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.
Answer:
(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]].
Answer: 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.
Answer: For A = [[-5, -3], [-1, -2]], one computes A^2 + 7A + 7I = 0. Multiplying the matrix equation on the right by A^{-1} gives A + 7I + 7A^{-1} = 0, so A^{-1} = -(1/7)(A + 7I) = -(1/7)A - I. Thus A^{-1} = [[-2/7, 3/7], [1/7, -5/7]].
Answer: The identity A·adj(A) = det(A)·I holds for any square matrix A. For the given A the determinant computes to det(A) = -217, so A·adj(A) = -217·I_3.
Answer: With A = [[-8,-4],[-5,-3]], det(A) = 4, so A·adj(A) = 4I. For n=2, adj(adj A) = A (since adj(adj A) = det(A)^{n-2} A and n-2=0). Also adj(A)^{-1} = (1/det A) A.
Answer: (AB)^{-1} = B^{-1}A^{-1}.
Reason: If A and B are invertible then (AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AA^{-1} = I and similarly (B^{-1}A^{-1})(AB)=I, so (AB)^{-1}=B^{-1}A^{-1}. (Numeric verification may be done by computing the inverses of A and B explicitly.)
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.
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.
Answer: adj(adj(A)) = 0 (the 3×3 zero matrix).
Reason: Let B = adj(A). Here det(B) = 0, so det(A)^2 = 0 ⇒ det(A)=0. For n=3 the identity adj(adj A) = det(A)·A then gives adj(adj A) = det(A)·A = 0·A = the zero matrix.
Answer: The final conclusion in the degraded card was only “Hence proved,” without the source matrix. This item remains unverified until the exact textbook matrix is restored.
No validation badge is shown for this item.
Answer: A = [[3, -1], [1, 2]].
Work: A = N·M^{-1} with M = [[5,3],[1,2]] (det 7) so M^{-1} = (1/7)[[2,-3],[-1,5]]. Multiply N = [[14,7],[7,7]] by M^{-1} to obtain A = [[3,-1],[1,2]].
Method: Premultiply by $A^{-1}$ and postmultiply by $B^{-1}$: $A^{-1}(AXB)B^{-1}=A^{-1}CB^{-1}$. Hence $X=A^{-1}CB^{-1}$.
The exact matrices $A$, $B$ and $C$ are missing from the text extract, so the numerical value of $X$ is not marked validated.
Answer: $A^2=\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}=A+2I$.
Therefore $A^2-A-2I=O$.
So the sequence of decoded row matrics is [8 5] [12 16]
The receiver reads the message as “HELP”.
So the sequence of decoded row matrics is [8 5] [12 16] The receiver reads the message as “HELP”.
Answer: The preserved working gives ranks $1$, $2$, $3$ and $3$ for the listed matrices.
Some matrix entries are absent from the text extract, so this combined item is not marked validated.
Answer: The preserved row-echelon conclusions give ranks $2$, $3$ and $3$.
Some starting matrices are absent from the text extract, so this combined item is not marked validated.
(i) A = [[2,1],[5,2]], det(A)=2·2−1·5=−1. A^{-1} = (1/det)[[2,−1],[−5,2]] = [[−2,1],[5,−2]].
(ii) A = [[1,1,0],[1,0,1],[6,2,3]]. det(A)=1. Using cofactors one finds
A^{-1} = [[−2,−3,1],[3,3,−1],[2,4,−1]].
(iii) A = [[1,2,3],[2,5,3],[1,0,8]], det(A)=−1. The adjugate is [[40,−16,−9],[−13,5,3],[−5,2,1]] so
A^{-1} = (1/−1)·adj(A) = [[−40,16,9],[13,−5,−3],[5,−2,−1]].
(i) 2x + 5y = -2, x + 2y = -3v
(ii) 2x – y = 8, 3x + 2y = -2
x = 2, y = -4
(iii) 2x + 3y – z = 9, x + y + z = 9, 3x – y – z = -1
|A| = 2(-1+1)-3(-1-3)-1(-1-3)
= 0 + 12 + 4 =16 ≠ 0 A -1 exists.
∴ x = 2, y = 3, z = 4
(iv) x + y + z – 2 = 0, 6x – 4y + 5z – 31 = 0, 5x + 2y + 2z = 13
AX = B
X = A -1 B
A = $\left[\begin{array}{ccc}
1 & 1 & 1 \\
6 & -4 & 5 \\
5 & 2 & 2
\end{array}\right]$
|A| = 1(-8-10)-1(12-25)+1(12+20)
= 18 + 13 +32 = 27
≠ 0
A -1 Exists
∴ x = 3, y = -2, z = 1
≠ 0 A -1 Exists ∴ x = 3, y = -2, z = 1
Find the products AB and BA and hence solve the system of equations x + y + 2z = 1, 3x + 2y + z = 7, 2x + y + 3z = 2v
AB = BA = 4I 3
∴ x = 2, y = 1, z = -1
AB = BA = 4I 3 ∴ x = 2, y = 1, z = -1
Let s = starting monthly salary, r = annual increment. Then after 3 years: s + 3r = 19800. After 9 years: s + 9r = 23400. In matrix form [1 3; 1 9][s; r] = [19800;23400]. Subtract equations: 6r = 3600 ⇒ r = 600. Then s = 19800 − 3·600 = 18000.
Starting salary = ₹18,000 per month; Annual increment = ₹600 per month.
Let m,w be work per day by one man and one woman. Total work = 1. (4m+4w)·3 =1 ⇒ 4m+4w = 1/3 ⇒ m+w =1/12. (2m+5w)·4 =1 ⇒ 2m+5w =1/4. Solve: from m = 1/12 − w; substitute: 2(1/12 − w)+5w =1/4 ⇒ 1/6 +3w =1/4 ⇒ 3w =1/12 ⇒ w=1/36, m=1/18. Times = reciprocals: 18 and 36 days.
One man alone: 18 days. One woman alone: 36 days.
Set profit equations (sell − buy): P: 2x −4y +5z =15000. Q: 3x + y −2z =1000. R: −x +3y + z =4000. Solve: from R, x = 3y + z −4000. Substitute into Q and P to get two linear equations in y,z: 10y + z =13000 and 2y +7z =23000. Solve: y=1000, z=3000, then x =3·1000+3000−4000 =2000.
x = ₹2000, y = ₹1000, z = ₹3000.
(i) 5x – 2y + 16 = 0, x + 3y – 7 = 0v
(ii) $\frac{3}{x}$ + 2y =12, $\frac{2}{x}$ + 3y = 13
Let $\frac{1}{x}$ = a
3a + 2b = 12
2a + 3b = 13
(iii) 3x + 3y – z = 1 1, 2x – y + 2z = 9, 4x + 3y + 2z = 25
Δ = $\left| \begin{matrix} 3 & 3 & -1 \\ 2 & -1 & 2 \\ 4 & 3 & 2 \end{matrix} \right| $
= 3(-2 – 6) -3 (4 – 8) -1(6 + 4)
= 3(-8) -3(-4) -1(10)
= -24 + 12 – 10 = -22 ≠ 0
Δ x = $\left| \begin{matrix} 11 & 3 & -1 \\ 9 & -1 & 2 \\ 25 & 3 & 2 \end{matrix} \right| $
= 11 (-2 – 6) – 3(18 – 50) – 1(27 + 25)
= 11(-8) -3(32) -1(52)
= -88 + 96 – 52 = -44
Δ y = $\left| \begin{matrix} 3 & 11 & -1 \\ 2 & 9 & 2 \\ 4 & 25 & 2 \end{matrix} \right| $
= 3(18 – 50) – 11(4 – 8) – 1(50 – 36)
= 3(32) -11(4) -1(14)
= -96 + 44 – 14 = -66
Δ x = $\left| \begin{matrix} 3 & 3 & 11 \\ 2 & -1 & 9 \\ 4 & 3 & 25 \end{matrix} \right| $
= 3(-25 – 27) – 3(50 – 36) + 11(6 + 4)
= 3(-52) -3(14) + 11(10)
= -156 – 42 + 110
= -88
3a – 4b – 2c = 1 ……….. (1)
a + 2b + c = 2 …………… (2)
2a – 5b – 4c = -1 ………….. (3)
Δ = $\left| \begin{matrix} 3 & -4 & -2 \\ 1 & 2 & 1 \\ 2 & -5 & -4 \end{matrix} \right| $
= 3(-8 + 5) + 4 (-4 – 2) – 2(-5 – 4)
= 3(-3) +4(-6) -2(-9)
= -9 – 24 + 18
= -15 ≠ 0
Δ a = $\left| \begin{matrix} 1 & -4 & -2 \\ 2 & 2 & 1 \\ -1 & -5 & -4 \end{matrix} \right| $
= 1(-8 + 5) + 4(-8 + 1) – 2(-10 + 2)
= 1(-3) + 4(-7) – 2(-8)
= -3 – 28 + 16
= -15
Δ b = $\left| \begin{matrix} 3 & 1 & -2 \\ 1 & 2 & 1 \\ 2 & -1 & -4 \end{matrix} \right| $
= 3(-8 + 1) – 1(-4 – 2) – 2(-1 – 4)
= 3(-7) -1(-6) – 2(-5)
= – 21 + 6 + 10 = -5
Δ c = $\left| \begin{matrix} 3 & -4 & 1 \\ 1 & 2 & 2 \\ 2 & -5 & -1 \end{matrix} \right| $
= 3(-2 + 10) + 4(-1 – 4) + 1 (-5 – 4)
= 24 – 20 – 9 = -5
Δ c = $\left| \begin{matrix} 3 & -4 & 1 \\ 1 & 2 & 2 \\ 2 & -5 & -1 \end{matrix} \right| $ = 3(-2 + 10) + 4(-1 – 4) + 1 (-5 – 4) = 24 – 20 – 9 = -5
Let x = correct, y = wrong. x + y = 100. Score: x − (1/4)y = 80. From first y = 100 − x. Substitute: x − (1/4)(100 − x) = 80 ⇒ x −25 + x/4 =80 ⇒ (5x/4) =105 ⇒ x = (4/5)·105 =84.
84 questions correct.
Let a litres of 50%, b litres of 25%. a + b = 10. Acid: 0.5a +0.25b = 4. Solve: multiply second by 4: 2a + b =16. Subtract a + b =10 ⇒ a =6, b =4.
6 litres of 50% solution and 4 litres of 25% solution.
Let rates (tank/min) be a for A and b for B (positive when filling). a + b = 1/10. If B runs in reverse rate is −b, then a − b = 1/30. Add: 2a = 1/10 +1/30 =4/30 =2/15 ⇒ a =1/15 ⇒ 15 min. Then b = 1/10 −1/15 =1/30 ⇒ 30 min.
Pump A alone: 15 minutes. Pump B alone: 30 minutes (when running correctly).
Let D,I,V be prices. Equations: 2D+3I+2V=150, 2D+2I+4V=200, 5D+4I+2V=250. Subtracting first from second: −I+2V=50 ⇒ I=2V−50. Subtract first from third: 3D+I=100. Substitute I into third: 3D+2V−50=100 ⇒3D+2V=150 ⇒ D=(150−2V)/3. Substitute into first and solve: V=30 ⇒ I=10 ⇒ D=30. Total for 3D+6I+6V = 3·30 +6·10 +6·30 =90+60+180=330.
Price per item: dosai = ₹30, idli = ₹10, vadai = ₹30. Bill for 3 dosai,6 idli,6 vadai = ₹330 ≤ ₹350, so yes (₹20 change).
As in id 1-28.
Refer to id 1-28 (handled together).
As in id 1-28.
Refer to id 1-28 (handled together).
As in id 1-28.
Refer to id 1-28: Prices D=30, I=10, V=30; bill = ₹330, so yes they can pay.
(i) 2x – 2y + 3z = 2, x + 2y – z = 3, 3x – y + 2z = 1v
Augmented matrix
Writing the equivalent equations from echelon from.
x – y + 2z = 3 …………. (1)
5y – 6z = -4 ………….. (2)
-z = -4
z = 4
(2) ⇒ 5y – 6z = -4
5y – 24 = -4
5y = -4 + 24
5y = 20
y = 4
(1) ⇒ x – y + 2z = 3
x – 4 + 8 = 3
x = 3 + 4 – 8
x = -1
∴ x = -1, y = 4, z = 4
(ii) 2x + 4y + 6z = 22, 3x + 8y + 5z = 27, -x + y + 2z = 2.
Augmented matrix
Writing the equivalent equations from echelon from.
x + 2y + 3z = 11 …………. (1)
y – 2z = -3 ………….. (2)
11z = 22
z = 2
(2) ⇒ y – 2z = -3
y – 4 = -3
y = -3 + 4
y = 1
(1) ⇒ x + 2y + 3z = 11
x + 2(1) + 3(2) = 11
x + 2 + 6 = 11
x = 11 – 8 = 3
∴ x = 3, y = 1, z = 2
x + 2 + 6 = 11 x = 11 – 8 = 3 ∴ x = 3, y = 1, z = 2
Given: f(x) = ax² + bx + c
In Remainder Theorem
f(-3) = 21
a(-3)² + b(-3) + c = 21
9a – 3b + c = 21 ……….. (1)
f(5) = 61
25a + 5b + c = 61 …………. (2)
f(1) = 9
a + b + c = 9 ………… (3)
Augmented matrix
Writing the equivalent equations from echelon from.
a + b + c = 9 …………. (1)
b + 2c = 5 ………….. (2)
-4c = -8
c = 2
(2) ⇒ b + 2c = 5
b + 4 = 5
b = 5 – 4
b = 1
(1) ⇒ a + b + c = 9
a + 1 + 2 = 9
a = 9 – 3
a = 6
a = 6, b = 1, c = 2
a = 9 – 3 a = 6 a = 6, b = 1, c = 2
Let amounts be x (6%), y (8%), z (9%). x + y + z = 65000. Annual income: 0.06x +0.08y +0.09z = 4800. Income from third is 600 more than second: 0.09z = 0.08y +600 ⇒ 0.08y −0.09z = −600. Multiply the income eqn by 100 to avoid decimals: 6x +8y +9z = 480000/100? Better: multiply by 100: 6x +8y +9z = 4800·100 = 480000 (but units inconsistent). Simpler: from 0.09z −0.08y = 600 ⇒ divide by 0.01: 9z −8y = 60000. Also 6x +8y +9z = 480000 (multiplying earlier by 10000? To avoid confusion, solve algebraically:) From third: 0.09z = 0.08y +600 ⇒ z = (0.08/0.09)y + 600/0.09 = (8/9)y + 6666.66... Non-integer indicates a nicer approach: Let incomes in rupees: income from x is 6% of x etc. Let X = x, Y = y, Z = z. Equations: X+Y+Z=65000. 0.06X+0.08Y+0.09Z=4800. 0.09Z −0.08Y =600. Solve the last two: multiply last by 100: 9Z −8Y =60000. Multiply income eqn by 100: 6X +8Y +9Z =480000. Add these two: 6X + (9Z+9Z) + (8Y−8Y) = 480000+60000 ⇒ 6X +18Z =540000 ⇒ divide 6: X +3Z =90000 ⇒ X = 90000 −3Z. Use X+Y+Z=65000 ⇒ (90000 −3Z) + Y + Z =65000 ⇒ Y −2Z = −25000 ⇒ Y = 2Z −25000. Substitute into 9Z −8Y =60000 ⇒ 9Z −8(2Z −25000) =60000 ⇒ 9Z −16Z +200000 =60000 ⇒ −7Z = −140000 ⇒ Z =20000. Then Y = 2·20000 −25000 =15000. X = 65000 − (20000+15000) =30000. Wait that gives X=30000 at 6% etc. Check incomes: 0.06·30000=1800; 0.08·15000=1200; 0.09·20000=1800; total =4800 and third income 1800 is 600 more than second 1200. So investments are 30000,15000,20000 corresponding to 6%,8%,9% respectively. (Order corrected.)
Investments: at 6% = ₹20,000; at 8% = ₹15,000; at 9% = ₹30,000.
Find a,b,c from three points. For (−6,8): 36a −6b + c = 8. For (−2,12): 4a −2b + c =12. For (3,8): 9a +3b + c =8. Subtract second from first: 32a −4b = −4 ⇒8a − b = −1. Subtract second from third: 5a +5b = −4 ⇒ a + b = −4/5. Solve: from a + b = −4/5 ⇒ b = −4/5 − a. Substitute into 8a − (−4/5 − a) = −1 ⇒ 8a + a +4/5 = −1 ⇒9a = −9/5 ⇒ a = −1/5. Then b = −4/5 −(−1/5) = −3/5. From 4a −2b + c =12 ⇒ 4(−1/5) −2(−3/5) + c =12 ⇒ (−4/5 +6/5) + c =12 ⇒ 2/5 + c =12 ⇒ c = 58/5 =11.6. Now evaluate y at x=7: y = a·49 + b·7 + c = (−1/5)·49 + (−3/5)·7 + 58/5 = (−49 −21 +58)/5 = (−12)/5 = −12/5 = −2.4. This is not 60; hence he will not meet at P(7,60).
No; the parabola determined by those three points does not pass through P(7,60).
(i) x – y + 2z = 2, 2x + y + 4z = 7, 4x – y + z = 4v
Matrix form
The system is consistent.
ρ(A) ρ[A|B] = 3 = n
it has unique solution.
Writing the equivalent equations from echelon form
x – y + 2z = 2 ………… (1)
3y = 3 ⇒ y = 1
-7z = -7
z = 1
(1)⇒ x – y + 2z = 2
x – 1 + 2 = 2
x = 1
∴ Solution is x = 1, y = 1, z = 1
(ii) 3x + y + z = 2, x – 3y + 2z = 1, 7x – y + 4z = 5
ρ(A) = 2 ρ[A | B] = 2
ρ(A) = ρ[A | B] = 2 < n
The system is consistent. It has infinitely many solution.
Writing the equivalent equations from echelon form.
x – 3y + 2z = 1 ………. (1)
10y – 5z = -1 ………. (2)
Put z = t.
(2) ⇒ 10y – 5z = -1
10y = -1 + 5z = 5t – 1
(iii) 2x + 2y + z = 5, x – y + z = 1, 3x + y + 2z = 4
ρ(A) = 2 ρ[A | B] = 3
ρ(A) ≠ ρ[A | B] = 2 < n
∴ The system is inconsistent. It has no solution.
(iv) 2x – y + z = 2, 6x – 3y + 3z = 6, 4x – 2y + 2z = 4
ρ(A) = 1 ρ[A | B] = 1
ρ(A) = ρ[A | B] = 1 < n.
∴ The system reduces into a single equation.
∴ It is consistent and has infinitely many solutions.
Writing the equivalent equations from echelon form
2x – y + z = 2
Put y = s, z = t
2x – s + t = 2
2x = 2 + s – t
x = $\frac {2+s-t}{2}$
(x, y, z) = ($\frac {2+s-t}{2}$, s, t) ∀ s, t ∈ R
2x = 2 + s – t x = $\frac {2+s-t}{2}$ (x, y, z) = ($\frac {2+s-t}{2}$, s, t) ∀ s, t ∈ R
Coefficient matrix A = [[k,1,1],[1,k,1],[1,1,k]]. det(A) = (k−1)^2(k+2).
• If det(A) ≠ 0 (i.e. (k−1)^2(k+2) ≠ 0) then the system has a unique solution. Hence unique solution for k ≠ 1 and k ≠ −2.
• If k = 1 then A = matrix of all ones, rank(A)=1 and the augmented column [2,2,2]^T is in the same one-dimensional space, so rank([A|b])=1. Therefore there are infinitely many solutions for k = 1.
• If k = −2 then det(A)=0 and adding the three equations gives 0 = 6 (since each left-hand side sums to 0 but RHS sums to 6), so the augmented system is inconsistent. Thus k = −2 gives no solution.
Final: (i) No solution: k = −2. (ii) Unique solution: k ≠ 1, −2. (iii) Infinitely many solutions: k = 1.
(i) no solution
(ii) a unique solution
(iii) an infinite number of solutions.v
Case (i)
If λ = 5, µ ≠ 9
ρ(A) = 2, ρ(A | B) = 3
ρ(A) ≠ ρ(A | B)
The system is inconsistent. It has no solution.
Case (ii)
If λ = 5, µ ≠ 9
ρ(A) = 3, ρ(A | B) = 3
ρ(A) = ρ(A | B) = 3 = n
The system is consistent. It has unique solution.
Case (iii)
If λ = 5, µ = 9
ρ(A) = 2, ρ(A | B) =2
ρ(A) = ρ(A | B) = 2 < n
The system is consistent. It has infinitely many solution.
ρ(A) = 2, ρ(A | B) =2 ρ(A) = ρ(A | B) = 2 < n The system is consistent. It has infinitely many solution.
(i) 3x + 2y + 7z = 0; 4x – 3y – 2z = 0; 5x + 9y + 23z = 0v
ρ(A) = 2 ρ[A | B] = 2
ρ(A) ρ[A | B] = 2 < n
The system is consistent. It has non trivial solution.
Writing the equivalent equations from echelon form
3x + 2y + 7z = 0 ………… (1)
-17y – 34z = 0 ……….. (2)
Put z = t
(2) ⇒ -17y = 34t
y = $\frac {34t}{-17}$ = -2t
(1) ⇒ 3x + 2(-2t) + 7t = 0
3x – 4t + 7t = 0
3x + 3t = 0
3x = -3t
x = -t
(x, y, z) (-t, -2t, t) ∀ t ∈ R
(ii) 2x + 3y – z = 0, x – y – 2z = 0, 3x + y + 3z = 0
ρ(A) = 3 ρ[A | B] = 3
ρ(A) = ρ[A | B] = 3
The system is consistent. It has trivial solution.
x = 0, y = 0, z = 0
ρ(A) = ρ[A | B] = 3 The system is consistent. It has trivial solution. x = 0, y = 0, z = 0
Coefficient matrix A = [[1,1,1],[0,3,0],[3,0,2]]. det(A) = 1·(3·2 −0·0) −1·(0·2 −0·3) +1·(0·0 −3·3) = 6 −0 −9 = −3 ≠ 0. Hence for any λ the non-homogeneous system has unique solution. The homogeneous system Ax = 0 has only the trivial solution because det(A) ≠ 0.
Unique solution for all λ (matrix invertible). A non-trivial solution (non-zero homogeneous) corresponds to λ making homogeneous system singular — none here unless λ also affects coefficients. For the homogeneous system (RHS 0) only trivial solution since coefficient matrix has nonzero determinant.
Let coefficients be a C6H12O6 + b O2 → c CO2 + d H2O. Equating atoms: C: 6a = c; H: 12a = 2d ⇒ d = 6a; O: 6a + 2b = 2c + d. Substitute c=6a and d=6a: 6a + 2b = 12a + 6a ⇒ 6a + 2b = 18a ⇒ 2b = 12a ⇒ b = 6a. Choose smallest integer a = 1 ⇒ b = 6, c = 6, d = 6. Hence C6H12O6 + 6O2 → 6CO2 + 6H2O.
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O
Property: det(adj(adj A)) = (det A)^{(n-1)^2}. Given (det A)^{(n-1)^2} = det A (nonzero). So (n-1)^2 = 1 ⇒ n-1 = 1 ⇒ n = 2. (Discarding n=0). Therefore n = 2. Hence option (3).
(A) A (B) B (C) I 3 (D) B Tv
(C) I
Hint:
AA T = A T A and B = A -1 A T
B T = (A -1 ) T (A T ) T = (A -1 ) T A
BB -1 = (A -1 A T ) (A -1 ) T A = A -1 (AA -1 ) T A
= A -1 (I) T A = A -1 A = I 3
I 3
det A = 3·2 − 5·1 = 1. For 2×2, det(adj X) = det X. So det(adj B) = det(adj(adj A)) = det A = 1. det C = det(3A) = 3^2 det A = 9·1 = 9. Hence |adj B|/|C| = 1/9. Option (2).
Answer: $A=\begin{bmatrix}4&2\\-1&1\end{bmatrix}$.
Let $M=\begin{bmatrix}1&-2\\1&4\end{bmatrix}$. Since $\det M=6$, $M^{-1}=\frac16\begin{bmatrix}4&2\\-1&1\end{bmatrix}$. Hence $A=6I\,M^{-1}=\begin{bmatrix}4&2\\-1&1\end{bmatrix}$.
(A) A -1 (B) $\frac{A^{-1}}{2}$ (C) 3A -1 (D) 2A -1v
(d) 2A -1
Hint:
|A| = 14 – 12 = 2
2A -1
For 2×2, det(adj M) = det M. So det(adj(AB)) = det(AB) = det A · det B. det A = 2·5 − 0·1 = 10. det B = 1·0 − 4·2 = −8. Hence det AB = 10·(−8) = −80. Option (2).
(A) 15 (B) 12 (C) 14 (D) 11v
(d) 11
Hint:
Since |adj A| = |A| n – 1 (n = 3)
1(-6 + 0) – x(-2) = 4²
-6 + 2x = 16
2x = 16 + 6 = 22
x = 11
11
Answer: The preserved key is $-1$.
The defining matrix relation is missing from the text extract, so this item is not marked validated.
(A) adj A = |A|A -1 (B) adj (AB) = (adj A)(adj B) (C) det A -1 = (det A) -1 (D) (ABC) -1 = C -1 B -1 A -1v
(b) adj (AB) = (adj A)(adj B)
Hint:
adj (AB) ≠ (adj A)(adj B)
adj (AB) = (adj A)(adj B)
Method: From $(AB)^{-1}=B^{-1}A^{-1}$, multiply on the right by $A$ to get $B^{-1}=(AB)^{-1}A$.
The preserved option key is $\begin{bmatrix}2&-5\\-3&8\end{bmatrix}$, but the source matrices are missing from the text extract, so the card is not marked validated.
Assume A^{-1}A^T is symmetric ⇒ A^{-1}A^T = (A^{-1}A^T)^T = A (A^{-1})^T = A A^{-T}. Multiplying on left by A and on right by A^{-1} leads to A^2 = A^T. Thus A^2 = A^T. Hence option (3). (This uses A invertible.)
v
(d) $\begin{bmatrix} 5 & -2 \\ 3 & -1 \end{bmatrix}$
Hint:
A -1 = $\begin{bmatrix} 5 & 3 \\ -2 & -1 \end{bmatrix}$
Since (A T ) -1 = (A -1 ) T = $\begin{bmatrix} 5 & -2 \\ 3 & -1 \end{bmatrix}$
D
Answer: The preserved key is $-\frac45$.
The source matrix relation is missing from the text extract, so this item is not marked validated.
Answer: The preserved key is $(\cos^2\frac{\theta}{2})A^T$.
The exact matrix equation is missing from the text extract, so this item is not marked validated.
Answer: The preserved key is $k=1$, using $\cos^2\theta+\sin^2\theta=1$.
The full condition involving $k$ is missing from the text extract, so this item is not marked validated.
From $\lambda A^{-1}=A$, multiplying both sides by $A$ gives $\lambda I=A^2$. Now $A^2=\begin{bmatrix}2&3\\5&-2\end{bmatrix}^2=\begin{bmatrix}4+15&6-6\\10-10&15+4\end{bmatrix}=\begin{bmatrix}19&0\\0&19\end{bmatrix}=19I$. Hence $\lambda I=19I$, so $\lambda=19$.
Method: Use $\operatorname{adj}(AB)=\operatorname{adj}(B)\operatorname{adj}(A)$.
The option key in the degraded card is not safely supported by the visible matrix entries, so this item is not marked validated.
Rows: r1 = [1,2,3,4], r2 = 2·r1, r3 = r1. All rows are scalar multiples of r1, so only one independent row ⇒ rank = 1. Option (1).
Answer: $x=e^{\Delta_1/\Delta_3}$ and $y=e^{\Delta_2/\Delta_3}$.
Taking logarithms gives $a\log x+b\log y=m$ and $c\log x+d\log y=n$. Applying Cramer's rule to $\log x$ and $\log y$ gives the stated exponential form.
(i) True: adj(A)^T = adj(A^T); if A is symmetric A^T=A so adj(A) is symmetric. (ii) True: for a diagonal matrix all off–diagonal cofactors vanish so adj(D) is diagonal. (iii) False: adj(λA)=λ^{n-1} adj(A), so adj(adj(λA))=(λ^{n-1})^{n-1} adj(adj A)=λ^{(n-1)^2} adj(adj A), not λ^{n}. (iv) True: A·adj(A)=adj(A)·A=|A|I. Hence (i),(ii),(iv) are correct.
Answer: (3) n. Explanation: A constant function maps every element of the domain to the same single element of the codomain. There are n choices for that common image, so there are n constant functions.
Non‑trivial solution exists iff determinant of coefficient matrix is 0. With rows R1=[1,1,1], R2=[1,cosθ,sinθ], R3=[1,sinθ,cosθ]. Subtract R1: det = det[[c-1,s-1],[s-1,c-1]] = (c-s)(c+s-2). Since c+s≤√2<2, second factor ≠0; so c−s=0 ⇒ cosθ=sinθ ⇒ θ=π/4 (in [0,π]).
Answer: The preserved key is $\lambda=7$, $\mu=-5$.
The augmented matrix is missing from the text extract, so this item is not marked validated.
Answer: (4) not a function. Explanation: The set contains both (2,c) and (2,d), so the element 2 in the domain is assigned two different images. That violates the definition of a function (which must assign exactly one image to each domain element).
For a 3×3 matrix adj(adj A) = (det A)^{3-2} A = (det A) A. Compute det A: expanding gives det A = 1. Hence adj(adj A)=1·A = A. Thus option (1).