Punjab Board Class 11 Mathematics

Chapter 4: Matrices and Determinants — Class 11 Maths Notes

Q: What is covered in Chapter 4 (Matrices and Determinants)?

This chapter covers Types of matrices, determinants, adjoint, inverse, Cramer's rule, row operations. The page includes theory and definitions, solved textbook examples, complete exercise solutions, MCQs and a formula sheet.

Q: How many marks does Chapter 4 carry in the Punjab Board paper?

Chapter 4 (Matrices and Determinants) carries approximately 13 marks in the Punjab Board Class 11 Mathematics paper.

Q: Are these notes for the new PCTB syllabus?

Yes. These notes follow the PCTB 2025 new book for Class 11 Mathematics, used by BISE Lahore, Gujranwala, Rawalpindi, Faisalabad, Multan, Sahiwal, Sargodha, Bahawalpur and D.G. Khan.

Q: Where are the exercise solutions for Chapter 4?

Use the exercise index on this page — every exercise of Chapter 4 has its own page with complete step-by-step solutions.

Matrices and determinants: operations, inverses, Cramer's rule and solving linear systems step by step. This page has the theory, solved textbook examples, every exercise solution, MCQs and a formula sheet for Chapter 4.

Exercise solutions — Chapter 4

Every exercise of Matrices and Determinants solved step by step.

Exercise 4.1 Exercise 4.2 Exercise 4.3

Mathematics 11 (Punjab Board)

Chapter 4: Matrices and Determinants — Theory

What Is a Matrix?

A matrix is a rectangular arrangement of numbers (or expressions) enclosed in brackets, written in rows and columns. Matricon ke har entry ko a_{ij} se denote karte hain, jahan i row number aur j column number batata hai.

A = [a_{ij}]_{m \times n}

A matrix with \(m\) rows and \(n\) columns is said to have order \(m \times n\). Order likhte waqt pehle rows, phir columns likhe jate hain — yehi order board exams mein sab se zyada ghalat hota hai, is liye isay yaad rakhen.

Key point: Order = (rows) × (columns). A \(3 \times 2\) matrix has 3 rows and 2 columns — never mix the two up.

Types of Matrices

Row matrix: sirf aik row hoti hai (order \(1 \times n\)). Column matrix: sirf aik column hota hai (order \(m \times 1\)).

Square matrix: rows aur columns barabar hote hain (order \(n \times n\)). Rectangular matrix: rows aur columns barabar nahi hote.

Diagonal matrix: square matrix jis mein sirf principal diagonal ke entries non-zero ho sakte hain, baqi sab zero. Scalar matrix: diagonal matrix jis ke tamam diagonal entries aik hi non-zero number hon. Identity (unit) matrix: diagonal matrix jis ke tamam diagonal entries 1 hon; ise \(I_n\) ya \(I\) likhte hain. Null (zero) matrix: jis ke tamam entries zero hon; ise \(O\) likhte hain.

Key point: Identity matrix \(I\) aisa behave karti hai jaise numbers mein 1: \(IA = AI = A\).

Equal Matrices and Transpose

Do matrices barabar (equal) hoti hain jab un ka order same ho aur un ke corresponding entries barabar hon. Transpose \(A^t\) rows aur columns ko aapas mein badal kar banaya jata hai: agar \(A\) ka order \(m \times n\) hai to \(A^t\) ka order \(n \times m\) hoga.

(A^t)_{ij} = a_{ji}

Key point: Transpose lene par order "flip" ho jata hai: \(m \times n \to n \times m\).

Matrix Addition, Subtraction and Scalar Multiplication

Addition aur subtraction sirf tab hi possible hai jab dono matrices ka order same ho. Har entry ko us ke corresponding entry ke sath add ya subtract karte hain.

[a_{ij}] + [b_{ij}] = [a_{ij} + b_{ij}], \qquad [a_{ij}] - [b_{ij}] = [a_{ij} - b_{ij}]

Scalar multiplication mein number \(k\) ko matrix ke har entry se multiply kar dete hain: \(kA = [k\,a_{ij}]\).

Key point: Addition ke liye same order zaroori hai — mukhtalif order wali matrices ko kabhi add ya subtract nahi kar sakte.

Matrix Multiplication

Do matrices \(A\) aur \(B\) ka product \(AB\) sirf tab define hota hai jab \(A\) ke columns ki tadad \(B\) ke rows ki tadad ke barabar ho. Agar \(A\) ka order \(m \times n\) aur \(B\) ka order \(n \times p\) hai, to \(AB\) ka order \(m \times p\) hoga.

(AB)_{ij} = \sum_{k=1}^{n} a_{ik}\,b_{kj}

Aasan lafzon mein: \(AB\) ka har entry \(A\) ki aik row aur \(B\) ke aik column ka "dot product" hota hai — row ke entries ko column ke corresponding entries se multiply kar ke add kar dete hain.

Key point: Aam tor par \(AB \neq BA\) — matrix multiplication commutative nahi hoti. Isi wajah se \((A+B)^2 = A^2 + 2AB + B^2\) likhna ghalat hai.

Laws of Matrix Algebra

Jab matrices indicated operations ke liye conformable hon, to ye qawaneen lagte hain: \(A + B = B + A\) (commutative for addition), \((A+B)+C = A+(B+C)\), \(A(BC) = (AB)C\) (associative for multiplication), \(A(B+C) = AB + AC\) (distributive), \(c(AB) = (cA)B = A(cB)\), aur \(IA = AI = A\), \(A + O = A\).

Key point: Addition commutative hai, multiplication nahi. Board mein aksar isi farq par sawal aata hai.

Determinants: Minors and Cofactors

Determinant sirf square matrix ka hota hai. Element \(a_{ij}\) ka minor \(M_{ij}\) woh determinant hai jo \(i\)-th row aur \(j\)-th column delete karne se banta hai. Cofactor \(A_{ij}\) is minor ko sign ke sath likha jata hai:

A_{ij} = (-1)^{i+j} M_{ij}

\(3 \times 3\) matrix ka determinant kisi bhi row ya column ke entries ko un ke cofactors se multiply kar ke add karne se milta hai:

|A| = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}

Expansion mein sign pattern \(+,-,+\); \(-,+,-\); \(+,-,+\) hota hai — isi pattern ki ghalti board mein sab se aam hai.

Key point: Cofactor ka sign \((-1)^{i+j}\) se aata hai. Pehli row expand karte waqt signs \(+,-,+\) hote hain.

Properties of Determinants

Ye properties "without expansion" wale sawalon ki jaan hain: \(|A^t| = |A|\); do rows ya columns interchange karne se determinant ka sign badal jata hai; do identical rows ya columns hon to determinant 0; kisi row ke tamam entries zero hon to determinant 0; aik row ko \(k\) se multiply karne se determinant \(k\) times ho jata hai; \(R_i \to R_i + kR_j\) se determinant ki value nahi badalti; triangular matrix ka determinant sirf diagonal entries ka product hota hai.

Key point: "Show without expansion that the determinant is 0" ka matlab hai: operations laga kar koi row ya column zero banao, ya do rows/columns identical banao.

Adjoint and Inverse of a Matrix

Matrix of cofactors ka transpose adjoint kehlata hai: \(\text{adj } A = [A_{ij}]^t\). Agar \(|A| \neq 0\) (non-singular matrix) ho to inverse is formula se milta hai:

A^{-1} = \frac{1}{|A|} \, \text{adj } A

Inverse ki tasdeeq \(AA^{-1} = A^{-1}A = I\) se hoti hai. Agar \(|A| = 0\) ho to matrix singular kehlati hai aur us ka inverse exist nahi karta.

Key point: Inverse nikalne se pehle hamesha determinant check karo — \(|A| = 0\) ho to inverse ka sawal wahin khatam.

Row Operations, Echelon Form and Rank

Elementary row operations: do rows ko interchange karna, kisi row ko non-zero number se multiply karna, ya aik row ka multiple doosri row mein add karna. In operations se matrix echelon form mein layi jati hai, jis mein har non-zero row ka leading entry 1 hota hai aur har agli row ka leading entry pechli se aage hota hai. Echelon form mein non-zero rows ki tadad matrix ka rank kehlati hai.

Inverse bhi row operations se nikal sakta hai: \([A \mid I]\) ko operations laga kar \([I \mid A^{-1}]\) banao — right side par inverse aa jayega.

Key point: Rank = echelon form mein non-zero rows ki ginti. Rank homogeneous aur non-homogeneous systems dono mein faisla-kun hai.

Solving Systems of Linear Equations

System \(AX = B\) ko teen tareeqon se solve karte hain. Matrix inversion: \(X = A^{-1}B\). Cramer's rule: har variable ka determinant (us ke column ko constants \(B\) se replace kar ke) ko \(|A|\) se divide karo, masalan \(x = D_x / D\). Echelon form: augmented matrix \([A \mid B]\) ko reduce kar ke back substitution karo.

Consistency ke usool: agar \(\text{rank}(A) = \text{rank}([A|B]) =\) variables ki tadad, to unique solution; agar dono ranks barabar magar variables se kam, to infinitely many solutions; agar ranks barabar na hon, to koi solution nahi (inconsistent system).

Homogeneous system \(AX = O\) hamesha consistent hota hai. Agar \(\text{rank}(A) =\) variables ki tadad, to sirf trivial solution \((0,0,0)\); agar rank kam ho, to non-trivial (infinite) solutions parameter \(t\) mein likhe jate hain.

Key point: Cramer's rule sirf tab lagao jab \(D = |A| \neq 0\) ho. \(D = 0\) ka matlab hai ya to infinite solutions ya no solution.

Applications: Transformations and Coding

Matrices se geometric transformations ki jati hain: x-axis par reflection ke liye \(\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\), y-axis ke liye \(\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\), aur line \(y = x\) ke liye \(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\) se multiply karte hain.

Coding mein message ke letters ko numbers (A=1, B=2, …, Z=26) mein badal kar columns banaye jate hain aur encoding matrix se multiply karte hain. Decode karne ke liye encoded matrix ko encoding matrix ke inverse se multiply karte hain.

Key point: Encode = \(A \times \text{message}\); decode = \(A^{-1} \times \text{coded message}\).

Mathematics 11 (Punjab Board)

Solved Examples

Textbook solved examples for Punjab Board Class 11 Mathematics, Chapter 4: Matrices and Determinants.

Example 1: If \(A = \begin{bmatrix} 1 & 0 & -1 & 2 \\ 3 & 1 & 2 & 5 \\ 0 & -2 & 1 & 6 \end{bmatrix}\) and \(B = \begin{bmatrix} 2 & -1 & 3 & 1 \\ 3 & 1 & -1 & 4 \\ 3 & 1 & 2 & -1 \end{bmatrix}\), show that \((A+B)^t = A^t + B^t\)

Pehle \(A + B\) nikal kar us ka transpose lenge, phir \(A\) aur \(B\) ke alag alag transpose nikal kar add karenge. Dono taraf same matrix aani chahiye.

A + B = \begin{bmatrix} 3 & -1 & 2 & 3 \\ 6 & 2 & 1 & 9 \\ 3 & -1 & 3 & 5 \end{bmatrix}, \qquad (A+B)^t = \begin{bmatrix} 3 & 6 & 3 \\ -1 & 2 & -1 \\ 2 & 1 & 3 \\ 3 & 9 & 5 \end{bmatrix}

A^t = \begin{bmatrix} 1 & 3 & 0 \\ 0 & 1 & -2 \\ -1 & 2 & 1 \\ 2 & 5 & 6 \end{bmatrix}, \quad B^t = \begin{bmatrix} 2 & 3 & 3 \\ -1 & 1 & 1 \\ 3 & -1 & 2 \\ 1 & 4 & -1 \end{bmatrix}, \quad A^t + B^t = \begin{bmatrix} 3 & 6 & 3 \\ -1 & 2 & -1 \\ 2 & 1 & 3 \\ 3 & 9 & 5 \end{bmatrix}

Both sides are equal, so \((A+B)^t = A^t + B^t\) is proved.

Answer: \((A+B)^t = A^t + B^t\). Proved.

Example 2: If \(A = \begin{bmatrix} 2 & -1 & 0 \\ 1 & 2 & -3 \\ 1 & 2 & -2 \end{bmatrix}\) and \(B = \begin{bmatrix} 2 & -2 & 3 \\ -1 & -4 & 6 \\ 0 & -5 & 5 \end{bmatrix}\), compute \(A^2B\)

Pehle \(A^2 = A \cdot A\) nikalenge, phir us result ko \(B\) se multiply karenge. Har entry row-column dot product se nikalti hai.

A^2 = \begin{bmatrix} 2 & -1 & 0 \\ 1 & 2 & -3 \\ 1 & 2 & -2 \end{bmatrix}\begin{bmatrix} 2 & -1 & 0 \\ 1 & 2 & -3 \\ 1 & 2 & -2 \end{bmatrix} = \begin{bmatrix} 3 & -4 & 3 \\ 1 & -3 & 0 \\ 2 & -1 & -2 \end{bmatrix}

A^2B = \begin{bmatrix} 3 & -4 & 3 \\ 1 & -3 & 0 \\ 2 & -1 & -2 \end{bmatrix}\begin{bmatrix} 2 & -2 & 3 \\ -1 & -4 & 6 \\ 0 & -5 & 5 \end{bmatrix} = \begin{bmatrix} 10 & -5 & 0 \\ 5 & 10 & -15 \\ 5 & 10 & -10 \end{bmatrix}

Answer: \(A^2B = \begin{bmatrix} 10 & -5 & 0 \\ 5 & 10 & -15 \\ 5 & 10 & -10 \end{bmatrix}\)

Example 3: If \(A = \begin{bmatrix} 2 & 0 & 1 \\ 1 & 4 & 2 \\ 3 & 0 & 6 \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & -1 \\ 1 & -2 & 3 \end{bmatrix}\), find \(AB\) and \(BA\) and show that \(AB \neq BA\)

Dono products alag alag nikal kar compare karenge. Ye example matrix multiplication ki non-commutative property ko zahir karta hai.

AB = \begin{bmatrix} 2 & 0 & 1 \\ 1 & 4 & 2 \\ 3 & 0 & 6 \end{bmatrix}\begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & -1 \\ 1 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 3 & -4 & 3 \\ 11 & 7 & 2 \\ 9 & -15 & 18 \end{bmatrix}

BA = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & -1 \\ 1 & -2 & 3 \end{bmatrix}\begin{bmatrix} 2 & 0 & 1 \\ 1 & 4 & 2 \\ 3 & 0 & 6 \end{bmatrix} = \begin{bmatrix} 1 & -4 & -1 \\ 4 & 12 & 2 \\ 9 & -8 & 15 \end{bmatrix}

Clearly \(AB \neq BA\).

Answer: \(AB = \begin{bmatrix} 3 & -4 & 3 \\ 11 & 7 & 2 \\ 9 & -15 & 18 \end{bmatrix}\), \(BA = \begin{bmatrix} 1 & -4 & -1 \\ 4 & 12 & 2 \\ 9 & -8 & 15 \end{bmatrix}\); hence \(AB \neq BA\).

Example 4: If \(A = \begin{bmatrix} 1 & -2 & 3 \\ -2 & 3 & 1 \\ 4 & -3 & 2 \end{bmatrix}\), evaluate the determinant \(|A|\)

Pehli row ke along expand karenge signs \(+,-,+\) ke sath. Har \(2 \times 2\) minor ko ehtiyat se solve karen.

|A| = 1\begin{vmatrix} 3 & 1 \\ -3 & 2 \end{vmatrix} - (-2)\begin{vmatrix} -2 & 1 \\ 4 & 2 \end{vmatrix} + 3\begin{vmatrix} -2 & 3 \\ 4 & -3 \end{vmatrix}

\[ = 1(6 + 3) + 2(-4 - 4) + 3(6 - 12) = 9 - 16 - 18 = -25 \]

Answer: \(|A| = -25\)

Example 5: If \(A = \begin{bmatrix} 1 & -2 & 3 \\ -2 & 3 & 1 \\ 4 & -3 & 2 \end{bmatrix}\), find the cofactors \(A_{12}\), \(A_{22}\), \(A_{32}\) and \(|A|\)

Cofactor \(A_{ij} = (-1)^{i+j}M_{ij}\). Teeno cofactors doosre column mein hain, is liye determinant bhi doosre column ke along expand kar ke verify kar sakte hain.

M_{12} = \begin{vmatrix} -2 & 1 \\ 4 & 2 \end{vmatrix} = -8, \quad M_{22} = \begin{vmatrix} 1 & 3 \\ 4 & 2 \end{vmatrix} = -10, \quad M_{32} = \begin{vmatrix} 1 & 3 \\ -2 & 1 \end{vmatrix} = 7

A_{12} = (-1)^{3}(-8) = 8, \quad A_{22} = (-1)^{4}(-10) = -10, \quad A_{32} = (-1)^{5}(7) = -7

|A| = a_{12}A_{12} + a_{22}A_{22} + a_{32}A_{32} = (-2)(8) + 3(-10) + (-3)(-7) = -16 - 30 + 21 = -25

Answer: \(A_{12} = 8\), \(A_{22} = -10\), \(A_{32} = -7\), \(|A| = -25\)

Example 6: Without expansion, show that \(\begin{vmatrix} x & a+x & b+c \\ x & b+x & c+a \\ x & c+x & a+b \end{vmatrix} = 0\)

\(C_2\) mein \(C_1\) add karne se har entry mein \((a+b+c+x)\) aa jayega. Phir common factor bahar nikalne par do columns identical ban jayenge.

Apply \(C_2 \to C_2 + C_1\):

\text{L.H.S} = \begin{vmatrix} x & a+b+c+x & b+c \\ x & a+b+c+x & c+a \\ x & a+b+c+x & a+b \end{vmatrix}

Take \(x\) common from \(C_1\) and \((a+b+c+x)\) common from \(C_2\):

= x(a+b+c+x)\begin{vmatrix} 1 & 1 & b+c \\ 1 & 1 & c+a \\ 1 & 1 & a+b \end{vmatrix}

Since \(C_1\) and \(C_2\) are identical, the determinant is \(0\). Hence L.H.S \(= 0 =\) R.H.S.

Answer: The determinant equals \(0\). Proved.

Example 7: If \(A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 1 & -1 & 1 \end{bmatrix}\), find \(A^{-1}\)

Saare 9 cofactors nikalenge, un ka transpose (adjoint) banayenge, aur \(A^{-1} = \frac{1}{|A|}\text{adj } A\) apply karenge.

\begin{aligned} A_{11} &= 3, & A_{12} &= 1, & A_{13} &= -2, \\ A_{21} &= -2, & A_{22} &= -1, & A_{23} &= 1, \\ A_{31} &= -4, & A_{32} &= -1, & A_{33} &= 2. \end{aligned}

\text{adj } A = \begin{bmatrix} 3 & -2 & -4 \\ 1 & -1 & -1 \\ -2 & 1 & 2 \end{bmatrix}, \qquad |A| = 1(3) + 0(1) + 2(-2) = -1

A^{-1} = \frac{1}{-1}\begin{bmatrix} 3 & -2 & -4 \\ 1 & -1 & -1 \\ -2 & 1 & 2 \end{bmatrix} = \begin{bmatrix} -3 & 2 & 4 \\ -1 & 1 & 1 \\ 2 & -1 & -2 \end{bmatrix}

Answer: \(A^{-1} = \begin{bmatrix} -3 & 2 & 4 \\ -1 & 1 & 1 \\ 2 & -1 & -2 \end{bmatrix}\)

Example 8: Reduce the matrix \(\begin{bmatrix} 2 & 3 & -1 & 9 \\ 1 & -1 & 2 & -3 \\ 3 & 1 & 3 & 2 \end{bmatrix}\) to row echelon and reduced row echelon form

Pehle \(R_1 \leftrightarrow R_2\) se leading 1 layenge, phir neeche zeros banayenge (echelon form). Us ke baad leading 1s ke upar bhi zeros bana kar reduced echelon form hasil karenge.

\(R_1 \leftrightarrow R_2\); \(R_2 \to R_2 - 2R_1\); \(R_3 \to R_3 - 3R_1\); \(R_2 \to \frac{1}{5}R_2\); \(R_3 \to R_3 - 4R_2\):

\begin{bmatrix} 1 & -1 & 2 & -3 \\ 0 & 1 & -1 & 3 \\ 0 & 0 & 1 & -1 \end{bmatrix} \quad \text{(row echelon form)}

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

\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -1 \end{bmatrix} \quad \text{(reduced row echelon form)}

Answer: Echelon form has rows \((1,-1,2,-3)\), \((0,1,-1,3)\), \((0,0,1,-1)\); reduced echelon form is \([I \mid (1, 2, -1)^t]\).

Example 9: If \(A = \begin{bmatrix} 2 & 5 & -1 \\ 3 & 4 & 2 \\ 1 & 2 & -2 \end{bmatrix}\), find the inverse using elementary row operations

\([A \mid I]\) augment kar ke row operations se left side ko \(I\) banayenge. Pehle \(|A| \neq 0\) check kar lenge taake pata ho ke inverse exist karta hai.

\(|A| = 2(-12) - 5(-8) - 1(2) = -24 + 40 - 2 = 14 \neq 0\), so \(A^{-1}\) exists.

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

\(R_1 \leftrightarrow R_3\); \(R_2 \to R_2 - 3R_1\); \(R_3 \to R_3 - 2R_1\); \(R_2 \to -\frac{1}{2}R_2\):

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

\(R_1 \to R_1 - 2R_2\); \(R_3 \to R_3 - R_2\); \(R_3 \to \frac{1}{7}R_3\):

\left[\begin{array}{ccc|ccc} 1 & 0 & 6 & 0 & 1 & -2 \\ 0 & 1 & -4 & 0 & -1/2 & 3/2 \\ 0 & 0 & 1 & 1/7 & 1/14 & -1/2 \end{array}\right]

\(R_1 \to R_1 - 6R_3\); \(R_2 \to R_2 + 4R_3\):

\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & -6/7 & 4/7 & 1 \\ 0 & 1 & 0 & 4/7 & -3/14 & -1/2 \\ 0 & 0 & 1 & 1/7 & 1/14 & -1/2 \end{array}\right]

Answer: \(A^{-1} = \begin{bmatrix} -6/7 & 4/7 & 1 \\ 4/7 & -3/14 & -1/2 \\ 1/7 & 1/14 & -1/2 \end{bmatrix}\)

Example 10: Find the rank of the matrix \(\begin{bmatrix} 1 & -1 & 2 & -3 \\ 2 & 0 & 7 & -7 \\ 3 & 1 & 12 & -11 \end{bmatrix}\)

Echelon form mein convert kar ke non-zero rows gin lenge.

\(R_2 \to R_2 - 2R_1\); \(R_3 \to R_3 - 3R_1\); \(R_2 \to \frac{1}{2}R_2\); \(R_3 \to R_3 - 4R_2\):

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

There are 2 non-zero rows.

Answer: Rank \(= 2\)

Example 11: Solve and explain consistent and inconsistent systems: (i) \(2x+5y-z=5\), \(3x+4y+2z=11\), \(x+2y-2z=-3\); (ii) \(x+2y-2z=1\), \(2x-y+7z=11\), \(3x+5y+4z=-3\); (iii) \(x-y+2z=1\), \(2x-6y+5z=7\), \(3x+5y+4z=-3\)

Augmented matrix ko echelon form mein laa kar solution nikalenge. Agar aakhri row \(0 = \text{non-zero}\) ban jaye to system inconsistent (koi solution nahi).

(i)

\left[\begin{array}{ccc|c} 2 & 5 & -1 & 5 \\ 3 & 4 & 2 & 11 \\ 1 & 2 & -2 & -3 \end{array}\right] \xrightarrow{R_1 \leftrightarrow R_3} \left[\begin{array}{ccc|c} 1 & 2 & -2 & -3 \\ 3 & 4 & 2 & 11 \\ 2 & 5 & -1 & 5 \end{array}\right]

\(R_2 \to R_2 - 3R_1\); \(R_3 \to R_3 - 2R_1\); \(R_2 \to -\frac{1}{2}R_2\); \(R_3 \to R_3 - R_2\); \(R_3 \to \frac{1}{7}R_3\); then back-eliminate:

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

Solution: \(x = -1\), \(y = 2\), \(z = 3\). The system is consistent with a unique solution.

Answer: (i) \(x = -1\), \(y = 2\), \(z = 3\) — consistent, unique solution.

(ii)

\left[\begin{array}{ccc|c} 1 & 2 & -2 & 1 \\ 2 & -1 & 7 & 11 \\ 3 & 5 & 4 & -3 \end{array}\right]

\(R_2 \to R_2 - 2R_1\); \(R_3 \to R_3 - 3R_1\); \(R_2 \to -\frac{1}{5}R_2\); \(R_3 \to R_3 + R_2\); \(R_3 \to \frac{5}{39}R_3\); then back-eliminate:

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

Solution: \(x = 7\), \(y = -4\), \(z = -1\). The system is consistent with a unique solution.

Answer: (ii) \(x = 7\), \(y = -4\), \(z = -1\) — consistent, unique solution.

(iii)

\left[\begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 2 & -6 & 5 & 7 \\ 3 & 5 & 4 & -3 \end{array}\right]

\(R_2 \to R_2 - 2R_1\); \(R_3 \to R_3 - 3R_1\); \(R_2 \to -\frac{1}{4}R_2\); \(R_3 \to R_3 - 8R_2\):

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

The last row reads \(0 = 4\), which is impossible. Hence the system has no solution — it is inconsistent.

Answer: (iii) No solution — inconsistent system.

Example 12: Solve by matrix inversion method: \(x_1 - 2x_2 + x_3 = -4\), \(2x_1 - 3x_2 + 2x_3 = -6\), \(2x_1 + 2x_2 + x_3 = 5\)

\(AX = B\) likh kar \(|A|\) nikalenge, phir cofactors se \(A^{-1}\) banayenge aur \(X = A^{-1}B\) compute karenge.

A = \begin{bmatrix} 1 & -2 & 1 \\ 2 & -3 & 2 \\ 2 & 2 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, \quad B = \begin{bmatrix} -4 \\ -6 \\ 5 \end{bmatrix}

\(|A| = -1 \neq 0\) (expand after \(R_2 \to R_2 - 2R_1\) along the new \(R_2\)). Cofactors: \(A_{11} = -7\), \(A_{12} = 2\), \(A_{13} = 10\), \(A_{21} = 4\), \(A_{22} = -1\), \(A_{23} = -6\), \(A_{31} = -1\), \(A_{32} = 0\), \(A_{33} = 1\).

A^{-1} = \frac{1}{-1}\begin{bmatrix} -7 & 4 & -1 \\ 2 & -1 & 0 \\ 10 & -6 & 1 \end{bmatrix} = \begin{bmatrix} 7 & -4 & 1 \\ -2 & 1 & 0 \\ -10 & 6 & -1 \end{bmatrix}

X = A^{-1}B = \begin{bmatrix} 7 & -4 & 1 \\ -2 & 1 & 0 \\ -10 & 6 & -1 \end{bmatrix}\begin{bmatrix} -4 \\ -6 \\ 5 \end{bmatrix} = \begin{bmatrix} -28+24+5 \\ 8-6+0 \\ 40-36-5 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}

Answer: \((x_1, x_2, x_3) = (1, 2, -1)\)

Example 13: Solve by Cramer's rule: \(3x_1 + x_2 - x_3 = -4\), \(x_1 + x_2 - 2x_3 = -4\), \(-x_1 + 2x_2 - x_3 = 1\)

\(|A|\) nikalenge, phir har variable ke column ko constants se replace kar ke teen determinants aur nikalenge.

|A| = \begin{vmatrix} 3 & 1 & -1 \\ 1 & 1 & -2 \\ -1 & 2 & -1 \end{vmatrix} = 3(3) - 1(-3) - 1(3) = 9

x_1 = \frac{1}{9}\begin{vmatrix} -4 & 1 & -1 \\ -4 & 1 & -2 \\ 1 & 2 & -1 \end{vmatrix} = \frac{-12-6+9}{9} = -1

x_2 = \frac{1}{9}\begin{vmatrix} 3 & -4 & -1 \\ 1 & -4 & -2 \\ -1 & 1 & -1 \end{vmatrix} = \frac{18-12+3}{9} = 1

x_3 = \frac{1}{9}\begin{vmatrix} 3 & 1 & -4 \\ 1 & 1 & -4 \\ -1 & 2 & 1 \end{vmatrix} = \frac{27+3-12}{9} = 2

Answer: \((x_1, x_2, x_3) = (-1, 1, 2)\)

Example 14: Solve by Gaussian elimination: \(x + 2y + z = 0\), \(2x + 3y + 4z = 0\), \(4x + 3y + 2z = 0\)

Homogeneous system hai. Augmented matrix ko row echelon form mein laa kar rank dekhenge — rank 3 hua to sirf trivial solution.

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

\(R_2 \to R_2 - 2R_1\); \(R_3 \to R_3 - 4R_1\); \(R_2 \to -R_2\); \(R_3 \to R_3 + 5R_2\); \(R_3 \to -\frac{1}{12}R_3\):

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

Back substitution: \(z = 0\), \(y = 0\), \(x = 0\). Rank \(= 3 =\) number of variables.

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

Example 15: Solve by Gaussian elimination: \(x_1 + x_2 + x_3 = 0\), \(x_1 - x_2 + 3x_3 = 0\), \(x_1 + 3x_2 - x_3 = 0\)

Yahan rank variables se kam niklegi, is liye infinite solutions honge. Free variable ko \(t\) rakhenge.

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

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

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

Rank \(= 2 < 3\). From the second row: \(x_2 = x_3\). 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 \(t\) — infinitely many solutions.

Example 16: The vertices of a triangle are \(A(2, 3)\), \(B(-1, 4)\) and \(C(3, -2)\). Find the vertices of the reflected triangle about the x-axis using a transformation matrix.

x-axis par reflection ki transformation matrix \(\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\) hai — y-coordinate ka sign badal jata hai.

A' = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}, \quad B' = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} -1 \\ 4 \end{bmatrix} = \begin{bmatrix} -1 \\ -4 \end{bmatrix}, \quad C' = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} 3 \\ -2 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}

Answer: \(A'(2, -3)\), \(B'(-1, -4)\), \(C'(3, 2)\).

Example 17: Using matrix \(A = \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix}\), encode the message "ATTACK", where A to Z are numbers 1 to 26

Message ko pairs mein divide karenge (AT, TA, CK), har pair ko \(2 \times 1\) column banayenge, aur encoding matrix \(A\) se left-multiply karenge.

\begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 20 \end{bmatrix} = \begin{bmatrix} 41 \\ 23 \end{bmatrix}, \quad \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix}\begin{bmatrix} 20 \\ 1 \end{bmatrix} = \begin{bmatrix} 22 \\ 61 \end{bmatrix}, \quad \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix}\begin{bmatrix} 3 \\ 11 \end{bmatrix} = \begin{bmatrix} 25 \\ 20 \end{bmatrix}

Answer: The coded message is \(\begin{bmatrix} 41 & 22 & 25 \\ 23 & 61 & 20 \end{bmatrix}\).

Mathematics 11 (Punjab Board)

MCQs: Chapter 4 – Matrices and Determinants

A matrix has 5 rows and 2 columns. What is its order?
1. \(2 \times 5\)
2. \(5 \times 2\)
3. \(10 \times 1\)
4. \(5 \times 5\)
Order hamesha (rows × columns) likha jata hai, is liye 5 rows aur 2 columns wali matrix ka order \(5 \times 2\) hoga.
A matrix in which every entry is zero is called a:
1. Identity matrix
2. Scalar matrix
3. Null matrix
4. Diagonal matrix
Jis matrix ke tamam entries zero hon usay null ya zero matrix kehte hain aur ise \(O\) se denote karte hain.
In the identity matrix \(I_3\), the entry \(a_{22}\) is equal to:
1. \(0\)
2. \(3\)
3. \(1\)
4. \(-1\)
Identity matrix ke principal diagonal ke tamam entries 1 hote hain, is liye \(a_{22} = 1\).
If \(A\) is a matrix of order \(2 \times 3\), then the order of \(A^t\) is:
1. \(2 \times 3\)
2. \(3 \times 2\)
3. \(2 \times 2\)
4. \(3 \times 3\)
Transpose lene par rows aur columns interchange ho jate hain, is liye \(2 \times 3\) matrix ka transpose \(3 \times 2\) ban jata hai.
Two matrices \(A\) and \(B\) can be added if and only if:
1. They have the same order
2. They have different orders
3. Both are square matrices
4. Both are \(2 \times 2\) matrices
Matrix addition ke liye dono matrices ka order bilkul same hona zaroori hai, warna corresponding entries ka koi wujood nahi hota.
The product \(AB\) of two matrices is defined when:
1. \(A\) and \(B\) have the same order
2. The number of columns of \(A\) equals the number of rows of \(B\)
3. The number of rows of \(A\) equals the number of columns of \(B\)
4. \(A\) and \(B\) are both square
Multiplication ke liye pehli matrix ke columns aur doosri matrix ke rows barabar hone chahiye — isi ko conformable for multiplication kehte hain.
If \(A\) is \(3 \times 4\) and \(B\) is \(4 \times 2\), then the order of \(AB\) is:
1. \(3 \times 2\)
2. \(4 \times 4\)
3. \(3 \times 4\)
4. \(2 \times 3\)
Product ka order (pehli matrix ki rows) × (doosri matrix ke columns) hota hai, yani \(3 \times 2\).
Matrix multiplication is, in general:
1. Commutative
2. Not commutative
3. Always commutative for square matrices
4. Commutative only for \(2 \times 2\) matrices
Aam tor par \(AB \neq BA\) hota hai, is liye matrix multiplication commutative nahi hoti — yahi Chapter 4 ka sab se important concept hai.
For conformable matrices \(A\) and \(B\), \((AB)^t\) equals:
1. \(A^t B^t\)
2. \(B^t A^t\)
3. \(AB\)
4. \(BA\)
Product ka transpose lene par order reverse ho jata hai: \((AB)^t = B^t A^t\).
The value of \(\begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix}\) is:
1. \(11\)
2. \(5\)
3. \(-5\)
4. \(8\)
\(2 \times 2\) determinant \(= (2)(4) - (3)(1) = 8 - 3 = 5\).
The cofactor \(A_{ij}\) of the element \(a_{ij}\) is defined as:
1. \(M_{ij}\)
2. \((-1)^i M_{ij}\)
3. \((-1)^{i+j} M_{ij}\)
4. \((-1)^j M_{ij}\)
Cofactor minor ko sign \((-1)^{i+j}\) ke sath likha jata hai, jahan \(i\) row aur \(j\) column number hai.
If two rows of a determinant are identical, then the value of the determinant is:
1. \(1\)
2. \(-1\)
3. \(0\)
4. \(2\)
Determinants ki property ke mutabiq do identical rows ya columns hon to determinant ki value hamesha 0 hoti hai.
If one row of a determinant is multiplied by \(k\), the determinant is multiplied by:
1. \(k^3\)
2. \(k\)
3. \(1/k\)
4. \(0\)
Aik row ko \(k\) se multiply karne par poori determinant ki value \(k\) times ho jati hai.
A square matrix \(A\) is called singular if:
1. \(|A| = 1\)
2. \(|A| = 0\)
3. \(|A| \neq 0\)
4. \(A = I\)
Jis matrix ka determinant 0 ho usay singular kehte hain aur aisi matrix ka inverse exist nahi karta.
For a non-singular matrix \(A\), \(A^{-1}\) equals:
1. \(|A| \cdot \text{adj } A\)
2. \(\frac{1}{|A|} \text{ adj } A\)
3. \(\frac{|A|}{\text{adj } A}\)
4. \(\text{adj } A\)
Inverse ka formula \(A^{-1} = \frac{1}{|A|} \text{adj } A\) hai — adjoint ko determinant se divide karte hain.
If \(A\) is a \(3 \times 3\) matrix, then \(|kA|\) equals:
1. \(k|A|\)
2. \(3k|A|\)
3. \(k^3|A|\)
4. \(k^2|A|\)
Har row se \(k\) common aata hai aur 3 rows hain, is liye \(|kA| = k^3|A|\). Aam tor par \(n \times n\) ke liye \(|kA| = k^n|A|\).
The rank of a matrix is:
1. The number of rows
2. The number of non-zero rows in its echelon form
3. Its determinant
4. Its order
Matrix ko echelon form mein laa kar non-zero rows gin lete hain — yehi tadad us ka rank hoti hai.
Cramer's rule can be applied to solve a system when:
1. \(D = 0\)
2. \(D \neq 0\)
3. The system is homogeneous
4. The matrix is singular
Cramer's rule mein \(D = |A|\) se divide karte hain, is liye \(D \neq 0\) hona zaroori hai.
In Cramer's rule for three variables, \(x\) equals:
1. \(\frac{D}{D_x}\)
2. \(\frac{D_x}{D}\)
3. \(D_x \cdot D\)
4. \(D - D_x\)
Har variable us ke apne determinant ko \(D\) se divide kar ke milta hai: \(x = D_x/D\), jahan \(D_x\) mein pehla column constants se replace hota hai.
A homogeneous system \(AX = O\) always has:
1. No solution
2. The trivial solution \((0, 0, 0)\)
3. Exactly one non-zero solution
4. Infinitely many solutions
Homogeneous system mein \(X = 0\) hamesha solution hota hai, jise trivial solution kehte hain.
A homogeneous system has a non-trivial solution if:
1. Rank of \(A\) = number of variables
2. Rank of \(A\) < number of variables
3. Rank of \(A\) > number of variables
4. \(|A| \neq 0\)
Jab rank variables ki tadad se kam ho to system ke paas trivial ke ilawa bhi solutions hote hain.
Reflecting the point \((a, b)\) over the x-axis gives:
1. \((-a, b)\)
2. \((a, -b)\)
3. \((b, a)\)
4. \((-a, -b)\)
x-axis par reflection mein sirf y-coordinate ka sign badalta hai, is liye \((a, b)\) se \((a, -b)\) ban jata hai.
In the matrix inversion method, the solution of \(AX = B\) is:
1. \(X = BA^{-1}\)
2. \(X = A^{-1}B\)
3. \(X = AB\)
4. \(X = B^{-1}A\)
Dono sides ko left se \(A^{-1}\) se multiply karne par \(X = A^{-1}B\) milta hai — order ka khayal rakhen, \(BA^{-1}\) ghalat hai.
If \(A\) is a \(2 \times 3\) matrix, then \(|A|\) is:
1. \(0\)
2. Not defined
3. \(6\)
4. \(-1\)
Determinant sirf square matrix ka hota hai, is liye \(2 \times 3\) (non-square) matrix ka determinant define nahi hota.
For square matrices \(A\) and \(B\), \((A+B)^2 = A^2 + 2AB + B^2\) holds:
1. Always
2. Only when \(AB = BA\)
3. Never
4. Only when \(A = B\)
\((A+B)^2 = A^2 + AB + BA + B^2\) hota hai, aur \(AB + BA = 2AB\) sirf tab jab \(AB = BA\) ho.

Mathematics 11 (Punjab Board)

Chapter 4: Matrices and Determinants — Formula Sheet

Matrix Operations

\[ [a_{ij}] + [b_{ij}] = [a_{ij} + b_{ij}] \qquad \text{(same order required)} \]

\[ [a_{ij}] - [b_{ij}] = [a_{ij} - b_{ij}] \]

\[ k[a_{ij}] = [k\,a_{ij}] \]

\[ \text{If } A \text{ is } m \times n \text{ and } B \text{ is } n \times p, \text{ then } AB \text{ is } m \times p \text{ with } (AB)_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj} \]

\[ A^2 = A \cdot A, \quad A^3 = A \cdot A \cdot A \]

Laws of Matrix Algebra

\[ A + B = B + A, \qquad (A + B) + C = A + (B + C) \]

\[ A(BC) = (AB)C, \qquad A(B + C) = AB + AC, \qquad (A + B)C = AC + BC \]

\[ IA = AI = A, \qquad A + O = A, \qquad c(AB) = (cA)B = A(cB) \]

\[ \text{In general } AB \neq BA \]

Transpose Properties

\[ (A^t)^t = A, \qquad (A + B)^t = A^t + B^t, \qquad (AB)^t = B^t A^t \]

Determinants: Minor, Cofactor, Expansion

\[ A_{ij} = (-1)^{i+j} M_{ij} \]

\[ |A| = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} \quad (3 \times 3, \text{ expand along any row or column}) \]

\[ \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \]

Properties of Determinants

\[ |A^t| = |A| \]

\[ \text{Interchanging two rows (or columns): } |A| \to -|A| \]

\[ \text{Two identical rows (or columns)} \Rightarrow |A| = 0 \]

\[ \text{A zero row (or column)} \Rightarrow |A| = 0 \]

\[ \text{Multiplying one row by } k \Rightarrow \text{determinant becomes } k|A| \]

\[ R_i \to R_i + kR_j \text{ does not change the determinant} \]

\[ |kA| = k^n|A| \quad \text{for an } n \times n \text{ matrix} \]

\[ |AB| = |A|\,|B|, \qquad |A^{-1}| = \frac{1}{|A|} \]

Adjoint and Inverse

\[ \text{adj } A = \text{transpose of the cofactor matrix } [A_{ij}] \]

\[ A^{-1} = \frac{1}{|A|} \, \text{adj } A \qquad (|A| \neq 0) \]

\[ AA^{-1} = A^{-1}A = I \]

\[ |A| = 0 \Rightarrow A \text{ is singular (no inverse exists)} \]

Rank and Echelon Form

\[ \text{Rank}(A) = \text{number of non-zero rows in echelon form} \]

\[ [A \mid I] \xrightarrow{\text{row operations}} [I \mid A^{-1}] \]

Solving \(AX = B\): Cramer's Rule

\[ x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad z = \frac{D_z}{D} \]

\[ D = |A|, \quad D_x = \text{replace 1st column of } A \text{ by } B, \text{ etc.} \]

Solving \(AX = B\): Matrix Inversion

\[ X = A^{-1}B \]

Consistency Rules

\[ \text{rank}(A) = \text{rank}([A|B]) = n \Rightarrow \text{unique solution} \]

\[ \text{rank}(A) = \text{rank}([A|B]) < n \Rightarrow \text{infinitely many solutions} \]

\[ \text{rank}(A) \neq \text{rank}([A|B]) \Rightarrow \text{no solution (inconsistent)} \]

\[ AX = O \text{ (homogeneous): always consistent; only trivial solution if rank}(A) = n \]

Transformation Matrices

\[ \text{Reflection over x-axis: } \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \quad \text{over y-axis: } \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}, \quad \text{over } y=x\text{: } \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \]

\[ \text{Encode: } A \times \text{message matrix}; \qquad \text{Decode: } A^{-1} \times \text{coded matrix} \]

Chapter 4 FAQs

What is covered in Chapter 4 (Matrices and Determinants)?

This chapter covers Types of matrices, determinants, adjoint, inverse, Cramer's rule, row operations. The page includes theory and definitions, solved textbook examples, complete exercise solutions, MCQs and a formula sheet.

How many marks does Chapter 4 carry in the Punjab Board paper?

Chapter 4 (Matrices and Determinants) carries approximately 13 marks in the Punjab Board Class 11 Mathematics paper.

Are these notes for the new PCTB syllabus?

Yes. These notes follow the PCTB 2025 new book for Class 11 Mathematics, used by BISE Lahore, Gujranwala, Rawalpindi, Faisalabad, Multan, Sahiwal, Sargodha, Bahawalpur and D.G. Khan.

Where are the exercise solutions for Chapter 4?

Use the exercise index on this page — every exercise of Chapter 4 has its own page with complete step-by-step solutions.