (B) $\tfrac13|\vec a||\vec b||\vec c|$ (C) $1$ (D) $-1$
$|\vec a||\vec b||\vec c|$
Ch 7Applications of Differential Calculus
5-Mark Questions
A particle moves according to the law s(t)=−t^3+4t^2−9t+2, t≥0. (i) When does the particle change direction? (ii) Find the total distance travelled in the first 4 seconds. (iii) Find the particle's acceleration each time the velocity is zero.
(i) Never (no real zeros of v). (ii) 36 units. (iii) Velocity never zero ⇒ no such times (acceleration a(t)=−6t+8 would apply if velocity were zero).
A ladder 17 m long is leaning against the wall. The base of the ladder is pulled away from the wall at 5 m/s. When the base is 8 m from the wall, (i) how fast is the top moving down? (ii) at what rate is the area of the triangle formed by ladder, wall and floor changing? Also: A police jeep approaching an intersection from the north is chasing a car moving east. When the jeep is 0.6 km north and the car 0.8 km east of the intersection the distance between them is increasing at 20 km/h. If the jeep's speed is 60 km/h at that instant, what is the speed of the car?
(Ladder) (i) dy/dt=−8/3 m/s. (ii) dA/dt=161/6 m^2/s. (Police) speed of car = 70 km/h.
2-Mark Questions
A particle moves along a straight line in such a way that after t seconds its distance from the origin is s(t)=t^3+2t^2 metres. (i) Find the average velocity between t=3 and t=6 seconds. (ii) Find the instantaneous velocities at t=3 and t=6 seconds.
(i) 81 m/s. (ii) v(3)=39 m/s, v(6)=132 m/s.
A camera is accidentally knocked off an edge of a cliff 400 ft high. The camera falls a distance s(t)=16t^2 in t seconds. (i) How long does the camera fall before it hits the ground? (ii) What is the average velocity with which the camera falls during the last 2 seconds? (iii) What is the instantaneous velocity of the camera when it hits the ground?
(i) 5 s. (ii) 128 ft/s. (iii) 160 ft/s.
If the volume of a cube of side length x is v = x³. Find the rate of change of the volume with respect to x when x = 5 units.
volume of a cube v = x³ Rate of change $\frac { dv }{ dx }$ = 3x² When x = 5 units, $\frac { dv }{ dx }$ = 3(5)² = 3(25) = 75 units.
1-Mark Questions (MCQ)
Sketch the graphs of the following functions (i) y = –$\frac { 1 }{ 3 }$ (x³ – 3x + 2) (ii) y = x $\sqrt { 4-x }$ (iii) y = $\frac { x^2+1 }{ x^2-4 }$ (iv) y = $\frac { 1 }{ 1+e^{-x} }$ (v) y = $\frac { x^3 }{ 24 }$ – log x
No point of inflection exists. No Horizontal asymptotes are possible, but the vertical asymptote is x = 0 (y-axis). ESCORTS Pivot Calculator
Ch 8Differentials and Partial Derivatives
5-Mark Questions
Find Δf and df for the function f for the indicated values of x, Δx and compare: (i) f(x) = x³ – 2x², x = 2, Δx = dx = 0.5 (ii) f(x) = x² + 2x + 3, x = -0.5, Δx = dx = 0.1
f(x) = f(-0.5) = (-0.5) 2 + 2(-0.5) + 3 = 0.25 – 1 + 3 = 3.25 – 1 = 2.25 So ∆ f = f(x + ∆x) – f(x) = 2.36 – 2.25 = 0.11
Evaluate $\lim _{(x, y) \rightarrow(1,2)}$ g(x, y), if the limit exists where g(x, y) = $\frac { 3x^2-xy }{ x^2+y^2+3 }$
$\lim _{(x, y) \rightarrow(1,2)}$ g(x, y) = $\lim _{(x, y) \rightarrow(1,2)}$ $\frac { 3x^2-xy }{ x^2+y^2+3 }$ = $\frac { 3(1)-1×2 }{ 1+4+3 }$ = $\frac { 1 }{ 8 }$
2-Mark Questions
Let f(x) = $\sqrt[3] { x }$. Find the linear approximation at x = 27. Use the linear approximation to approximate $\sqrt[3] { 27.2 }$
We know that f(x 0 + Δx) = f(x 0 ) + f'(x 0 ) Δx ∴ Approximate value of $\sqrt[3] { 27.2 }$ = 3.0074
Use the linear approximation to find approximate values of (i) 123^{1/3} (ii) 15^{1/4} (iii) 26^{1/3}.
(i) 4.973333... (ii) 1.96875 (iii) 2.962963...
Find the linear approximation at x = 27. Use the linear approximation to approximate 27^{2/3}. (interpreted as f(x)=x^{2/3})
Linearization L(x)=9 + (2/9)(x-27). Hence 27^{2/3}=9 (exact). For nearby x (e.g. 26) L(26)=9-2/9=8.777... .
1-Mark Questions (MCQ)
If v (x, y) = log($\frac{x^2+y^2}{x+y}$) Prove that x $\frac{\partial v}{\partial x}$ + y$\frac{\partial u}{\partial y}$ = 1.
Hence Proved
Ch 9Applications of Integration
5-Mark Questions
Explain the right-end and mid-point Riemann rules and remarks about geometric interpretation of the Riemann integral.
Right-end rule: choose sample points ξ_i = x_i (right endpoints); then ∫_a^b f(x) dx = lim_{n→∞} Σ_{i=1}^n f(x_i)Δx_i when limit exists. Mid-point rule: choose ξ_i = (x_{i-1}+x_i)/2; then ∫_a^b f(x) dx = lim_{n→∞} Σ f(midpoint)Δx_i. Remarks: (1) F(x)=∫_a^x f(u) du defines an antiderivative. (2) If f≥0 on [a,b], the integral equals the geometric area under the curve. (3) If f≤0 the integral equals negative of the geometric area. (4) If f changes sign, split [a,b] into subintervals where f has constant sign and add/subtract areas accordingly.
State Bernoulli's formula for integration by parts repeated application.
Bernoulli's formula: ∫ u dv = uv − ∫ v du; applying repeatedly gives ∫ u dv = uv − u'v_1 + u''v_2 − ... alternating signs, where v_k are successive antiderivatives of v.
2-Mark Questions
Find an approximate value of ∫_1^5 x dx by applying the left-end rule with partition {1,2,3,4,5}.
10
Find an approximate value of ∫_1^5 x dx by applying the right-end rule with partition {1,2,3,4,5}.
14
Find an approximate value of ∫_1^5 x dx by applying the mid-point rule with partition {1,2,3,4,5}.
12
1-Mark Questions (MCQ)
The value of $int_{-4}^{4}$ $left[ an^{-1} rac { x^2 }{ x^4+1 }+ an^{-1} rac { x^4+1 }{ x^2 } ight]$ dx is (A) $pi$ (B) $2pi$ (C) $3pi$ (D) $4pi$
$4pi$
Ch 10Ordinary Differential Equations
5-Mark Questions
For each of the following differential equations, determine its order and degree (if exists): (i) dy/dx + x y = cot x. (ii) (d^3 y/dx^3)^2 − (d^2 y/dx^2)^3 + 5(dy/dx)^4 = 0. (iii) (contains sin of derivatives — non‑polynomial). (iv) (highest derivative order 7, appears linearly). (v) (highest derivative order 3, appears linearly). (vi) (equation linear in highest derivative of order 2). (vii) (highest derivative order 2 appears linearly). (viii) (contains cos of derivative — non‑polynomial). (ix) (contains an integral — non‑algebraic in derivatives). (x) x e^{x} y (dy/dx) = ... (derivative appears polynomially).
(i) order = 1, degree = 1. (ii) order = 3, degree = 2. (iii) order = 2, degree does not exist (non‑polynomial because of sin). (iv) order = 7, degree = 1 (highest derivative assumed 7 and appears linearly). (v) order = 3, degree = 1. (vi) order = 2, degree = 1. (vii) order = 2, degree = 1. (viii) order = 2, degree does not exist (contains cos — non‑polynomial in derivatives). (ix) order = 1, degree does not exist (contains an integral — not algebraic in derivatives). (x) order = 1, degree = 1 (derivative appears to first power).
Assume a spherical raindrop evaporates at a rate proportional to its surface area. Form a differential equation for the rate of change of its radius r(t).
Let V=(4/3)π r^3 and surface area S=4π r^2. dV/dt = -k S = -4π k r^2 ⇒ 4π r^2 dr/dt = -4π k r^2 ⇒ dr/dt = -k.
2-Mark Questions
Express each physical statement as a differential equation: (i) Radium decays at a rate proportional to the amount Q present. (ii) Population P increases at a rate proportional to P(500000−P). (iii) For a substance, rate of change of vapour pressure P wrt temperature T is proportional to P and inversely proportional to T^2. (iv) A sum grows at 8% per year (continuous compounding) and receives continuously an additional ₹400 per year.
(i) \(\dfrac{dQ}{dt}=-kQ\), \(k>0\). (ii) \(\dfrac{dP}{dt}=kP(500000-P)\). (iii) \(\dfrac{dP}{dT}=k\dfrac{P}{T^2}\). (iv) If A(t) is amount, \(\dfrac{dA}{dt}=0.08A+400\).
Form the differential equation of all straight lines touching the circle x^2+y^2=r^2.
For line y=mx+c to be tangent to circle: c^2=r^2(1+m^2). Using m=y' and c=y-xy' gives differential equation: (y-xy')^2 = r^2(1+(y')^2).
Find the differential equation of the family of circles passing through the origin and having their centres on the x‑axis.
Differential equation: y^2 - x^2 = 2 x y y' .
1-Mark Questions (MCQ)
The degree of the differential equation $y=x\left(1+\dfrac{dy}{dx}+\dfrac1{2!}\left(\dfrac{dy}{dx}\right)^2+\dfrac1{3!}\left(\dfrac{dy}{dx}\right)^3+\cdots\right)$ is (A) $2$ (B) $3$ (C) $1$ (D) $4$
$1$
Ch 11Probability Distributions
5-Mark Questions
A six sided die is marked '1' on one face, '3' on two of its faces, and '5' on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find (i) the probability mass function (ii) the cumulative distribution function (iii) P(0 ≤ X < 4) (iv) P(X ≥ 6). (Interpretation: faces values are 1,3,5 with multiplicities 1,2,3.)
(i) Possible sums 2,4,6,8,10 with probabilities 1/36, 4/36, 10/36, 12/36, 9/36 respectively. (ii) CDF: F(x)=0 (x<2); =1/36 for 2≤x<4; =5/36 for 4≤x<6; =15/36 for 6≤x<8; =27/36 for 8≤x<10; =1 for x≥10. (iii) P(0≤X<4)=P(X=2)=1/36. (iv) P(X≥6)=(10+12+9)/36=31/36.
The probability density function of X is given by f(x)=k x^3 e^{-x} for x>0 (0 otherwise). Find (i) k (ii) the distribution function (iii) P(X<3) (iv) P(X≤5) (v) P(X≤4).
(i) k=1/6. (ii) F(x)=1−e^{-x}(1+x+x^2/2+x^3/6) for x>0. (iii) P(X<3)=1−13e^{-3}≈0.3528. (iv) P(X≤5)=1−39.3333 e^{-5}≈0.7348. (v) P(X≤4)=1−23.6667 e^{-4}≈0.5665.
2-Mark Questions
Suppose X is the number of tails occurred when three fair coins are tossed once simultaneously. Find the values of the random variable X and number of points in its inverse images.
X ∈ {0,1,2,3}. |X^{-1}(0)|=1, |X^{-1}(1)|=3, |X^{-1}(2)|=3, |X^{-1}(3)|=1.
An urn contains 5 mangoes and 4 apples. Three fruits are taken at random. If the number of apples taken is a random variable, then find the values of the random variable and number of points in its inverse images.
X ∈ {0,1,2,3}. Counts: 0→10, 1→40, 2→30, 3→4 (total 84).
Two balls are chosen randomly from an urn containing 6 red and 8 black balls. Suppose that we win `15 for each red ball selected and we lose `10 for each black ball selected. If X denotes the winning amount, find the values of X and number of points in its inverse images.
Possible X: −20, 5, 30. Counts: X=30 (2 red):15 outcomes; X=5 (1 red,1 black):48 outcomes; X=−20 (0 red):28 outcomes (total 91).
1-Mark Questions (MCQ)
For a binomial distribution with $n=5$, $P(X=1)=0.4096$ and $P(X=2)=0.2048$. Find $p$, the mean and the variance.
$p=\dfrac15,\ q=\dfrac45$; mean $=1$, variance $=\dfrac45$.
Ch 12Discrete Mathematics
5-Mark Questions
Determine whether ∗ is a binary operation on the sets given below. (i) a ∗ b = a/b on ℝ. (ii) a ∗ b = min(a,b) on A = {1,2,3,4,5}. (iii) a ∗ b = a + b on ℝ.
(i) Not a binary operation on ℝ. (ii) Binary operation on A. (iii) Binary operation on ℝ.
Let A = {1,2,3,4,5}. Check whether the usual multiplication is a binary operation on A.
Yes, multiplication is not closed on A in general; (counterexample) so it is not a binary operation on A.
2-Mark Questions
On ℤ (or the intended integer set), define ∗ by m ∗ n = m + n for all m,n. Is ∗ binary on ℤ ?
Yes. ∗ is a binary operation on ℤ; it is closed, commutative and associative.
Let ∗ be defined on ℝ by a ∗ b = a + b + ab − 7. Is ∗ binary on ℝ ? If so, find 3 ∗ (−7).
Yes; 3 ∗ (−7) = −32.
Write sentences in symbolic form using p = '19 is a prime number' and q = 'all the angles of a triangle are equal'. (i) 19 is not a prime number and all the angles are equal. (ii) 19 is a prime or all the angles are not equal. (iii) 19 is a prime and all the angles are equal. (iv) 19 is not a prime number.
(i) ¬p ∧ q. (ii) p ∨ ¬q. (iii) p ∧ q. (iv) ¬p.
1-Mark Questions (MCQ)
Determine truth values: (i) If 6+2=5 then the milk is white. (ii) China is in Europe or 3 is an integer. (iii) It is not true that 5+5=9 or Earth is a planet. (iv) 11 is a prime and all sides of a rectangle are equal.
(i) True. (ii) True. (iii) True. (iv) False.
Frequently asked questions
- For the given $3\times3$ adjoint matrix in the textbook, compute $|\operatorname{adj}(A)|$.
- Answer: The preserved working gives $|\operatorname{adj}(A)|=2(24-0)+4(-6-14)+2(0+24)=48-80+48=16$. The source matrix itself is missing from the text extract, so this card is not marked validated. <div
- For the given $3\times3$ adjoint matrix in the textbook, compute $|\operatorname{adj} A|$.
- Answer: The preserved working gives $|\operatorname{adj} A|=0+2(36-18)+0=36$. The source matrix itself is missing from the text extract, so this card is not marked validated. <div
- Find the adjoint of the matrices: (i) $-\begin{bmatrix}3&4\\6&2\end{bmatrix}$ (ii) $\begin{bmatrix}2&3&1\\3&4&1\\3&7&2\end{bmatrix}$. The third matrix in this textbook item needs source-text restoration before validation.
- Answer: (i) $\operatorname{adj}(A)=\begin{bmatrix}-2&4\\6&-3\end{bmatrix}$. (ii) $\operatorname{adj}(A)=\begin{bmatrix}1&1&-1\\-3&1&1\\9&-5&-1\end{bmatrix}$. The missing third matrix is intentionally left unverified here; no answer is marked validated until the exact textbook entries are restored. <div
- Find the inverse (if it exists) of the following: (i) −[[2 4][1 3]] (ii) [[5 1 1][1 5 1][1 1 5]] (iii) [[2 3 1][3 4 1][3 7 2]]
- (i) A^{-1} = [[-3/2, 2], [1/2, -1]]. (ii) A^{-1} = [[3/14, -1/28, -1/28], [-1/28, 3/14, -1/28], [-1/28, -1/28, 3/14]]. (iii) Using det(A)=2 and adj(A) from earlier, A^{-1} = (1/2) · [[1,1,-1],[-3,1,1],[9,-5,-1]] = [[1/2,1/2,-1/2],[-3/2,1/2,1/2],[9/2,-5/2,-1/2]].
These important questions are selected from the Samacheer Kalvi Class 12 Maths textbook book-back exercises to help you revise the most useful questions. Mark weightage (5/2/1) follows the usual exam pattern and may vary by exam — always check your latest syllabus and question pattern. Open each chapter for the complete set of questions and answers.