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Punjab Board Class 11 Mathematics

Exercise 4.3 Solutions — Matrices and Determinants

Complete worked solutions for Exercise 4.3 from Chapter 4 (Matrices and Determinants), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 4.3 Solutions

Worked solutions for Punjab Board Class 11 Mathematics, Chapter 4: Matrices and Determinants.

Question 1: Find the inverses of the following matrices by using row operations

Row operations se inverse nikalne ke liye matrix ke sath identity matrix augment karte hain \([A \mid I]\), phir operations laga kar left side ko identity banate hain. Jab left side \(I\) ban jaye to right side par \(A^{-1}\) aa jata hai.

(i) \(A = \begin{bmatrix} 2 & 6 & -3 \\ 0 & -2 & 0 \\ -2 & 5 & 6 \end{bmatrix}\)

\[ \left[\begin{array}{ccc|ccc} 2 & 6 & -3 & 1 & 0 & 0 \\ 0 & -2 & 0 & 0 & 1 & 0 \\ -2 & 5 & 6 & 0 & 0 & 1 \end{array}\right] \]

\(R_3 \to R_3 + R_1\), \(R_1 \to \frac{1}{2}R_1\), \(R_2 \to -\frac{1}{2}R_2\):

\[ \left[\begin{array}{ccc|ccc} 1 & 3 & -3/2 & 1/2 & 0 & 0 \\ 0 & 1 & 0 & 0 & -1/2 & 0 \\ 0 & 11 & 3 & 1 & 0 & 1 \end{array}\right] \]

\(R_1 \to R_1 - 3R_2\), \(R_3 \to R_3 - 11R_2\), \(R_3 \to \frac{1}{3}R_3\):

\[ \left[\begin{array}{ccc|ccc} 1 & 0 & -3/2 & 1/2 & 3/2 & 0 \\ 0 & 1 & 0 & 0 & -1/2 & 0 \\ 0 & 0 & 1 & 1/3 & 11/6 & 1/3 \end{array}\right] \]

\(R_1 \to R_1 + \frac{3}{2}R_3\):

\[ \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 17/4 & 1/2 \\ 0 & 1 & 0 & 0 & -1/2 & 0 \\ 0 & 0 & 1 & 1/3 & 11/6 & 1/3 \end{array}\right] \]
Answer: \(A^{-1} = \begin{bmatrix} 1 & 17/4 & 1/2 \\ 0 & -1/2 & 0 \\ 1/3 & 11/6 & 1/3 \end{bmatrix}\)

(ii) \(A = \begin{bmatrix} 1 & 2 & -1 \\ 0 & -2 & 8 \\ 1 & 0 & 2 \end{bmatrix}\)

\[ \left[\begin{array}{ccc|ccc} 1 & 2 & -1 & 1 & 0 & 0 \\ 0 & -2 & 8 & 0 & 1 & 0 \\ 1 & 0 & 2 & 0 & 0 & 1 \end{array}\right] \]

\(R_3 \to R_3 - R_1\), \(R_3 \to R_3 - R_2\), \(R_2 \to -\frac{1}{2}R_2\):

\[ \left[\begin{array}{ccc|ccc} 1 & 2 & -1 & 1 & 0 & 0 \\ 0 & 1 & -4 & 0 & -1/2 & 0 \\ 0 & 0 & -5 & -1 & -1 & 1 \end{array}\right] \]

\(R_1 \to R_1 - 2R_2\), \(R_3 \to -\frac{1}{5}R_3\), \(R_1 \to R_1 - 7R_3\), \(R_2 \to R_2 + 4R_3\):

\[ \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & -2/5 & -2/5 & 7/5 \\ 0 & 1 & 0 & 4/5 & 3/10 & -4/5 \\ 0 & 0 & 1 & 1/5 & 1/5 & -1/5 \end{array}\right] \]
Answer: \(A^{-1} = \begin{bmatrix} -2/5 & -2/5 & 7/5 \\ 4/5 & 3/10 & -4/5 \\ 1/5 & 1/5 & -1/5 \end{bmatrix}\)

(iii) \(A = \begin{bmatrix} 1 & 6 & 2 \\ 2 & 13 & 0 \\ 0 & -1 & 1 \end{bmatrix}\)

\[ \left[\begin{array}{ccc|ccc} 1 & 6 & 2 & 1 & 0 & 0 \\ 2 & 13 & 0 & 0 & 1 & 0 \\ 0 & -1 & 1 & 0 & 0 & 1 \end{array}\right] \]

\(R_2 \to R_2 - 2R_1\), \(R_1 \to R_1 - 6R_2\), \(R_3 \to R_3 + R_2\), \(R_3 \to -\frac{1}{3}R_3\):

\[ \left[\begin{array}{ccc|ccc} 1 & 0 & 26 & 13 & -6 & 0 \\ 0 & 1 & -4 & -2 & 1 & 0 \\ 0 & 0 & 1 & 2/3 & -1/3 & -1/3 \end{array}\right] \]

\(R_1 \to R_1 - 26R_3\), \(R_2 \to R_2 + 4R_3\):

\[ \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & -13/3 & 8/3 & 26/3 \\ 0 & 1 & 0 & 2/3 & -1/3 & -4/3 \\ 0 & 0 & 1 & 2/3 & -1/3 & -1/3 \end{array}\right] \]
Answer: \(A^{-1} = \begin{bmatrix} -13/3 & 8/3 & 26/3 \\ 2/3 & -1/3 & -4/3 \\ 2/3 & -1/3 & -1/3 \end{bmatrix}\)

Common mistake: Row operations mein fraction arithmetic ghalat hona aam hai — har step ke baad left side check karen ke woh identity ki taraf barh rahi hai. (Note: part (i) ke printed key mein \(17/4\) ki jagah \(17/2\) likha hai; upar di gayi value verify-shuda durust hai.)

Question 2: Find the rank of the following matrices

Rank nikalne ke liye matrix ko row operations se echelon form mein lao. Jo rows poori tarah zero ban jayein unhein ignore kar do — baqi non-zero rows ki ginti hi rank hai.

(i) \(A = \begin{bmatrix} 1 & -1 & 3 & 1 \\ -2 & -6 & 1 & -1 \\ 3 & 1 & 4 & -2 \end{bmatrix}\)

\(R_2 \to R_2 + 2R_1\), \(R_3 \to R_3 - 3R_1\):

\[ \begin{bmatrix} 1 & -1 & 3 & 1 \\ 0 & -8 & 7 & 1 \\ 0 & 4 & -5 & -5 \end{bmatrix} \]

\(R_2 \to -\frac{1}{8}R_2\), \(R_3 \to R_3 - 4R_2\):

\[ \begin{bmatrix} 1 & -1 & 3 & 1 \\ 0 & 1 & -7/8 & -1/8 \\ 0 & 0 & -3/2 & -9/2 \end{bmatrix} \]

Three non-zero rows, so rank = 3.

Answer: Rank \(= 3\)

(ii) \(A = \begin{bmatrix} 1 & -2 & 3 \\ 2 & -4 & 6 \\ -1 & 0 & 2 \\ 0 & 1 & -1 \end{bmatrix}\)

\(R_2 \to R_2 - 2R_1\), \(R_3 \to R_3 + R_1\):

\[ \begin{bmatrix} 1 & -2 & 3 \\ 0 & 0 & 0 \\ 0 & -2 & 5 \\ 0 & 1 & -1 \end{bmatrix} \]

\(R_2 \leftrightarrow R_4\), \(R_3 \to R_3 + 2R_2\):

\[ \begin{bmatrix} 1 & -2 & 3 \\ 0 & 1 & -1 \\ 0 & 0 & 3 \\ 0 & 0 & 0 \end{bmatrix} \]

Three non-zero rows, so rank = 3.

Answer: Rank \(= 3\)

(iii) \(A = \begin{bmatrix} 3 & -1 & 3 & 0 & 1 \\ 1 & 2 & -1 & -3 & -2 \\ 2 & 3 & 4 & 2 & 1 \\ 2 & 5 & -2 & -3 & 3 \end{bmatrix}\)

\(R_1 \leftrightarrow R_2\); then \(R_2 \to R_2 - 3R_1\), \(R_3 \to R_3 - 2R_1\), \(R_4 \to R_4 - 2R_1\):

\[ \begin{bmatrix} 1 & 2 & -1 & -3 & -2 \\ 0 & -7 & 6 & 9 & 7 \\ 0 & -1 & 6 & 8 & 5 \\ 0 & 1 & 0 & 3 & 7 \end{bmatrix} \]

\(R_2 \leftrightarrow R_4\); then \(R_3 \to R_3 + R_2\), \(R_4 \to R_4 + 7R_2\):

\[ \begin{bmatrix} 1 & 2 & -1 & -3 & -2 \\ 0 & 1 & 0 & 3 & 7 \\ 0 & 0 & 6 & 11 & 12 \\ 0 & 0 & 6 & 30 & 56 \end{bmatrix} \]

\(R_4 \to R_4 - R_3\):

\[ \begin{bmatrix} 1 & 2 & -1 & -3 & -2 \\ 0 & 1 & 0 & 3 & 7 \\ 0 & 0 & 6 & 11 & 12 \\ 0 & 0 & 0 & 19 & 44 \end{bmatrix} \]

Four non-zero rows, so rank = 4.

Answer: Rank \(= 4\)

Question 3: Solve the following systems of linear equations by Cramer's rule

Cramer's rule mein pehle coefficient matrix ka determinant \(D\) nikalte hain. Phir har variable ke liye us ke column ko constants se replace kar ke \(D_1, D_2, D_3\) nikalte hain. Aakhir mein \(x = D_1/D\), \(y = D_2/D\), \(z = D_3/D\).

(i) \(2x + y - z = 1\), \(x - y + 2z = 3\), \(3x + 2y + z = 4\)

\[ D = \begin{vmatrix} 2 & 1 & -1 \\ 1 & -1 & 2 \\ 3 & 2 & 1 \end{vmatrix} = 2(-1-4) - 1(1-6) - 1(2+3) = -10 \]
\[ D_1 = \begin{vmatrix} 1 & 1 & -1 \\ 3 & -1 & 2 \\ 4 & 2 & 1 \end{vmatrix} = -5 + 5 - 10 = -10, \quad D_2 = \begin{vmatrix} 2 & 1 & -1 \\ 1 & 3 & 2 \\ 3 & 4 & 1 \end{vmatrix} = -10 + 5 + 5 = 0 \]
\[ D_3 = \begin{vmatrix} 2 & 1 & 1 \\ 1 & -1 & 3 \\ 3 & 2 & 4 \end{vmatrix} = -20 + 5 + 5 = -10 \]
\[ x = \frac{-10}{-10} = 1, \quad y = \frac{0}{-10} = 0, \quad z = \frac{-10}{-10} = 1 \]
Answer: \(x = 1\), \(y = 0\), \(z = 1\)

(ii) \(x_1 + 2x_2 - 3x_3 = 0\), \(4x_1 - x_2 + x_3 = 5\), \(-2x_1 + 3x_2 + 2x_3 = 3\)

\[ D = \begin{vmatrix} 1 & 2 & -3 \\ 4 & -1 & 1 \\ -2 & 3 & 2 \end{vmatrix} = -5 - 20 - 30 = -55 \]
\[ D_1 = \begin{vmatrix} 0 & 2 & -3 \\ 5 & -1 & 1 \\ 3 & 3 & 2 \end{vmatrix} = -14 - 54 = -68, \quad D_2 = \begin{vmatrix} 1 & 0 & -3 \\ 4 & 5 & 1 \\ -2 & 3 & 2 \end{vmatrix} = 7 - 66 = -59 \]
\[ D_3 = \begin{vmatrix} 1 & 2 & 0 \\ 4 & -1 & 5 \\ -2 & 3 & 3 \end{vmatrix} = -18 - 44 = -62 \]
\[ x_1 = \frac{68}{55}, \quad x_2 = \frac{59}{55}, \quad x_3 = \frac{62}{55} \]
Answer: \(x_1 = \frac{68}{55}\), \(x_2 = \frac{59}{55}\), \(x_3 = \frac{62}{55}\)

(iii) \(2x_1 - x_2 + x_3 = 1\), \(x_1 + 2x_2 + 2x_3 = 2\), \(x_1 - 2x_2 - x_3 = 1\)

\[ D = \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & 2 \\ 1 & -2 & -1 \end{vmatrix} = 4 - 3 - 4 = -3 \]
\[ D_1 = \begin{vmatrix} 1 & -1 & 1 \\ 2 & 2 & 2 \\ 1 & -2 & -1 \end{vmatrix} = 2 - 4 - 6 = -8, \quad D_2 = \begin{vmatrix} 2 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 1 & -1 \end{vmatrix} = -8 + 3 - 1 = -6 \]
\[ D_3 = \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & 2 \\ 1 & -2 & 1 \end{vmatrix} = 12 - 1 - 4 = 7 \]
\[ x_1 = \frac{8}{3}, \quad x_2 = 2, \quad x_3 = -\frac{7}{3} \]
Answer: \(x_1 = \frac{8}{3}\), \(x_2 = 2\), \(x_3 = -\frac{7}{3}\)

Common mistake: \(D_1, D_2, D_3\) mein ghalat column replace karna. \(D_1\) ke liye pehla column, \(D_2\) ke liye doosra, \(D_3\) ke liye teesra column constants se badlo.

Question 4: Solve the following systems of linear equations by matrix inversion method

System ko \(AX = B\) ki form mein likhenge. Phir \(A^{-1} = \frac{1}{|A|}\text{adj } A\) nikal kar \(X = A^{-1}B\) se solution hasil karenge. Adjoint cofactor matrix ka transpose hota hai.

(i) \(x - 2y + z = -1\), \(3x + y - 2z = 4\), \(y - z = 1\)

\(A = \begin{bmatrix} 1 & -2 & 1 \\ 3 & 1 & -2 \\ 0 & 1 & -1 \end{bmatrix}\), \(B = \begin{bmatrix} -1 \\ 4 \\ 1 \end{bmatrix}\).

\[ |A| = 1(1) + 2(-3) + 3 = -2 \neq 0 \]

Cofactors: \(A_{11} = 1\), \(A_{12} = 3\), \(A_{13} = 3\), \(A_{21} = -1\), \(A_{22} = -1\), \(A_{23} = -1\), \(A_{31} = 3\), \(A_{32} = 5\), \(A_{33} = 7\).

\[ \text{adj } A = \begin{bmatrix} 1 & -1 & 3 \\ 3 & -1 & 5 \\ 3 & -1 & 7 \end{bmatrix}, \quad X = A^{-1}B = \frac{1}{-2}\begin{bmatrix} 1 & -1 & 3 \\ 3 & -1 & 5 \\ 3 & -1 & 7 \end{bmatrix}\begin{bmatrix} -1 \\ 4 \\ 1 \end{bmatrix} = \frac{1}{-2}\begin{bmatrix} -2 \\ -2 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \]
Answer: \(x = 1\), \(y = 1\), \(z = 0\)

(ii) \(2x_1 + x_2 - 3x_3 = 0\), \(x_1 + 3x_2 - 2x_3 = 0\), \(-3x_1 - x_2 + 2x_3 = 4\)

\(A = \begin{bmatrix} 2 & 1 & -3 \\ 1 & 3 & -2 \\ -3 & -1 & 2 \end{bmatrix}\), \(B = \begin{bmatrix} 0 \\ 0 \\ 4 \end{bmatrix}\).

\[ |A| = 2(4) - 1(-4) - 3(8) = -12 \neq 0 \]

Cofactor matrix \(= \begin{bmatrix} 4 & 4 & 8 \\ 1 & -5 & -1 \\ 7 & 1 & 5 \end{bmatrix}\), so \(\text{adj } A = \begin{bmatrix} 4 & 1 & 7 \\ 4 & -5 & 1 \\ 8 & -1 & 5 \end{bmatrix}\).

\[ X = \frac{1}{-12}\begin{bmatrix} 4 & 1 & 7 \\ 4 & -5 & 1 \\ 8 & -1 & 5 \end{bmatrix}\begin{bmatrix} 0 \\ 0 \\ 4 \end{bmatrix} = \frac{1}{-12}\begin{bmatrix} 28 \\ 4 \\ 20 \end{bmatrix} = \begin{bmatrix} -7/3 \\ -1/3 \\ -5/3 \end{bmatrix} \]
Answer: \(x_1 = -\frac{7}{3}\), \(x_2 = -\frac{1}{3}\), \(x_3 = -\frac{5}{3}\)

(iii) \(x + y = 2\), \(2x - z = 1\), \(2y - 3z = -1\)

\(A = \begin{bmatrix} 1 & 1 & 0 \\ 2 & 0 & -1 \\ 0 & 2 & -3 \end{bmatrix}\), \(B = \begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}\).

\[ |A| = 1(2) - 1(-6) + 0 = 8 \neq 0 \]

Cofactors: \(A_{11} = 2\), \(A_{12} = 6\), \(A_{13} = 4\), \(A_{21} = 3\), \(A_{22} = -3\), \(A_{23} = -2\), \(A_{31} = -1\), \(A_{32} = 1\), \(A_{33} = -2\).

\[ X = \frac{1}{8}\begin{bmatrix} 2 & 3 & -1 \\ 6 & -3 & 1 \\ 4 & -2 & -2 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix} = \frac{1}{8}\begin{bmatrix} 8 \\ 8 \\ 8 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \]
Answer: \(x = 1\), \(y = 1\), \(z = 1\)

Question 5: Solve the following systems by reducing their augmented matrices to echelon form and reduced echelon form

Augmented matrix \([A \mid B]\) banao, row operations se echelon form mein lao, phir back substitution se values nikalo. Reduced echelon form tak le jane par solution direct parha ja sakta hai.

(i) \(x_1 + 2x_2 - 2x_3 = -1\), \(2x_1 + 3x_2 + x_3 = 1\), \(5x_1 + 4x_2 - 3x_3 = -1\)

\[ \left[\begin{array}{ccc|c} 1 & 2 & -2 & -1 \\ 2 & 3 & 1 & 1 \\ 5 & 4 & -3 & -1 \end{array}\right] \]

\(R_2 \to R_2 - 2R_1\), \(R_3 \to R_3 - 5R_1\), \(R_2 \to -R_2\), \(R_3 \to R_3 + 6R_2\), \(R_3 \to -\frac{1}{23}R_3\):

\[ \left[\begin{array}{ccc|c} 1 & 2 & -2 & -1 \\ 0 & 1 & -5 & -3 \\ 0 & 0 & 1 & 14/23 \end{array}\right] \]

Back substitution: \(x_3 = \frac{14}{23}\); \(x_2 = -3 + 5\cdot\frac{14}{23} = \frac{1}{23}\); \(x_1 = -1 - 2\cdot\frac{1}{23} + 2\cdot\frac{14}{23} = \frac{3}{23}\).

Answer: \(x_1 = \frac{3}{23}\), \(x_2 = \frac{1}{23}\), \(x_3 = \frac{14}{23}\)

(ii) \(x + 2y + z = 2\), \(2x + y + 2z = 3\), \(2x + 3y - z = 7\)

\[ \left[\begin{array}{ccc|c} 1 & 2 & 1 & 2 \\ 2 & 1 & 2 & 3 \\ 2 & 3 & -1 & 7 \end{array}\right] \]

\(R_2 \to R_2 - 2R_1\), \(R_3 \to R_3 - 2R_1\), \(R_3 \to 3R_3 - R_2\), \(R_3 \to -\frac{1}{9}R_3\), \(R_2 \to -\frac{1}{3}R_2\):

\[ \left[\begin{array}{ccc|c} 1 & 2 & 1 & 2 \\ 0 & 1 & 0 & 1/3 \\ 0 & 0 & 1 & -10/9 \end{array}\right] \]

Reduced form via \(R_1 \to R_1 - 2R_2 - R_3\) gives \(x = \frac{22}{9}\), \(y = \frac{1}{3}\), \(z = -\frac{10}{9}\).

Answer: \(x = \frac{22}{9}\), \(y = \frac{1}{3}\), \(z = -\frac{10}{9}\)

(iii) \(x_1 + 4x_2 + x_3 = 2\), \(2x_1 + 2x_2 - 2x_3 = 9\), \(3x_1 + x_2 - x_3 = 12\)

\[ \left[\begin{array}{ccc|c} 1 & 4 & 1 & 2 \\ 2 & 2 & -2 & 9 \\ 3 & 1 & -1 & 12 \end{array}\right] \]

\(R_2 \to R_2 - 2R_1\), \(R_3 \to R_3 - 3R_1\), \(R_2 \to -\frac{1}{6}R_2\), \(R_3 \to R_3 + 11R_2\), \(R_3 \to \frac{3}{10}R_3\):

\[ \left[\begin{array}{ccc|c} 1 & 4 & 1 & 2 \\ 0 & 1 & 2/3 & -5/6 \\ 0 & 0 & 1 & -19/20 \end{array}\right] \]

Back substitution: \(x_3 = -\frac{19}{20}\); \(x_2 = -\frac{5}{6} - \frac{2}{3}\left(-\frac{19}{20}\right) = -\frac{1}{5}\); \(x_1 = 2 - 4\left(-\frac{1}{5}\right) - \left(-\frac{19}{20}\right) = \frac{15}{4}\).

Check in all three equations: \(\frac{15}{4} + 4\left(-\frac{1}{5}\right) - \frac{19}{20} = 2\) ✓; \(2\left(\frac{15}{4}\right) + 2\left(-\frac{1}{5}\right) - 2\left(-\frac{19}{20}\right) = 9\) ✓; \(3\left(\frac{15}{4}\right) - \frac{1}{5} + \frac{19}{20} = 12\) ✓.

Answer: \(x_1 = \frac{15}{4}\), \(x_2 = -\frac{1}{5}\), \(x_3 = -\frac{19}{20}\)

Common mistake: Back substitution mein fraction arithmetic ghalat hona. Hamesha final values ko original equations mein rakh kar verify karen. (Note: part (iii) ke printed key ki values teenon equations ko satisfy nahi kartin; upar di gayi values verify-shuda durust hain.)

Question 6: Solve the following systems of homogeneous linear equations by using Gaussian elimination method

Homogeneous system mein constants sab zero hote hain. Echelon form banao aur rank dekho: agar rank = variables ki tadad to sirf trivial solution \((0,0,0)\); agar rank kam ho to aik variable ko parameter \(t\) rakho aur baqi us mein express karo.

(i) \(x + 4y - 2z = 0\), \(2x + y + 5z = 0\), \(5x + 2y + 8z = 0\)

\[ \left[\begin{array}{ccc|c} 1 & 4 & -2 & 0 \\ 2 & 1 & 5 & 0 \\ 5 & 2 & 8 & 0 \end{array}\right] \]

\(R_2 \to R_2 - 2R_1\), \(R_3 \to R_3 - 5R_1\), \(R_3 \to R_3 - \frac{18}{7}R_2\):

\[ \left[\begin{array}{ccc|c} 1 & 4 & -2 & 0 \\ 0 & -7 & 9 & 0 \\ 0 & 0 & -36/7 & 0 \end{array}\right] \]

Rank \(= 3 =\) number of variables, so only the trivial solution.

Answer: \(x = 0\), \(y = 0\), \(z = 0\) (trivial solution)

(ii) \(x_1 + 4x_2 + 2x_3 = 0\), \(2x_1 + x_2 - 3x_3 = 0\), \(3x_1 + 2x_2 - 4x_3 = 0\)

\[ \left[\begin{array}{ccc|c} 1 & 4 & 2 & 0 \\ 2 & 1 & -3 & 0 \\ 3 & 2 & -4 & 0 \end{array}\right] \]

\(R_2 \to R_2 - 2R_1\), \(R_3 \to R_3 - 3R_1\), \(R_2 \to -\frac{1}{7}R_2\), \(R_3 \to R_3 + 10R_2\):

\[ \left[\begin{array}{ccc|c} 1 & 4 & 2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

Rank \(= 2 < 3\), so infinitely many solutions. Let \(x_3 = t\): then \(x_2 = -t\) and \(x_1 = 2t\).

Answer: \(x_1 = 2t\), \(x_2 = -t\), \(x_3 = t\) for any real \(t\)

(iii) \(x_1 + 2x_2 - x_3 = 0\), \(x_1 - x_2 + 5x_3 = 0\), \(2x_1 + x_2 + 4x_3 = 0\)

\[ \left[\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 1 & -1 & 5 & 0 \\ 2 & 1 & 4 & 0 \end{array}\right] \]

\(R_2 \to R_2 - R_1\), \(R_3 \to R_3 - 2R_1\), \(R_3 \to R_3 - R_2\):

\[ \left[\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & -3 & 6 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

Rank \(= 2 < 3\), so infinitely many solutions. Let \(x_3 = t\): then \(x_2 = 2t\) and \(x_1 = -3t\).

Answer: \(x_1 = -3t\), \(x_2 = 2t\), \(x_3 = t\) for any real \(t\)

Question 7: A triangle has vertices at \(A(4, 1)\), \(B(-2, 5)\) and \(C(0, -3)\). Find the vertices of the reflected triangle over the y-axis using a transformation matrix.

y-axis par reflection ke liye transformation matrix \(\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\) hoti hai. Har vertex ko column matrix bana kar is se multiply karen — x-coordinate ka sign badal jayega, y-coordinate same rahega.

Transformation matrix for reflection over the y-axis: \(T = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\).

\[ A' = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 4 \\ 1 \end{bmatrix} = \begin{bmatrix} -4 \\ 1 \end{bmatrix}, \quad B' = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} -2 \\ 5 \end{bmatrix} = \begin{bmatrix} 2 \\ 5 \end{bmatrix}, \quad C' = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 0 \\ -3 \end{bmatrix} = \begin{bmatrix} 0 \\ -3 \end{bmatrix} \]
Answer: The reflected vertices are \(A'(-4, 1)\), \(B'(2, 5)\), \(C'(0, -3)\).

Question 8: The point \(A\) is mapped to \((30, 20, -5)\) by the scaling matrix \(P = \text{diag}(5, -5, -5)\). Find the coordinates of \(A\).

Mapping ka matlab \(P \cdot A = A'\) hai. Diagonal matrix ke sath multiplication har coordinate ko us ke diagonal entry se multiply kar deta hai, is liye teenon coordinates alag alag equations se mil jayenge.

\[ \begin{bmatrix} 5 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & -5 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 30 \\ 20 \\ -5 \end{bmatrix} \]
\[ 5x = 30 \implies x = 6, \qquad -5y = 20 \implies y = -4, \qquad -5z = -5 \implies z = 1 \]

Common mistake: Sign ka khayal rakhen — doosre aur teesre diagonal entries \(-5\) hain. (Note: printed key mein \(x = -6\) likha hai, jo di gayi matrix ke sath durust nahi baithta; sahi value \(x = 6\) hai.)

Answer: \(A = (6, -4, 1)\)

Question 9: Find the equation of the image of the curve with equation \(y = x^2\) under the transformation with associated matrix \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\)

Transformation \(\begin{bmatrix} x' \\ y' \end{bmatrix} = T\begin{bmatrix} x \\ y \end{bmatrix}\) deti hai. Hum \(T^{-1}\) nikal kar \(x\) aur \(y\) ko \(x', y'\) mein express karenge, phir \(y = x^2\) mein substitute karenge.

\(\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}\), so \(x' = x + 2y\) and \(y' = 3x + 4y\).

Since \(|T| = 4 - 6 = -2\):

\[ T^{-1} = \frac{1}{-2}\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix} \]
\[ x = -2x' + y', \qquad y = \frac{3x' - y'}{2} \]

Substituting in \(y = x^2\):

\[ \frac{3x' - y'}{2} = (-2x' + y')^2 = (y' - 2x')^2 \]
Answer: The image has equation \(\frac{3x' - y'}{2} = (y' - 2x')^2\).

Question 10: Use the matrix \(A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & -1 & 0 \\ 0 & 0 & 2 \end{bmatrix}\) to encode the message "KEEP IT UP", where A to Z correspond to the numbers 1 to 26

Pehle letters ko numbers mein badlo (K=11, E=5, P=16, I=9, T=20, U=21). Phir inhein 3-3 ke groups mein column matrices banao (aakhri group mein 0 se pad karo). Aakhir mein encoding matrix \(A\) se har column ko multiply karo.

K = 11, E = 5, P = 16, I = 9, T = 20, U = 21. Groups: \(\begin{bmatrix} 11 \\ 5 \\ 5 \end{bmatrix}\) (KEE), \(\begin{bmatrix} 16 \\ 9 \\ 20 \end{bmatrix}\) (PIT), \(\begin{bmatrix} 21 \\ 16 \\ 0 \end{bmatrix}\) (UP, padded).

\[ A\begin{bmatrix} 11 \\ 5 \\ 5 \end{bmatrix} = \begin{bmatrix} 11 \\ 22-5 \\ 10 \end{bmatrix} = \begin{bmatrix} 11 \\ 17 \\ 10 \end{bmatrix}, \quad A\begin{bmatrix} 16 \\ 9 \\ 20 \end{bmatrix} = \begin{bmatrix} 16 \\ 32-9 \\ 40 \end{bmatrix} = \begin{bmatrix} 16 \\ 23 \\ 40 \end{bmatrix}, \quad A\begin{bmatrix} 21 \\ 16 \\ 0 \end{bmatrix} = \begin{bmatrix} 21 \\ 42-16 \\ 0 \end{bmatrix} = \begin{bmatrix} 21 \\ 26 \\ 0 \end{bmatrix} \]

Common mistake: Grouping ya padding ghalat hona. (Note: printed key ki encoded matrix di gayi \(A\) aur message se match nahi karti; upar wali calculation verify-shuda durust hai.)

Answer: The encoded message matrix is \(\begin{bmatrix} 11 & 16 & 21 \\ 17 & 23 & 26 \\ 10 & 40 & 0 \end{bmatrix}\).

Question 11: Decode the message \(\begin{bmatrix} 11 & 25 & 22 \\ 20 & 10 & 14 \\ 43 & 41 & 41 \end{bmatrix}\) that was encoded using matrix \(A = \begin{bmatrix} 1 & 1 & -1 \\ 1 & 0 & 1 \\ 2 & 1 & 1 \end{bmatrix}\), where 1 to 26 correspond to A to Z, and 27 represents a space

Decode karne ke liye pehle encoding matrix \(A\) ka inverse nikalenge (\(A^{-1} = \frac{1}{|A|}\text{adj } A\)), phir usay coded matrix se multiply karenge. Jo numbers aayenge unhein wapas letters mein badal lenge.

\(|A| = 1(0-1) - 1(1-2) - 1(1-0) = -1\). Cofactors: \(C_{11} = -1\), \(C_{12} = 1\), \(C_{13} = 1\), \(C_{21} = -2\), \(C_{22} = 3\), \(C_{23} = 1\), \(C_{31} = 1\), \(C_{32} = -2\), \(C_{33} = -1\).

\[ A^{-1} = \frac{1}{-1}\begin{bmatrix} -1 & -2 & 1 \\ 1 & 3 & -2 \\ 1 & 1 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & -1 \\ -1 & -3 & 2 \\ -1 & -1 & 1 \end{bmatrix} \]

Decoded matrix \(= A^{-1} \times\) (coded matrix):

\[ \begin{bmatrix} 1 & 2 & -1 \\ -1 & -3 & 2 \\ -1 & -1 & 1 \end{bmatrix}\begin{bmatrix} 11 & 25 & 22 \\ 20 & 10 & 14 \\ 43 & 41 & 41 \end{bmatrix} = \begin{bmatrix} 8 & 4 & 9 \\ 15 & 27 & 18 \\ 12 & 6 & 5 \end{bmatrix} \]

Converting to letters: column 1: 8, 15, 12 → H, O, L; column 2: 4, 27, 6 → D, (space), F; column 3: 9, 18, 5 → I, R, E.

Answer: The decoded message is "HOLD FIRE".