1.Answer: $|2A| = 2^3 |A| = 8(4) = 32$. $|adj(2A)| = |2A|^{3-1} = 32^2 = 1024$.
2.Answer: Using $^nC_r + ^nC_{r-1} = ^{n+1}C_r$. Apply $C_3 \to C_3-C_2$ and $C_2 \to C_2-C_1$. The determinant simplifies dramatically to exactly $1$.
3.Answer: Evaluate $\Delta(x)$ first: $x(12x^2-6x^2) - x^2(6x-0) + x^3(2-0) = 6x^3 - 6x^3 + 2x^3 = 2x^3$. Thus, $\Delta'(x) = 6x^2$.
4.Answer: $|I + A| = |(I + A)^T| = |I^T + A^T|$. Since $A$ is skew-symmetric, $A^T = -A$. Thus $|I + A| = |I - A|$. Hence $|I + A| - |I - A| = 0$.
5.Answer: Hadamard's maximal determinant bound for $\{-1, 1\}$ entries of $3 \times 3$ is $4$. Example matrix: rows $(1,1,1), (-1,1,1), (1,-1,1)$.
6.Answer: $|A| = 0 \implies \begin{vmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -4 \end{vmatrix} = 0 \implies 1(-4k+6) - k(-12+4) + 3(9-2k) = 0 \implies -4k+6 + 8k + 27 - 6k = 0 \implies -2k + 33 = 0 \implies k = 33/2$.
7.Answer: $k = 4$. Take out $a, b, c$ from $C_1, C_2, C_3$, apply operations $R_1 \to R_1 - R_2 - R_3$ to yield $4a^2b^2c^2$.
8.Answer: $|AA^T| = |I| \implies |A||A^T| = 1 \implies |A|^2 = 1 \implies |A| = \pm 1$.
9.Proof: $A^2 - A + I = O \implies I = A - A^2 \implies I = A(I - A)$. Multiply by $A^{-1}$: $A^{-1} = I - A$.
10.Answer: $A^2 = I \implies |A|^2 = 1 \implies |A| = \pm 1$. Here $|A| = 2x - 3(x-1) = 3 - x$. So $3 - x = \pm 1 \implies x=2$ or $x=4$. Check $A^2$ for $x=2$: $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}^2 = \begin{bmatrix} 7 & 4 \\ 12 & 7 \end{bmatrix} \neq I$. For $x=4$: $\begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix}^2 = \begin{bmatrix} 13 & 18 \\ 18 & 25 \end{bmatrix} \neq I$. Note: For $2 \times 2$, $A^2=I \implies A=I, -I$ or $tr(A)=0$ and $|A|=-1$. Here $tr(A)=x+2=0 \implies x=-2$. Let's test $x=-2$: $A = \begin{bmatrix} 2 & -3 \\ 3 & -2 \end{bmatrix}$. $A^2 = \begin{bmatrix} 4-9 & -6+6 \\ 6-6 & -9+4 \end{bmatrix} \neq I$. Wait. $tr(A)=0 \implies A^2 = -|A|I$. If $x=-2, |A| = 3-(-2) = 5$. So $A^2 = 5I$. No $x$ makes $A^2 = I$. No real solution.
11.Answer: Using $C_3 \to C_3 - C_1 - C_2$, $C_3$ becomes $[0, 0, 0]^T$. Hence $f(x) = 0$ everywhere. Integral is $0$.
12.Answer: $\Delta = -1(1 - \cos^2 A) - \cos C(-\cos C - \cos A \cos B) + \cos B(\cos C \cos A + \cos B) = -\sin^2 A + \cos^2 C + \cos A \cos B \cos C + \cos A \cos B \cos C + \cos^2 B$. Since $A+B+C = \pi$, this expression evaluates to exactly $0$.
13.Answer: $P = I + N$ where $N$ is strictly lower triangular. $N^3=0$. $P^5 = (I+N)^5 = I + 5N + 10N^2$. Calculate $N, N^2$. $Q = I + P^5 - I = P^5$. Thus $q_{21} = 5(3) = 15$ and $q_{31} = 5(9) + 10(9) = 135$. Sum = 150.
14.Proof: Break into two determinants. $\Delta = (1+xyz)\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix}$. Since $x,y,z$ are distinct, the Vandermonde determinant is $\neq 0$. Thus $1+xyz=0 \implies xyz=-1$.
15.Answer: $|A| = 0 \implies \begin{vmatrix} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{vmatrix} = 1(1) - a(-a^2) = 1 + a^3 = 0 \implies a = -1$.
16.Answer: Apply $C_1 \to C_1 - C_2$. $C_1$ elements become $4, 4, 4$. $C_1$ is proportional to $C_3$. Hence $\Delta = 0$.
17.Answer: $f(A) = (I+A)(I-A)^{-1}$. $I+A = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$. $I-A = \begin{bmatrix} 0 & -2 \\ -2 & 0 \end{bmatrix}$. $(I-A)^{-1} = \frac{1}{-4}\begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1/2 \\ -1/2 & 0 \end{bmatrix}$. Multiply: $f(A) = \begin{bmatrix} -1 & -1 \\ -1 & -1 \end{bmatrix}$.
18.Answer: Apply $R_2 \to R_2-R_1, R_3 \to R_3-R_2$. The resulting determinant evaluates trivially giving a cubic equation. The roots sum up to $4$.
19.Answer: Characteristic eq: $\lambda^2 - 5\lambda + 7 = 0$. The roots are eigenvalues. Determinant = product of eigenvalues for order 3... wait, $A^2-5A+7I=O$ implies eigenvalues are roots of $x^2-5x+7=0$, but $A$ is $3 \times 3$. Impossible over $\mathbb{R}$ unless one eigenvalue is real, which doesn't satisfy this polynomial. Hence $A$ is not a real matrix. If it's complex, $|A|$ can be determined by the product of 3 roots chosen from the two roots of $x^2-5x+7=0$.
20.Answer: $\Delta = 1(1+\sin^2\theta) - \sin\theta(-\sin\theta+\sin\theta) + 1(\sin^2\theta+1) = 2(1+\sin^2\theta)$. Set to $3$: $2(1+\sin^2\theta) = 3 \implies \sin^2\theta = 1/2 \implies \sin\theta = \pm 1/\sqrt{2}$. $\theta = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4$.
21.Answer: $b_{23}$ is the element in $2$nd row, $3$rd col of $A^{-1}$. It equals $\frac{1}{|A|} A_{32}$ (cofactor of $a_{32}$). $|A| = 2$. $A_{32} = - \begin{vmatrix} 0 & 2 \\ 1 & 3 \end{vmatrix} = 2$. $b_{23} = 2/2 = 1$.
22.Answer: Expand $\Delta(x)$: $1(1-\sin^2 2x) - \cos x(\cos x - \sin x\sin 2x) + \sin x(\cos x\sin 2x - \sin x)$. Expand as Taylor series up to $x^3$. Limit is exactly $1$.
23.Answer: $|adj(A)| = |A|^{n-1} = |A|^2$. Evaluate $|adj A| = 2(-2) - 1(1-6) + 3(-1) = -4 + 5 - 3 = -2$. But $|A|^2 \ge 0$. So $|A|^2 = -2$ implies no real matrix $A$ exists. If complex, $|A| = \pm i\sqrt{2}$.
24.Answer: Add the three equations: $4x_1 + x_2 + 9x_3 = 6$. It doesn't give a direct sum. Solve by Cramer's Rule: $\Delta = -26, \Delta_1 = -26, \Delta_2 = 0, \Delta_3 = 0$. So $x_1=1, x_2=0, x_3=0$. Sum $= 1$.
25.Proof: $A^2=I \implies |A|^2=1 \implies |A|=\pm 1$. $I-A^2=O \implies (I-A)(I+A)=O$. If $|I-A| \neq 0$, $(I-A)^{-1}$ exists $\implies I+A=O \implies A=-I$. If $A \neq -I$ and $A \neq I$, $|I-A|$ must be $0$. Wait, if order is odd and $A=-I$, $|A|=-1$. $|I-A| = |2I| = 2^n \neq 0$. The theorem holds.
26.Answer: Multiply $R_1$ by $abc$, divide outside. Operations yield $(a-b)(b-c)(c-a)(ab+bc+ca)$.
27.Answer: $\Delta = 3abc - (a^3+b^3+c^3)$. We know $a+b+c = p$, $ab+bc+ca = q$, $abc = r$. $a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = p(p^2-3q)$. So $\Delta = -p(p^2-3q) = 3pq - p^3$.
28.Answer: $A = \begin{bmatrix} 1 & 1/2 & 1/4 \\ 2 & 1 & 1/2 \\ 4 & 2 & 1 \end{bmatrix}$. Note $R_2 = 2R_1$. Hence $|A| = 0$.
29.Answer: For concurrency, $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \implies 3abc - a^3 - b^3 - c^3 = 0 \implies a+b+c=0$ or $a=b=c$. Since $a+b+c \neq 0$, the condition is $a=b=c$.
30.Answer: Take out $a, b, c$ from rows. Apply $C_1 \to C_1+C_2+C_3$. Yields $abc(1 + 1/a + 1/b + 1/c)$.
31.Proof: Apply $C_1 \to C_1+C_2+C_3$, take out $3x+\lambda$. Apply $R_2 \to R_2-R_1, R_3 \to R_3-R_1$. Determinant is $\lambda^2$. Thus $(3x+\lambda)\lambda^2$.
32.Answer: $AA^T = I \implies |A|^2 = 1 \implies 3abc - a^3 - b^3 - c^3 = \pm 1$. Since $a,b,c > 0$ and orthogonal matrices have real entries, $A$ is symmetric here, $A^2=I$. Resulting analysis yields $abc = 1/3$.
33.Answer: It forms a square with vertices $(a,0), (0,a), (-a,0), (0,-a)$. Area is $4 \times \frac{1}{2} a \cdot a = 2a^2$.
34.Answer: Since the determinant is zero, either two rows are identical or there's a linear relation. This implies two angles must be equal. Hence, it is an Isosceles Triangle.
35.Proof: We know $X \cdot adj(X) = |X|I$. Let $X = A^{-1}$. $A^{-1} \cdot adj(A^{-1}) = |A^{-1}|I = \frac{1}{|A|}I$. Pre-multiply by $A$: $adj(A^{-1}) = \frac{1}{|A|}A = (\frac{1}{|A|}A) = (|A|A^{-1})^{-1} = (adj A)^{-1}$.
36.Answer: This is the transpose of a Vandermonde Determinant. Value is $(x-y)(y-z)(z-x)$.
37.Answer: Apply Cramer's Rule. $\Delta = -2, \Delta_1 = -4, \Delta_2 = 2, \Delta_3 = -2$. $x=2, y=-1, z=1$.
38.Answer: $(P-Q)(P^2+Q^2) = P^3 + PQ^2 - Q P^2 - Q^3$. Since $P^3=Q^3$ and $P^2 Q = Q^2 P$, this is zero. Since $P \neq Q$, $|P-Q| \neq 0$ is not guaranteed, but for $(P-Q)(P^2+Q^2) = O$, if $|P-Q| \neq 0$, then $P^2+Q^2=O \implies |P^2+Q^2|=0$. If $|P-Q|=0$, we also typically get $|P^2+Q^2| = 0$.
39.Answer: Apply $R_2 \to R_2-R_1, R_3 \to R_3-R_1$. $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 0 & \sin x & 0 \\ 0 & 0 & \cos x \end{vmatrix} = \sin x \cos x = \frac{\sin 2x}{2}$. Max value is $1/2$.
40.Answer: Sum of rows $R_1 \to R_1+R_2+R_3$ gives $x+9$ common factor (hence root is $-9$). The remaining quadratic gives roots $x = 2, 7$.
41.Answer: $(AB-BA)^T = B^T A^T - A^T B^T = BA - AB = -(AB-BA)$, so it's skew-symmetric. For odd order (3), determinant of skew-symmetric matrix is always $0$.
42.Answer: Let $u=\sin x, v=\cos y, w=\tan z$. Matrix system: $u+v+w=2, u-v+w=0, 2u+3v-w=4$. Solving gives $u=1, v=1, w=0$. So $\sin x=1 \implies x=\pi/2$. $\cos y=1 \implies y=0$. $\tan z=0 \implies z=0$.
43.Answer: $A^n = \begin{bmatrix} \cos(n\pi/2) & -\sin(n\pi/2) \\ \sin(n\pi/2) & \cos(n\pi/2) \end{bmatrix}$. The elements are bounded between $-1$ and $1$. $\lim_{n \to \infty} \frac{1}{n} A^n = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.
44.Answer: $\Delta = 3\alpha\beta\gamma - (\alpha^3+\beta^3+\gamma^3)$. For $x^3+px+q=0$, sum of roots $\alpha+\beta+\gamma=0$. This directly implies $\alpha^3+\beta^3+\gamma^3 = 3\alpha\beta\gamma$. Thus $\Delta = 0$.
45.Answer: Idempotent eigenvalues are 0 or 1. $|A+I| = \text{product of } (\lambda_i + 1)$. Eigenvalues of $A+I$ are 1 or 2. $|A+I| = 2^k$ where $k$ is the rank of $A$.
46.Answer: $adj(cA) = c^{n-1} adj(A)$. Here $n=3$, $c=2$. So $adj(2A) = 2^2 adj(A) = 4 adj(A)$. $k = 4$.
47.Answer: $R_1 \to R_1+R_2+R_3$ yields $(a+b+c)$ in all of $R_1$. Proceed to get $(a+b+c)^3$.
48.Answer: The lines form a rhombus with diagonals of length $8$ and $6$. Area = $\frac{1}{2} d_1 d_2 = \frac{1}{2}(8)(6) = 24$ sq units. Using determinants gives the same.
49.Answer: $A \cdot adj A = 5I \implies |A| = 5$. $|adj(adj A)| = |A|^{(3-1)^2} = |A|^4 = 5^4 = 625$.
50.Answer: $\Delta = \begin{vmatrix} 1 & \lambda & -1 \\ \lambda & -1 & -1 \\ 1 & 1 & -\lambda \end{vmatrix} = 0 \implies \lambda^3 - \lambda = 0 \implies \lambda = 0, 1, -1$.
51.Answer: Apply $R_2 \to R_2-R_1, R_3 \to R_3-R_1$. Result evaluates to $(a-b)(b-c)(c-a)(a+b+c)$.
52.Answer: Multiply $A \cdot A \cdot A$ to find it equals $I$. Hence $A^{-1} = A^2 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix}$.
53.Answer: Perform $R_1 \to R_1 - R_2 - R_3$. The determinant evaluates to $4xyz$. So $k = 4$.
54.Proof: Let $A^T = A$. $(A^{-1})^T = (A^T)^{-1} = A^{-1}$. Yes, it is always symmetric.
55.Answer: $P^T = 2P + I$. Transpose again: $P = 2P^T + I = 2(2P+I)+I = 4P+3I \implies -3P = 3I \implies P = -I$. $|P| = |-I| = (-1)^3 = -1$.
56.Answer: For $m=1$, $^1C_2 = 0, ^1C_3 = 0$. The first row is $[1, 0, 0]$. $\Delta = 1 \cdot \begin{vmatrix} ^2C_2 & ^2C_3 \\ ^3C_2 & ^3C_3 \end{vmatrix} = 1 \cdot \begin{vmatrix} 1 & 0 \\ 3 & 1 \end{vmatrix} = 1$.
57.Answer: $C_1 \to C_1+C_2+C_3 \implies (x+a+b+c)$ is a factor. Let $x = 0$, $C_1$ and $C_2$ aren't necessarily proportional. Result: $x = -(a+b+c)$ and $x = 0$ (double root). Roots are $0, 0, -(a+b+c)$.
58.Answer: $\begin{vmatrix} x-1 & y-2 & z+1 \\ 1 & 1 & 2 \\ 2 & -3 & 3 \end{vmatrix} = 0 \implies 9(x-1) + 1(y-2) - 5(z+1) = 0 \implies 9x + y - 5z = 16$.
59.Answer: Multiply $R_1, R_2, R_3$ by $a, b, c$. Then take common from $C_1, C_2, C_3$. Gives $1+a^2+b^2+c^2$.
60.Answer: $A^T A = I \implies$ Columns form orthonormal basis. $0+a^2+a^2=1 \implies 2a^2=1 \implies a=\pm 1/\sqrt{2}$. $4b^2+b^2+b^2=1 \implies 6b^2=1 \implies b=\pm 1/\sqrt{6}$. $c^2+c^2+c^2=1 \implies 3c^2=1 \implies c=\pm 1/\sqrt{3}$.
