FIRST MID TERM TEST - 2026
Standard XII
MATHEMATICS
Reg. No.
I. Choose the correct answer.
10 x 1 = 10
1. If \( A^T A^{-1} \) is symmetric, then \( A^2 = \)
Solution:
Since \( A^T A^{-1} \) is symmetric, \( (A^T A^{-1})^T = A^T A^{-1} \).\( (A^{-1})^T (A^T)^T = A^T A^{-1} \implies (A^T)^{-1} A = A^T A^{-1} \).
Multiplying on the left by \( A^T \) and right by \( A \):
\( A = (A^T)^2 A^{-1} A \implies A^2 = (A^T)^2 \).
Correct Answer: c) \( (A^T)^2 \)
2. If \( (AB)^{-1} = \begin{bmatrix} 12 & -17 \\ -19 & 27 \end{bmatrix} \) and \( A^{-1} = \begin{bmatrix} 1 & -1 \\ -2 & 3 \end{bmatrix} \), then \( B^{-1} = \)
Solution:
We know \( (AB)^{-1} = B^{-1} A^{-1} \implies B^{-1} = (AB)^{-1} A \).First find \( A = (A^{-1})^{-1} = \frac{1}{3 - 2} \begin{bmatrix} 3 & 1 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 2 & 1 \end{bmatrix} \).
Now, \( B^{-1} = \begin{bmatrix} 12 & -17 \\ -19 & 27 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 36 - 34 & 12 - 17 \\ -57 + 54 & -19 + 27 \end{bmatrix} = \begin{bmatrix} 2 & -5 \\ -3 & 8 \end{bmatrix} \).
Correct Answer: a) \( \begin{bmatrix} 2 & -5 \\ -3 & 8 \end{bmatrix} \)
3. A is a non-singular matrix of order 2 and \( |A| = 5 \), then \( |\text{adj } A| = \)
Solution:
For a square matrix of order \( n \), \( |\text{adj } A| = |A|^{n-1} \).Here \( n = 2 \), so \( |\text{adj } A| = |A|^{2-1} = |A|^1 = 5 \).
Correct Answer: a) 5
4. If \( A = \begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix} \), \( B = \text{adj } A \) and \( C = 3A \), then \( \frac{|\text{adj } B|}{|C|} = \)
Solution:
\( |A| = (3)(2) - (5)(1) = 6 - 5 = 1 \).\( |\text{adj } B| = |\text{adj}(\text{adj } A)| = |A|^{(2-1)^2} = 1^1 = 1 \).
\( |C| = |3A| = 3^2 |A| = 9(1) = 9 \).
Therefore, \( \frac{|\text{adj } B|}{|C|} = \frac{1}{9} \).
Correct Answer: a) \( \frac{1}{9} \)
5. If \( z \) is a non-zero complex number such that \( 2i z^2 = \bar{z} \), then \( |z| \) is
Solution:
Taking modulus on both sides:\( |2i z^2| = |\bar{z}| \implies |2i| |z|^2 = |z| \).
Since \( |2i| = 2 \) and \( z \neq 0 \), we divide by \( |z| \):
\( 2 |z| = 1 \implies |z| = \frac{1}{2} \).
Correct Answer: a) \( \frac{1}{2} \)
6. The principal argument of \( \frac{3}{-1 + i} \) is
Solution:
\( \arg\left(\frac{3}{-1+i}\right) = \arg(3) - \arg(-1+i) \).\( \arg(3) = 0 \).
For \( -1+i \), it lies in the second quadrant: \( \theta = \pi - \tan^{-1}\left(\left|\frac{1}{-1}\right|\right) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).
So, \( \arg = 0 - \frac{3\pi}{4} = -\frac{3\pi}{4} \).
Correct Answer: b) \( \frac{-3\pi}{4} \)
7. The value of \( \left( \frac{1 + \sqrt{3}i}{1 - \sqrt{3}i} \right)^{10} \) is
Solution:
\( 1 + \sqrt{3}i = 2 e^{i\pi/3} \), and \( 1 - \sqrt{3}i = 2 e^{-i\pi/3} \).\( \frac{1 + \sqrt{3}i}{1 - \sqrt{3}i} = e^{i(2\pi/3)} \).
\( \left(e^{i(2\pi/3)}\right)^{10} = e^{i(20\pi/3)} = e^{i(6\pi + 2\pi/3)} = e^{i(2\pi/3)} = \text{CIS} \frac{2\pi}{3} \).
Correct Answer: a) \( \text{CIS} \frac{2\pi}{3} \)
8. Which of the following is not correct?
(Here \( z_1 \) and \( z_2 \) are any two complex numbers)
Solution:
By triangle inequality, \( |z_1 - z_2| \ge ||z_1| - |z_2|| \). It is incorrect to claim that \( |z_1 - z_2| \le |z_1| - |z_2| \).Correct Answer: c) \( |z_1 - z_2| \le |z_1| - |z_2| \)
9. If \( \alpha, \beta \) and \( \gamma \) are the roots of \( x^3 + px^2 + qx + r = 0 \), then \( \sum \frac{1}{\alpha} \) is
Solution:
From Vieta's formulas:\( \alpha\beta + \beta\gamma + \gamma\alpha = q \)
\( \alpha\beta\gamma = -r \)
\( \sum \frac{1}{\alpha} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{q}{-r} = \frac{-q}{r} \).
Correct Answer: a) \( \frac{-q}{r} \)
10. The number of real numbers in \( [0, 2\pi] \) satisfying \( \sin^4 x - 2\sin^2 x + 1 = 0 \) is
Solution:
The equation simplifies as \( (\sin^2 x - 1)^2 = 0 \implies \sin^2 x = 1 \implies \sin x = \pm 1 \).In \( [0, 2\pi] \):
\( \sin x = 1 \implies x = \frac{\pi}{2} \)
\( \sin x = -1 \implies x = \frac{3\pi}{2} \)
There are 2 real solutions.
Correct Answer: a) 2
Part - II
II. Answer any four questions. (Q.No.16 is compulsory)
4 x 2 = 8
11. Prove that \( A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \) is orthogonal.
Solution / Proof:
A matrix \( A \) is orthogonal if \( A A^T = I \).\( A^T = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \)
\( A A^T = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \)
\( A A^T = \begin{bmatrix} \cos^2\theta + \sin^2\theta & \cos\theta\sin\theta - \sin\theta\cos\theta \\ \sin\theta\cos\theta - \cos\theta\sin\theta & \sin^2\theta + \cos^2\theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \).
Hence, \( A \) is orthogonal.
12. Find the rank of \( A = \begin{bmatrix} 1 & -2 & -1 & 0 \\ 3 & -6 & -3 & 1 \end{bmatrix} \)
Solution:
Order of matrix \( A \) is \( 2 \times 4 \). Therefore, \( \text{rank}(A) \le \min(2, 4) = 2 \).Applying elementary row operations:
\( R_2 \to R_2 - 3R_1 \)
\( A \sim \begin{bmatrix} 1 & -2 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \)
The matrix is in row-echelon form and contains 2 non-zero rows.
Rank of A, \( \rho(A) = 2 \).
13. If \( A^{-1} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \), then find \( A \).
Solution:
We know that \( A = (A^{-1})^{-1} \).For a \( 2 \times 2 \) matrix \( M = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), \( M^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \).
Here, \( |A^{-1}| = (1)(4) - (2)(3) = 4 - 6 = -2 \).
\( A = (A^{-1})^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix} \).
14. Find the square roots of \( 4 + 3i \)
Solution:
Formula: \( \sqrt{a + ib} = \pm \left( \sqrt{\frac{|z| + a}{2}} + i \frac{b}{|b|} \sqrt{\frac{|z| - a}{2}} \right) \)Here \( a = 4, b = 3 \). \( |z| = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \).
Since \( b > 0 \), \( \frac{b}{|b|} = 1 \).
\( \sqrt{4 + 3i} = \pm \left( \sqrt{\frac{5 + 4}{2}} + i \sqrt{\frac{5 - 4}{2}} \right) = \pm \left( \frac{3}{\sqrt{2}} + \frac{i}{\sqrt{2}} \right) \).
15. Construct a cubic equation with roots 1, 2 and 3
Solution:
Let roots be \( \alpha = 1, \beta = 2, \gamma = 3 \).\( \sum_1 = \alpha + \beta + \gamma = 1 + 2 + 3 = 6 \)
\( \sum_2 = \alpha\beta + \beta\gamma + \gamma\alpha = (1)(2) + (2)(3) + (3)(1) = 2 + 6 + 3 = 11 \)
\( \sum_3 = \alpha\beta\gamma = (1)(2)(3) = 6 \)
Cubic equation form: \( x^3 - (\sum_1)x^2 + (\sum_2)x - \sum_3 = 0 \)
\( x^3 - 6x^2 + 11x - 6 = 0 \)
16. If \( (x_1 + iy_1)(x_2 + iy_2)(x_3 + iy_3)\cdots(x_n + iy_n) = a + ib \), show that \( (x_1^2 + y_1^2)(x_2^2 + y_2^2)(x_3^2 + y_3^2)\cdots(x_n^2 + y_n^2) = a^2 + b^2 \)
Solution:
Given \( \prod_{r=1}^n (x_r + iy_r) = a + ib \).Taking modulus on both sides:
\( \left| \prod_{r=1}^n (x_r + iy_r) \right| = |a + ib| \implies \prod_{r=1}^n |x_r + iy_r| = |a + ib| \)
\( \prod_{r=1}^n \sqrt{x_r^2 + y_r^2} = \sqrt{a^2 + b^2} \)
Squaring on both sides:
\( (x_1^2 + y_1^2)(x_2^2 + y_2^2)\cdots(x_n^2 + y_n^2) = a^2 + b^2 \). (Proved)
Part - III
III. Answer any four questions. (Q.No.22 is compulsory)
4 x 3 = 12
17. If \( A = \begin{bmatrix} 8 & -4 \\ -5 & 3 \end{bmatrix} \), verify that \( A(\text{adj } A) = (\text{adj } A)A = |A| I_2 \)
Solution:
\( |A| = (8)(3) - (-4)(-5) = 24 - 20 = 4 \).\( |A| I_2 = 4 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} \).
\( \text{adj } A = \begin{bmatrix} 3 & 4 \\ 5 & 8 \end{bmatrix} \).
\( A(\text{adj } A) = \begin{bmatrix} 8 & -4 \\ -5 & 3 \end{bmatrix} \begin{bmatrix} 3 & 4 \\ 5 & 8 \end{bmatrix} = \begin{bmatrix} 24-20 & 32-32 \\ -15+15 & -20+24 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} \).
\( (\text{adj } A)A = \begin{bmatrix} 3 & 4 \\ 5 & 8 \end{bmatrix} \begin{bmatrix} 8 & -4 \\ -5 & 3 \end{bmatrix} = \begin{bmatrix} 24-20 & -12+12 \\ 40-40 & -20+24 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} \).
Thus verified: \( A(\text{adj } A) = (\text{adj } A)A = |A| I_2 \).
18. Show that \( \left(\frac{19+9i}{5-3i}\right)^{15} - \left(\frac{8+i}{1+2i}\right)^{15} \) is purely imaginary.
Solution:
Simplify first term: \( \frac{19+9i}{5-3i} = \frac{(19+9i)(5+3i)}{25+9} = \frac{95+57i+45i-27}{34} = \frac{68+102i}{34} = 2 + 3i \).Simplify second term: \( \frac{8+i}{1+2i} = \frac{(8+i)(1-2i)}{1+4} = \frac{8-16i+i+2}{5} = \frac{10-15i}{5} = 2 - 3i \).
Let \( z = (2+3i)^{15} - (2-3i)^{15} \).
Take conjugate: \( \bar{z} = \overline{(2+3i)^{15}} - \overline{(2-3i)^{15}} = (2-3i)^{15} - (2+3i)^{15} = -z \).
Since \( \bar{z} = -z \), \( z \) is purely imaginary. (Proved)
19. If \( |z| = 1 \), show that \( 2 \le |z^2 - 3| \le 4 \)
Solution:
By triangle inequality: \( ||z_1| - |z_2|| \le |z_1 + z_2| \le |z_1| + |z_2| \).Let \( z_1 = z^2 \) and \( z_2 = -3 \).
Since \( |z| = 1 \), \( |z^2| = |z|^2 = 1^2 = 1 \).
Lower bound: \( |z^2 - 3| \ge ||z^2| - |-3|| = |1 - 3| = |-2| = 2 \).
Upper bound: \( |z^2 - 3| \le |z^2| + |-3| = 1 + 3 = 4 \).
Hence, \( 2 \le |z^2 - 3| \le 4 \).
20. If \( \omega \neq 1 \) is a cube root of unity, show that \( (1+\omega)(1+\omega^2)(1+\omega^4)(1+\omega^8)\cdots(1+\omega^{2^n}) = 1 \)
Solution:
We know \( 1 + \omega + \omega^2 = 0 \implies 1 + \omega = -\omega^2 \) and \( 1 + \omega^2 = -\omega \).Also \( \omega^3 = 1 \).
For powers: \( \omega^4 = \omega \implies 1+\omega^4 = 1+\omega = -\omega^2 \).
\( \omega^8 = \omega^2 \implies 1+\omega^8 = 1+\omega^2 = -\omega \).
Terms pair up as: \( (1+\omega)(1+\omega^2) = (-\omega^2)(-\omega) = \omega^3 = 1 \).
Every adjacent pair evaluates to 1. Thus, the product of all pairs is 1. (Proved)
21. Show that if \( p,q,r \) are rational, the roots of the equation \( x^2 + 2px + p^2 - q^2 - 2pr - r^2 = 0 \) are rational.
Solution:
To show roots are rational, the discriminant \( \Delta = B^2 - 4AC \) must be a perfect square.Here, \( A = 1, B = 2p, C = p^2 - q^2 - 2pr - r^2 \).
\( \Delta = (2p)^2 - 4(1)(p^2 - q^2 - 2pr - r^2) \)
\( \Delta = 4p^2 - 4p^2 + 4q^2 + 8pr + 4r^2 = 4(q^2 + 2pr + r^2) = [2(q+r)]^2 \).
Since \( p, q, r \) are rational, \( \Delta \) is a perfect square, hence the roots are rational.
22. Solve \( 2x + 5y = -2 \), \( x + 2y = -3 \) by matrix inversion method.
Solution:
Matrix form \( AX = B \):\( \begin{bmatrix} 2 & 5 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -2 \\ -3 \end{bmatrix} \)
\( |A| = (2)(2) - (5)(1) = 4 - 5 = -1 \neq 0 \).
\( A^{-1} = \frac{1}{-1} \begin{bmatrix} 2 & -5 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 5 \\ 1 & -2 \end{bmatrix} \).
\( X = A^{-1}B = \begin{bmatrix} -2 & 5 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} -2 \\ -3 \end{bmatrix} = \begin{bmatrix} 4 - 15 \\ -2 + 6 \end{bmatrix} = \begin{bmatrix} -11 \\ 4 \end{bmatrix} \).
\( x = -11, y = 4 \)
Part - IV
IV. Answer all the questions.
4 x 5 = 20
23. a) Solve \( 3x + 3y - z = 11 \), \( 2x - y + 2z = 9 \), \( 4x + 3y + 2z = 25 \) by Cramer's rule.
Solution:
\( \Delta = \begin{vmatrix} 3 & 3 & -1 \\ 2 & -1 & 2 \\ 4 & 3 & 2 \end{vmatrix} = 3(-2-6) - 3(4-8) - 1(6+4) = -24 + 12 - 10 = -22 \).\( \Delta_x = \begin{vmatrix} 11 & 3 & -1 \\ 9 & -1 & 2 \\ 25 & 3 & 2 \end{vmatrix} = 11(-2-6) - 3(18-50) - 1(27+25) = -88 + 96 - 52 = -44 \).
\( \Delta_y = \begin{vmatrix} 3 & 11 & -1 \\ 2 & 9 & 2 \\ 4 & 25 & 2 \end{vmatrix} = 3(18-50) - 11(4-8) - 1(50-36) = -96 + 44 - 14 = -66 \).
\( \Delta_z = \begin{vmatrix} 3 & 3 & 11 \\ 2 & -1 & 9 \\ 4 & 3 & 25 \end{vmatrix} = 3(-25-27) - 3(50-36) + 11(6+4) = -156 - 42 + 110 = -88 \).
\( x = \frac{\Delta_x}{\Delta} = \frac{-44}{-22} = 2 \), \( y = \frac{\Delta_y}{\Delta} = \frac{-66}{-22} = 3 \), \( z = \frac{\Delta_z}{\Delta} = \frac{-88}{-22} = 4 \).
\( x = 2, y = 3, z = 4 \)
(OR)
23. b) Solve the equation \( z^3 + 8i = 0 \)
Solution:
\( z^3 = -8i = 8(-i) = 8 \left( \cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right) \right) \).Add \( 2k\pi \): \( z^3 = 8 \left( \cos\left(2k\pi - \frac{\pi}{2}\right) + i\sin\left(2k\pi - \frac{\pi}{2}\right) \right) \).
Using De Moivre's theorem: \( z = 2 \left( \cos\left(\frac{4k-1}{6}\pi\right) + i\sin\left(\frac{4k-1}{6}\pi\right) \right) \) for \( k = 0, 1, 2 \).
For \( k = 0 \): \( z_1 = 2 \text{cis}\left(-\frac{\pi}{6}\right) = \sqrt{3} - i \).
For \( k = 1 \): \( z_2 = 2 \text{cis}\left(\frac{3\pi}{6}\right) = 2 \text{cis}\left(\frac{\pi}{2}\right) = 2i \).
For \( k = 2 \): \( z_3 = 2 \text{cis}\left(\frac{7\pi}{6}\right) = -\sqrt{3} - i \).
24. a) If \( z = x + iy \) is a complex number such that \( \text{Im}\left(\frac{2z+1}{iz+1}\right) = 0 \), show that the locus of \( z \) is \( 2x^2 + 2y^2 + x - 2y = 0 \).
Solution:
Substitute \( z = x + iy \):\( \frac{2(x+iy)+1}{i(x+iy)+1} = \frac{(2x+1) + i(2y)}{(1-y) + ix} \).
Multiply numerator and denominator by conjugate \( (1-y) - ix \):
Denominator: \( (1-y)^2 + x^2 \).
Numerator imaginary part: \( 2y(1-y) - x(2x+1) = 2y - 2y^2 - 2x^2 - x \).
Since \( \text{Im}\left(\frac{2z+1}{iz+1}\right) = 0 \):
\( 2y - 2y^2 - 2x^2 - x = 0 \implies 2x^2 + 2y^2 + x - 2y = 0 \). (Proved)
(OR)
24. b) If the equations \( x^2 + px + q = 0 \) and \( x^2 + p'x + q' = 0 \) have a common root, show that it must be equal to \( \frac{pq'-p'q}{q-q'} \) or \( \frac{q-q'}{p'-p} \).
Solution:
Let \( \alpha \) be the common root.\( \alpha^2 + p\alpha + q = 0 \) ---(1)
\( \alpha^2 + p'\alpha + q' = 0 \) ---(2)
Subtracting (2) from (1):
\( (p - p')\alpha + (q - q') = 0 \implies \alpha = \frac{q - q'}{p' - p} \).
Eliminating \( \alpha^2 \) using cross-multiplication on (1) & (2):
\( \frac{\alpha^2}{pq' - p'q} = \frac{\alpha}{q - q'} = \frac{1}{p' - p} \implies \alpha = \frac{pq' - p'q}{q - q'} \).
Hence, common root is \( \frac{pq'-p'q}{q-q'} \) or \( \frac{q-q'}{p'-p} \). (Proved)
25. a) If \( 2 + i \) and \( 3 - \sqrt{2} \) are roots of the equation \( x^6 - 13x^5 + 62x^4 - 126x^3 + 65x^2 + 127x - 140 = 0 \), solve the equation.
Solution:
Complex and irrational roots occur in conjugate pairs:Roots are \( 2+i, 2-i \) and \( 3-\sqrt{2}, 3+\sqrt{2} \).
Factor 1: \( [x - (2+i)][x - (2-i)] = (x-2)^2 + 1 = x^2 - 4x + 5 \).
Factor 2: \( [x - (3-\sqrt{2})][x - (3+\sqrt{2})] = (x-3)^2 - 2 = x^2 - 6x + 7 \).
Combined factor: \( (x^2 - 4x + 5)(x^2 - 6x + 7) = x^4 - 10x^3 + 36x^2 - 58x + 35 \).
Dividing original polynomial by this product gives remaining factor: \( x^2 - 3x - 4 = 0 \).
\( (x - 4)(x + 1) = 0 \implies x = 4, -1 \).
Roots: \( 2 \pm i, 3 \pm \sqrt{2}, 4, -1 \)
(OR)
25. b) If \( A = \begin{bmatrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & 1 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{bmatrix} \), find \( AB \) and \( BA \) and hence solve the system of equations \( x + y + 2z = 1 \), \( 3x + 2y + z = 7 \), \( 2x + y + 3z = 2 \).
Solution:
\( AB = \begin{bmatrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} = 4I_3 \).\( BA = 4I_3 \).
Thus, \( B^{-1} = \frac{1}{4} A \).
The given system in matrix form is \( B X = C \), where \( C = \begin{bmatrix} 1 \\ 7 \\ 2 \end{bmatrix} \).
\( X = B^{-1}C = \frac{1}{4} \begin{bmatrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 7 \\ 2 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} -5 + 7 + 6 \\ 7 + 7 - 10 \\ 1 + 7 + 2 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 8 \\ 4 \\ 10 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ \frac{5}{2} \end{bmatrix} \).
\( x = 2, y = 1, z = \frac{5}{2} \)
26. a) Determine the values of \( \lambda \) for which the following system of equations \( x + y + 3z = 0 \), \( 4x + 3y + \lambda z = 0 \), \( 2x + y + 2z = 0 \) has (i) a unique solution (ii) a non-trivial solution.
Solution:
Coefficient determinant \( \Delta = \begin{vmatrix} 1 & 1 & 3 \\ 4 & 3 & \lambda \\ 2 & 1 & 2 \end{vmatrix} \).\( \Delta = 1(6 - \lambda) - 1(8 - 2\lambda) + 3(4 - 6) = 6 - \lambda - 8 + 2\lambda - 6 = \lambda - 8 \).
(i) For unique solution (trivial solution \( x=0, y=0, z=0 \)):
\( \Delta \neq 0 \implies \lambda - 8 \neq 0 \implies \mathbf{\lambda \neq 8} \).
(ii) For non-trivial solution:
\( \Delta = 0 \implies \lambda - 8 = 0 \implies \mathbf{\lambda = 8} \).
(OR)
26. b) If \( \cos\alpha + \cos\beta + \cos\gamma = \sin\alpha + \sin\beta + \sin\gamma = 0 \), show that:
i) \( \cos 3\alpha + \cos 3\beta + \cos 3\gamma = 3\cos(\alpha + \beta + \gamma) \)
ii) \( \sin 3\alpha + \sin 3\beta + \sin 3\gamma = 3\sin(\alpha + \beta + \gamma) \)
i) \( \cos 3\alpha + \cos 3\beta + \cos 3\gamma = 3\cos(\alpha + \beta + \gamma) \)
ii) \( \sin 3\alpha + \sin 3\beta + \sin 3\gamma = 3\sin(\alpha + \beta + \gamma) \)
Solution:
Let \( a = e^{i\alpha}, b = e^{i\beta}, c = e^{i\gamma} \).\( a + b + c = (\cos\alpha + \cos\beta + \cos\gamma) + i(\sin\alpha + \sin\beta + \sin\gamma) = 0 + i(0) = 0 \).
Algebraic Identity: If \( a + b + c = 0 \), then \( a^3 + b^3 + c^3 = 3abc \).
Substitute values:
\( (e^{i\alpha})^3 + (e^{i\beta})^3 + (e^{i\gamma})^3 = 3 e^{i\alpha} e^{i\beta} e^{i\gamma} \)
\( e^{i3\alpha} + e^{i3\beta} + e^{i3\gamma} = 3 e^{i(\alpha + \beta + \gamma)} \)
Expand using Euler's formula:
\( (\cos 3\alpha + i\sin 3\alpha) + (\cos 3\beta + i\sin 3\beta) + (\cos 3\gamma + i\sin 3\gamma) = 3[\cos(\alpha+\beta+\gamma) + i\sin(\alpha+\beta+\gamma)] \).
Equating real parts: \( \cos 3\alpha + \cos 3\beta + \cos 3\gamma = 3\cos(\alpha + \beta + \gamma) \). (Proved)
Equating imaginary parts: \( \sin 3\alpha + \sin 3\beta + \sin 3\gamma = 3\sin(\alpha + \beta + \gamma) \). (Proved)
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