While solving linear systems of equations, a new notation was introduced to reduce the amount of writing. The word matrix was first used by the English mathematician James Sylvester (1814–1897); Arthur Cayley (1821–1895) developed the theory of matrices and used them in linear transformations. Today matrices are used in high-speed computers and many other disciplines. The concept of determinants was used by Chinese and Japanese mathematicians, and the Japanese mathematician Seki Kowa (1642–1708) and the German mathematician Gottfried Wilhelm Leibniz (1646–1716) are credited with the invention of determinants. G. Cramer (1704–1752) employed determinants successfully for solving systems of linear equations.
A rectangular array of numbers enclosed by a pair of brackets is called a matrix, such as [[2, −1, 3], [−5, 4, 7]] (two rows, three columns) or the 4×3 matrix on page 44. The numbers in rows or columns are the entries or elements of the matrix. Generally, a bracketed rectangular array of m·n elements aij (i = 1, 2, …, m; j = 1, 2, …, n), arranged in m rows and n columns, is called an m by n matrix (written m×n); m×n is called the order of the matrix. Matrices are usually represented by capital letters A, B, C, X, Y, and small letters or double subscripts a11, a12, … indicate the entries. We write A = [aij]m×n, where aij is the element of the i-th row and j-th column — for example, 7 is at the a23 position in [[2, −1, 3], [−5, 4, 7]].
📌 Note (p. 46): the elements of matrices need not always be numbers, but in this study we take them from R or C (real or complex numbers). A matrix is called a real matrix if all of its elements are real.
Row matrix (row vector): a matrix with only one row, i.e. a 1×n matrix [a11 a12 … a1n]; e.g. [1 −1 3 4] is a row matrix with 4 columns. Column matrix (column vector): a matrix with only one column, i.e. an m×1 matrix; e.g. [[2], [−1], [3]] is a column matrix with 3 rows. Rectangular matrix: if m ≠ n, the matrix is rectangular — e.g. [[2, 3, 1], [−1, 0, 4]] (2×3) and the 4×3 matrix on page 46. Square matrix: if m = n, the matrix has the same number of rows and columns; e.g. [0], [[2, 5], [−1, 6]] and the 3×3 matrix on page 46 are square matrices of orders 1, 2 and 3.
خطی مساواتوں کے نظام حل کرتے وقت لکھنے کی مقدار کم کرنے کے لیے ایک نئی علامت متعارف ہوئی، جسے میٹرکس کہتے ہیں۔ یہ لفظ سب سے پہلے انگریز ریاضی دان جیمز سلویرسٹر نے استعمال کیا، اور آرتھر کیلی نے میٹرکس کا نظریہ تیار کیا۔
اعداد کی مستطیلی ترتیب جو قوسین میں بند ہو، میٹرکس کہلاتی ہے۔ اس میں افقی لائنیں قطاریں (rows) اور عمودی لائنیں کالم (columns) کہلاتی ہیں، اور اندر کے اعداد عناصر (entries) کہلاتے ہیں۔ m قطاروں اور n کالموں والی میٹرکس کو m×n میٹرکس کہتے ہیں، اور m×n اس کا آرڈر ہے۔ aij کا مطلب ہے i-ویں قطار اور j-ویں کالم کا عنصر۔
📌 نوٹ: عناصر ہمیشہ اعداد نہیں ہوتے، لیکن اس مطالعے میں ہم انہیں حقیقی یا مختلط اعداد سے لیں گے۔ جس میٹرکس کے تمام عناصر حقیقی ہوں، وہ حقیقی میٹرکس کہلاتی ہے۔
صرف ایک قطار والی میٹرکس قطار میٹرکس، صرف ایک کالم والی کالم میٹرکس، m ≠ n والی مستطیلی میٹرکس، اور m = n والی مربع میٹرکس کہلاتی ہے۔
For a square matrix A of order n, the entries a11, a22, …, ann form the principal diagonal (also called the leading or main diagonal), and a1n, a2,n−1, …, an1 form the secondary diagonal. In the 4×4 example on page 47, the principal diagonal is a11, a22, a33, a44 and the secondary diagonal is a14, a23, a32, a41.
Diagonal matrix: a square matrix with aij = 0 for all i ≠ j and at least one aij ≠ 0 for i = j — some diagonal entries may be zero, but not all; e.g. [7], [[1, 0, 0], [0, 2, 0], [0, 0, 5]] and the 4×4 matrix on page 47. Scalar matrix: aij = 0 for i ≠ j and aij = k (some non-zero scalar) for all i = j; e.g. [[7, 0], [0, 7]] and the 3×3 and 4×4 matrices of a and 3 on page 47. Unit (identity) matrix: aij = 0 for i ≠ j and aij = 1 for i = j, denoted In or I; e.g. I3 = [[1, 0, 0], [0, 1, 0], [0, 0, 1]]. Null (zero) matrix: every element zero, denoted Om×n; null matrices may be of any order.
📌 Note (p. 48): O may be used to denote a null matrix of any order if there is no confusion.
Equal matrices: two matrices of the same order are equal if their corresponding entries are equal — A = B iff aij = bij for all i, j. Transpose: if A is m×n, the n×m matrix obtained by interchanging rows and columns is the transpose of A, denoted At: At = [a′ij]n×m where a′ij = aji. So the 2nd row of B becomes the 2nd column of Bt, and so on — rows and columns swap places.
مربع میٹرکس میں a11, a22, …, ann اصلی قطر (principal diagonal) بناتے ہیں، اور مخالف کونے والی قطر ثانوی قطر (secondary diagonal) کہلاتی ہے۔
ڈائیگنل میٹرکس: قطر کے باہر سب صفر، قطر پر کم از کم ایک غیر صفر۔ اسکیلر میٹرکس: قطر کے باہر صفر، قطر پر سب برابر غیر صفر عدد k۔ یونٹ (شناختی) میٹرکس: قطر پر سب 1 اور باہر صفر، جسے In لکھتے ہیں۔ نل میٹرکس: تمام عناصر صفر، جسے O لکھتے ہیں۔
دو میٹرکس مساوی ہیں اگر ان کا آرڈر ایک ہو اور ہر متعلقہ عنصر برابر ہو۔ ٹرانسپوز: قطاروں اور کالموں کو آپس میں بدلنے سے بننے والی میٹرکس At کہلاتی ہے؛ یعنی پہلی قطار پہلا کالم بن جاتی ہے۔
4.2.1 Addition: two matrices are conformable for addition if they are of the same order. The sum A + B of m×n matrices A = [aij] and B = [bij] is the m×n matrix C = [cij] with cij = aij + bij — add corresponding entries, nothing fancier. 4.2.2 Subtraction: A − B = A + (−B) = [aij − bij] — subtract each entry of B from the corresponding entry of A.
Example 1 (pp. 49–50). For A = [[1, 0, −1, 2], [3, 1, 2, 5], [0, −2, 1, 6]] and B = [[2, −1, 3, 1], [1, 3, −1, 4], [3, 1, 2, −1]], A + B = [[3, −1, 2, 3], [4, 4, 1, 9], [3, −1, 3, 5]], so (A + B)t = [[3, 4, 3], [−1, 4, −1], [2, 1, 3], [3, 9, 5]]. Transposing A and B separately and adding gives the same matrix — hence (A + B)t = At + Bt.
4.2.3 Scalar multiplication: if A is m×n and k a real or complex number, kA = [kaij] — multiply every entry by k; the order stays m×n. 📌 Note (p. 50): if n is a positive integer, A + A + … (n terms) = nA.
4.2.4 Multiplication: A and B are conformable for the product AB if the number of columns of A equals the number of rows of B. If A is 2×3 and B is 3×2, AB is 2×2, with cij = sum of products of the i-th row of A with the j-th column of B — e.g. c21 = a21b11 + a22b21 + a23b31 (see Figure (A) on page 50). BA is then 3×3, a different order altogether. 📌 Note 1 (p. 51): in general, AB ≠ BA. 📌 Note 2: if AB is defined for A of order m×n and B of order n×p, AB has order m×p — the middle n “cancels”. Example 2 (p. 51): for A = [[2, −1, 0], [1, 2, −3], [1, 2, −2]] and B = [[2, −2, 3], [−1, −4, 6], [0, −5, 5]], first A² = A·A = [[3, −4, 3], [1, −3, 0], [2, −1, −2]], then A²B = [[10, −5, 0], [5, 10, −15], [5, 10, −10]]. Powers of square matrices are defined as A² = A×A, A³ = A×A×A, and Aⁿ = A×A×… (n factors).
جمع: دو میٹرکس تبھی جمع ہو سکتی ہیں جب ان کا آرڈر ایک ہو؛ متعلقہ عناصر جمع کریں۔ تفریق: A − B = [aij − bij]۔ اسکیلر ضرب: kA میں ہر عنصر کو k سے ضرب دیں، آرڈر وہی رہتا ہے۔
ضرب: AB تبھی ممکن ہے جب A کے کالموں کی تعداد B کی قطاروں کے برابر ہو۔ جواب کا عنصر cij، A کی i-ویں قطار اور B کے j-ویں کالم کے متعلقہ عناصر کی ضربوں کا مجموعہ ہے۔ یاد رکھیں: عام طور پر AB ≠ BA — ترتیب بدلنے سے جواب بدل جاتا ہے، اور کبھی آرڈر بھی مختلف ہو جاتا ہے۔
مثال ۱ سے (A + B)t = At + Bt ثابت ہوتا ہے۔ مثال ۲ میں پہلے A² نکالا گیا، پھر A²B = [[10, −5, 0], [5, 10, −15], [5, 10, −10]] حاصل ہوا۔
If A, B and C are conformable for the indicated sum or product and c, d are scalars, then: (i) Commutative w.r.t. addition: A + B = B + A. (ii) Associative w.r.t. addition: (A + B) + C = A + (B + C). (iii) Associative of scalar multiplication: (cd)A = c(dA). (iv) Existence of additive identity: A + O = O + A = A (O is the null matrix; the book’s side-note adds that A is a square matrix here). (v) Existence of multiplicative identity: IA = AI = A (I is the unit/identity matrix). (vi) Distributive w.r.t. scalar multiplication: (a) c(A + B) = cA + cB, (b) (c + d)A = cA + dA. (vii) Associative w.r.t. multiplication: A(BC) = (AB)C. (viii) Left distributive: A(B + C) = AB + AC. (ix) Right distributive: (A + B)C = AC + BC. (x) c(AB) = (cA)B = A(cB).
Notice what is missing from this list: commutativity of multiplication. That is deliberate — it is the one familiar number-rule that matrices break.
Example 3 (pp. 52–53). For A = [[2, 0, 1], [1, 4, 2], [3, 0, 6]] and B = [[1, −1, 0], [2, 3, −1], [1, −2, 3]], AB = [[3, −4, 3], [11, 7, 2], [9, −15, 18]] while BA = [[1, −4, −1], [4, 12, 2], [9, −8, 15]]. Hence AB ≠ BA. 📌 Note (p. 53): matrix multiplication is not commutative in general.
اگر A، B، C مناسب آرڈر کی ہوں اور c، d اسکیلر ہوں تو: جمع میں ترتیب بدل سکتے ہیں (A + B = B + A)، جمع اور ضرب میں انجمنی قانون (associative) درست ہے، نل میٹرکس O جمع کی شناخت ہے (A + O = A)، شناختی میٹرکس I ضرب کی شناخت ہے (IA = AI = A)، اور تقسیمی قوانین (distributive) بھی درست ہیں: A(B + C) = AB + AC اور (A + B)C = AC + BC۔
اس فہرست میں جو چیز غائب ہے وہ ضرب کی تبدیلی (commutativity) ہے — یہی وہ عام قاعدہ ہے جو میٹرکس توڑ دیتی ہیں۔ مثال ۳ میں AB اور BA دونوں نکال کر دکھایا گیا ہے کہ وہ برابر نہیں ہیں۔
The determinants of square matrices of order n ≥ 3 can be written by a pattern; for n = 3, if A = [[a11, a12, a13], [a21, a22, a23], [a31, a32, a33]], the determinant of A is |A| = the same array with vertical bars instead of brackets. (A matrix whose determinant is zero is called singular; otherwise it is non-singular — the terms are used from section 4.5 onward.)
4.4.1 Minor and cofactor. For a square matrix A of order n, the minor of an element aij, denoted Mij, is the determinant formed by deleting the i-th row and j-th column of A (or |A|). Example: for a 3×3 matrix, M12 = |a21 a23; a31 a33| (delete row 1, column 2). The cofactor Aij = (−1)i+j Mij — the minor with a checkerboard sign: A12 = (−1)3 M12 = −M12.
4.4.2 Determinant of order 3. |A| = ai1Ai1 + ai2Ai2 + ai3Ai3 for any row i, or a1jA1j + a2jA2j + a3jA3j for any column j — expand along any row or column; the answer is the same. In minor form, expanding along the second column gives |A| = −a12M12 + a22M22 − a32M32, and along the first row |A| = a11M11 − a12M12 + a13M13, i.e. |A| = a11(a22a33 − a23a32) − a12(a21a33 − a23a31) + a13(a21a32 − a22a31) = a11a22a33 + a12a23a31 + a13a21a32 − a11a23a32 − a12a21a33 − a13a22a31 — the six-term expansion for order 3.
Example 4 (p. 55). For A = [[1, −2, 3], [−2, 3, 1], [4, −3, 2]], expanding along row 1: |A| = 1·|3 1; −3 2| − (−2)·|−2 1; 4 2| + 3·|−2 3; 4 −3| = 1(6+3) + 2(−4−4) + 3(6−12) = 9 − 16 − 18 = −25. Example 5: for the same A, M12 = |−2 1; 4 2| = −8, M22 = |1 3; 4 2| = −10, M32 = |1 3; −2 1| = 7, so A12 = 8, A22 = −10, A32 = −7, and expanding down column 2: |A| = a12A12 + a22A22 + a32A32 = (−2)(8) + 3(−10) + (−3)(−7) = −16 − 30 + 21 = −25 — same answer, as it must be.
مربع میٹرکس کا دترمیننٹ |A| اسی ترتیب کو عمودی خطوں میں لکھ کر ظاہر کیا جاتا ہے۔ جس میٹرکس کا دترمیننٹ صفر ہو وہ سنگولر، ورنہ نان سنگولر کہلاتی ہے۔
مائنر Mij: عنصر aij کی قطار اور کالم حذف کر کے باقی دترمیننٹ۔ کوفیکٹر Aij = (−1)i+j Mij — یعنی شطرنجی علامت کے ساتھ مائنر۔
آرڈر ۳ کا دترمیننٹ کسی بھی قطار یا کالم کے ساتھ توسیع (expansion) سے نکلتا ہے: ہر عنصر کو اس کے کوفیکٹر سے ضرب دے کر جمع کریں۔ مثال ۴ اور ۵ میں ایک ہی میٹرکس کا دترمیننٹ دو طریقوں سے −25 آیا ہے — یہی اس طریقے کی جانچ ہے۔
i. For a square matrix A, |A| = |At| — transposing does not change the determinant. ii. If two rows or two columns are interchanged, the determinant becomes −|A|. iii. If A has two identical rows or two identical columns, |A| = 0. iv. If all entries of a row (or column) are zero, |A| = 0. v. If the entries of a row (or column) are multiplied by k ∈ R, the determinant becomes k|A| — a common factor can be pulled out of one row or column at a time. vi. If each entry of a row (or column) consists of two terms, the determinant splits as a sum of two determinants. vii. If any row (column) is multiplied by a non-zero k and added to another row (column), the determinant does not change — the workhorse for “without expansion” proofs. viii. If a matrix is in triangular form, its determinant is the product of the entries on its main diagonal.
📌 Note (p. 56): triangular matrices are defined on page 61 (see section 4.6 below).
Example 6 (p. 56). Without expansion, show |x a+x b+c; x b+x c+a; x c+x a+b| = 0. Add column 3 to column 2: the second column becomes [a+b+c+x, a+b+c+x, a+b+c+x] in every row. Now take x common from column 1 and (a+b+c+x) common from column 2: the determinant becomes x(a+b+c+x) times a determinant whose first two columns are both [1, 1, 1] — identical columns, so by property (iii) it is 0. Hence the whole determinant is 0, without ever expanding.
دترمیننٹ کی اہم خصوصیات: ٹرانسپوز سے دترمیننٹ نہیں بدلتا؛ دو قطاریں یا کالم بدلنے سے علامت بدل جاتی ہے؛ دو ایک جیسی قطاریں یا کالم ہوں تو دترمیننٹ صفر؛ پوری قطار صفر ہو تو دترمیننٹ صفر؛ ایک قطار سے مشترکہ عدد باہر نکال سکتے ہیں؛ ایک قطار کا کسی عدد سے ضرب شدہ حصہ دوسری قطار میں جمع کرنے سے دترمیننٹ نہیں بدلتا؛ اور تکونی (triangular) میٹرکس کا دترمیننٹ اس کے قطر کے عناصر کی ضرب ہے۔
مثال ۶ میں انہی خصوصیات سے، بغیر توسیع کے، دترمیننٹ صفر ثابت کیا گیا ہے: کالم جمع کرنے سے دو کالم ایک جیسے ہو گئے، اور ایک جیسے کالموں کا دترمیننٹ صفر ہوتا ہے۔
Let A be 3×3. The matrix of cofactors of A is [[A11, A12, A13], [A21, A22, A23], [A31, A32, A33]], and adj A is its transpose: adj A = [[A11, A21, A31], [A12, A22, A32], [A13, A23, A33]] — cofactor matrix, then transpose; forgetting the transpose is the classic mistake here.
Inverse of a square matrix of order n ≥ 3: let A be non-singular (|A| ≠ 0). If there is a matrix B with AB = BA = In, B is the multiplicative inverse of A, denoted A−1, of order n×n, with AA−1 = In and A−1A = In. If A is non-singular, A−1 = (1/|A|) adj A.
Example 7 (pp. 57–58). Find A−1 for A = [[1, 0, 2], [0, 2, 1], [1, −1, 1]]. Cofactors: A11 = 3, A12 = 1, A13 = −2, A21 = −2, A22 = −1, A23 = 1, A31 = −4, A32 = −1, A33 = 2. So the cofactor matrix is [[3, 1, −2], [−2, −1, 1], [−4, −1, 2]] and adj A = [[3, −2, −4], [1, −1, −1], [−2, 1, 2]]. |A| = a11A11 + a12A12 + a13A13 = 3 + 0 − 4 = −1. Hence A−1 = (1/−1) adj A = [[−3, 2, 4], [−1, 1, 1], [2, −1, −2]].
ایڈجوائنٹ (adj A): پہلے تمام عناصر کے کوفیکٹرز کی میٹرکس بنائیں، پھر اس کا ٹرانسپوز لیں — یہی adj A ہے۔ ٹرانسپوز بھولنا اس سوال میں سب سے عام غلطی ہے۔
اگر |A| ≠ 0 ہو تو میٹرکس نان سنگولر ہے اور اس کا معکوس موجود ہے: A−1 = (1/|A|) adj A۔ معکوس کی پہچان یہ ہے کہ AA−1 = A−1A = I ہو۔
مثال ۷ میں A = [[1, 0, 2], [0, 2, 1], [1, −1, 1]] کے لیے |A| = −1 اور A−1 = [[−3, 2, 4], [−1, 1, 1], [2, −1, −2]] حاصل ہوا۔
4.6 Elementary row operations. A system of linear equations is reduced to a simpler equivalent system by: (i) interchanging two equations, (ii) multiplying an equation by a non-zero number, (iii) adding a multiple of one equation to another. The corresponding row operations on matrices are: (i) interchanging two rows, (ii) multiplying a row by a non-zero number, (iii) adding a multiple of one row to another row. Notations: interchanging Ri and Rj is Ri ↔ Rj; k times Ri is kRi → Ri′; adding k times Rj to Ri is Ri + kRj → Ri′ (Ri′ is the new row). 📌 Note: matrices A and B are equivalent if B can be obtained by applying, in turn, a finite number of row operations on A; we write A ~ B (the book’s symbol), and if A ~ B then B ~ A.
Upper triangular: a square matrix with all elements below the principal diagonal zero (aij = 0 for i > j). Lower triangular: all elements above the diagonal zero (aij = 0 for i < j). A matrix is triangular if it is upper or lower triangular; e.g. [[1, 2, 3], [0, 1, 4], [0, 0, 6]] is upper triangular and the 4×4 matrix on page 61 is lower triangular. 📌 Remember: diagonal matrices are both upper and lower triangular.
4.7 Echelon and reduced echelon forms. In any non-zero row, the first non-zero entry is the leading entry. An m×n matrix is in echelon form if: (i) the number of zeros before the leading entry is greater than in the preceding row; (ii) every non-zero row precedes every zero row (if any); (iii) the leading entry in each row is 1. It is in reduced (row) echelon form if, in addition, whenever the leading entry in Ri lies in Cj, all other entries of Cj are zero. Example 8 (pp. 62–63): [[2, 3, −1, 9], [1, −1, 2, −3], [3, 1, 3, 2]] reduces to echelon form [[1, −1, 2, −3], [0, 1, −1, 3], [0, 0, 1, −1]] and to reduced echelon form [[1, 0, 0, 1], [0, 1, 0, 2], [0, 0, 1, −1]].
Inverse by row operations: if A is non-singular, applying elementary row operations to A⋮I in succession until A becomes I leaves I⋮A−1. Example 9 (pp. 63–64): for A = [[2, 5, −1], [3, 4, 2], [1, 2, −2]], |A| = 14 ≠ 0, and reducing [A⋮I] gives A−1 = [[−6/7, 4/7, 1], [4/7, −3/14, −1/2], [1/7, 1/14, −1/2]]. Rank: for a non-zero matrix A, if r is the number of non-zero rows when reduced to echelon form, r is the rank of A. Example 10: [[1, −1, 2, −3], [2, 0, 7, −7], [3, 1, 12, −11]] reduces to two non-zero rows, so its rank is 2.
ابتدائی قطار عملیں: (۱) دو قطاریں بدلنا، (۲) کسی قطار کو غیر صفر عدد سے ضرب دینا، (۳) ایک قطار کا کسی عدد سے ضرب شدہ حصہ دوسری قطار میں جمع کرنا۔ ان عملوں سے حاصل ہونے والی میٹرکس مساوی (equivalent) کہلاتی ہے۔
بالائی تکونی: قطر کے نیچے سب صفر۔ زیریں تکونی: قطر کے اوپر سب صفر۔ ڈائیگنل میٹرکس دونوں طرح کی تکونی ہوتی ہے۔
ایشلون صورت: ہر قطار کی پہلی غیر صفر قیمت (leading entry) 1 ہو، صفر والی قطاریں آخر میں ہوں، اور ہر قطار میں اس سے پہلی قطار سے زیادہ صفر ہوں۔ مختصر ایشلون صورت میں leading entry والے کالم کے باقی تمام عناصر بھی صفر ہوتے ہیں۔
قطار عملوں سے معکوس: [A⋮I] پر قطار عملیں لگا کر A کو I بنائیں تو دائیں طرف A−1 بن جاتا ہے۔ رینک: ایشلون صورت میں غیر صفر قطاروں کی تعداد ہی میٹرکس کا رینک ہے۔
Three linear equations in three variables, a1x + b1y + c1z = d1, a2x + b2y + c2z = d2, a3x + b3y + c3z = d3, form a system of non-homogeneous linear equations if the constant terms d1, d2, d3 are not all zero. A system is consistent if it has a unique solution or infinitely many solutions, and inconsistent if it has no solution. Three methods follow.
4.8.1 Reduced echelon form: (i) convert to the augmented matrix, (ii) convert to reduced echelon form, (iii) solve by back substitution. Example 11 (pp. 65–67) shows all three behaviours: (i) 2x + 5y − z = 5, 3x + 4y + 2z = 11, x + 2y − 2z = −3 gives x = −1, y = 2, z = 3 — unique solution, consistent. (ii) x + y + 2z = 1, 2x − y + 7z = 11, 3x + 5y + 4z = −3 reduces to x + 3z = 4, y − z = −3, 0z = 0 — z arbitrary, so x = 4 − 3t, y = t − 3, z = t — infinitely many solutions, consistent. (iii) x − y + 2z = 1, 2x − 6y + 5z = 7, 3x + 5y + 4z = −3 reduces to a row 0z = 4 — impossible, so no solution, inconsistent.
📌 Note (p. 67) — the rank test: the system has a unique solution if rank(coefficient matrix) = rank(augmented matrix) = number of variables; infinitely many solutions if the ranks are equal but less than the number of variables; and no solution if the ranks differ.
4.8.2 Matrix inversion method: write AX = B; if |A| ≠ 0, X = A−1B. Example 12 (pp. 67–69): x1 − 2x2 + x3 = −4, 2x1 − 3x2 + 2x3 = −6, 2x1 + 2x2 + x3 = 5 gives |A| = −1, A−1 = [[7, −4, 1], [−2, 1, 0], [−10, 6, −1]], so X = (1, 2, −1).
4.8.3 Cramer's rule: for AX = B with |A| ≠ 0, x1 = |A1|/|A|, x2 = |A2|/|A|, x3 = |A3|/|A|, where Ai is A with its i-th column replaced by B — the book derives this from X = (1/|A|) adj A · B. Example 13 (pp. 70–71): 3x1 + x2 − x3 = −4, x1 + x2 − 2x3 = −4, −x1 + 2x2 − x3 = 1 gives |A| = 9, x1 = −9/9 = −1, x2 = 9/9 = 1, x3 = 18/9 = 2, so the solution set is {(−1, 1, 2)}.
تین متغیروں کی تین خطی مساواتیں، جن کی مستقل قدریں سب صفر نہ ہوں، غیر ہوموجینیس نظام کہلاتی ہیں۔ نظام مطابق (consistent) ہے اگر اس کا ایک یا لامحدود حل ہو، اور نامطابق (inconsistent) ہے اگر کوئی حل نہ ہو۔
تین طریقے: (۱) افزودہ میٹرکس کو مختصر ایشلون صورت میں لا کر پیچھے سے حل کریں، (۲) AX = B سے X = A−1B نکالیں، (۳) کریمر کا قاعدہ: ہر متغیر کے لیے اس کے کالم کو مستقل قدروں سے بدل کر دترمیننٹ لیں اور |A| سے تقسیم کریں۔
رینک کی جانچ: اگر گتانک میٹرکس اور افزودہ میٹرکس کا رینک برابر ہو اور متغیروں کی تعداد کے برابر ہو تو واحد حل؛ برابر ہو مگر کم ہو تو لامحدود حل؛ اور اگر رینک مختلف ہوں تو کوئی حل نہیں۔ مثال ۱۱ میں تینوں صورتیں دکھائی گئی ہیں۔
4.9 Homogeneous systems have all constant terms zero: a11x1 + a12x2 + a13x3 = 0, and so on. Such a system is always consistent, since x1 = x2 = x3 = 0 always satisfies it. The solution (0, 0, 0) is the trivial solution; any other solution is a non-trivial solution.
4.9.1 Gaussian elimination: a systematic method, named after Carl Friedrich Gauss, of performing row operations on the augmented matrix to transform it into row-echelon form, then determining the solution by back substitution. Example 14 (p. 72): x + 2y + z = 0, 2x + 3y + 4z = 0, 4x + 3y + 2z = 0 reduces to rank 3 = number of variables, so only the trivial solution (0, 0, 0). Example 15 (p. 73): x1 + x2 + x3 = 0, x1 − x2 + 3x3 = 0, x1 + 3x2 − x3 = 0 reduces to rank 2 < 3, giving x2 = x3, x1 = −2x3, i.e. (x1, x2, x3) = (−2t, t, t) for any real t. 📌 Rule I (p. 74): a homogeneous system has only the trivial solution if rank of A = number of variables. 📌 Rule II: it has a non-trivial solution if rank of A < number of variables.
4.10 Applications. Matrices manipulate images in graphic design (scaling, rotation, reflection), encode and decode messages in cryptography, model earthquake waves in seismic analysis, and represent relationships in social network analysis. A transformation (reflection) matrix represents reflection across a mirror line: across the x-axis multiply by [[1, 0], [0, −1]]; across the y-axis by [[−1, 0], [0, 1]]; across y = x by [[0, 1], [1, 0]]. Example 16 (pp. 74–75): triangle A(2, 3), B(−1, 4), C(3, −2) reflected over the x-axis gives A′(2, −3), B′(−1, −4), C′(3, 2). Coding: a message is converted to a specific form using a code; to decode, multiply the coded matrix by the inverse of the given matrix. Example 17 (pp. 75–76): with A = [[1, 2], [3, 1]] and A–Z = 1–26, “ATTACK” splits as AT, TA, CK → [[1], [20]], [[20], [1]], [[3], [11]] → coded [[41], [23]], [[22], [61]], [[25], [20]].
ہوموجینیس نظام میں تمام مستقل قدریں صفر ہوتی ہیں، اس لیے (0, 0, 0) ہمیشہ حل ہے — یہ معمولی حل (trivial solution) ہے۔ اس کے علاوہ کوئی حل غیر معمولی حل کہلاتا ہے۔
گاؤسی اخراج: افزودہ میٹرکس پر قطار عملیں لگا کر اسے قطار ایشلون صورت میں لائیں، پھر پیچھے سے حل کریں۔ قاعدہ: رینک = متغیروں کی تعداد ہو تو صرف معمولی حل؛ رینک کم ہو تو غیر معمولی حل بھی موجود ہیں۔
عملی استعمال: گرافک ڈیزائن میں تصویر کو گھمانا/پلٹنا، خفیہ پیغامات کو کوڈ کرنا اور ڈی کوڈ کرنا۔ عکس (reflection) کے لیے: x-محور پر [[1, 0], [0, −1]]، y-محور پر [[−1, 0], [0, 1]]، اور y = x پر [[0, 1], [1, 0]] سے ضرب دیں۔ کوڈ کرنے کے لیے پیغام کے اعداد کو میٹرکس سے ضرب دیں، ڈی کوڈ کے لیے اس کے معکوس سے۔
Below are all questions of Exercises 4.1, 4.2 and 4.3 from Unit 4 of the official book (printed pages 53, 58–60 and 76–77), each solved step by step. The book has no multiple-choice section, no review exercise and no inquisitive/numerical section in this unit, so none is invented here. Counting rule: one numbered question = one answer block; its parts are solved inside that block. That gives 7 + 10 + 11 = 28 blocks, covering 77 parts in total (Ex. 4.1: 18 parts; Ex. 4.2: 36 parts; Ex. 4.3: 23 parts).
⚠️ Three answers worth a second look: Ex. 4.1 Q2 shows (A−B)−C ≠ A−(B−C) — subtraction of matrices is not associative. Ex. 4.2 Q1(ii) gives the neat value 9ab². Ex. 4.3 Q3(ii) and Q4(ii) have fraction answers (68/55, 59/55, 62/55 and −8/9, 10/9, 11/9) — they check in all three equations, so the fractions are genuine, not arithmetic slips. Ex. 4.3 Q10 needs padding: the message has 10 characters but the coding matrix is 3×3, so the last triple is padded with two spaces (27) — flagged in the answer, since the book does not state a padding rule.
(i) I3 is 3×3 and A is 3×4, so I3A is defined and is 3×4. Its (i, j) entry is Σk=1..3 (I3)ik·akj. Now (I3)ik is 1 when k = i and 0 otherwise, so the sum collapses to the single term aij. Every entry is unchanged, hence I3A = A.
(ii) A is 3×4 and I4 is 4×4, so AI4 is 3×4. Its (i, j) entry is Σk=1..4 aik·(I4)kj, and (I4)kj is 1 only when k = j, so the sum is aij. Hence AI4 = A. In words: multiplying by the identity on either side leaves the matrix untouched.
Subtract entry by entry.
(i) A − B = [[0−2, −1−1, 2−(−1)], [3−1, 2−2, 1−4], [−1−(−1), 0−2, 4−1]] = [[-2, −2, 3], [2, 0, −3], [0, −2, 3]].
(ii) B − C = [[2−1, 1−0, −1−(−2)], [1−(−1), 2−5, 4−0], [−1−3, 2−4, 1−(−1)]] = [[1, 1, 1], [2, −3, 4], [−4, −2, 2]].
(iii) (A − B) − C = [[−2−1, −2−0, 3−(−2)], [2−(−1), 0−5, −3−0], [0−3, −2−4, 3−(−1)]] = [[-3, −2, 5], [3, −5, −3], [−3, −6, 4]].
(iv) A − (B − C) = [[0−1, −1−1, 2−1], [3−2, 2−(−3), 1−4], [−1−(−4), 0−(−2), 4−2]] = [[-1, −2, 1], [1, 5, −3], [3, 2, 2]].
⚠️ Note (iii) ≠ (iv): matrix subtraction is not associative — where you put the brackets changes the answer, unlike addition.
(i) First AB = [[i(−i) + 2i(2i), i·1 + 2i·1], [1(−i) + (−i)(2i), 1·1 + (−i)(1)]] = [[1 − 4, 3i], [2 − i, 1 − i]] = [[−3, 3i], [2 − i, 1 − i]] (using i² = −1). Then (AB)C = [[−3(2i) + 3i(−i), −3·1 + 3i·i], [(2 − i)(2i) + (1 − i)(−i), (2 − i)·1 + (1 − i)i]] = [[−6i + 3, −6], [1 + 3i, 3]]. On the other side, BC = [[(−i)(2i) + 1(−i), (−i)(1) + 1·i], [2i(2i) + 1(−i), 2i·1 + 1·i]] = [[2 − i, 0], [−4 − i, 3i]], and A(BC) = [[i(2 − i) + 2i(−4 − i), 2i(3i)], [1(2 − i) + (−i)(−4 − i), (−i)(3i)]] = [[3 − 6i, −6], [1 + 3i, 3]]. Both sides agree, so (AB)C = A(BC) ✓.
(ii) B + C = [[i, 2], [i, 1 + i]]. Then A(B + C) = [[i·i + 2i·i, i·2 + 2i(1 + i)], [1·i + (−i)i, 1·2 + (−i)(1 + i)]] = [[−3, −2 + 4i], [1 + i, 3 − i]]. Meanwhile AC = [[i(2i) + 2i(−i), i·1 + 2i·i], [1(2i) + (−i)(−i), 1·1 + (−i)i]] = [[0, −2 + i], [−1 + 2i, 2]], so AB + AC = [[−3, 3i − 2 + i], [2 − i − 1 + 2i, 1 − i + 2]] = [[−3, −2 + 4i], [1 + i, 3 − i]]. Both sides agree, so A(B + C) = AB + AC ✓.
The scalar identities fail because matrix multiplication is not commutative: AB need not equal BA. Expanding with the distributive laws (which do hold):
(i) (A + B)² = (A + B)(A + B) = A² + AB + BA + B². This equals A² + 2AB + B² only if BA = AB, which is false in general. Counterexample: A = [[1, 1], [0, 0]], B = [[0, 1], [0, 0]] gives AB = [[0, 1], [0, 0]] but BA = [[0, 0], [0, 0]], so (A + B)² ≠ A² + 2AB + B².
(ii) (A − B)² = A² − AB − BA + B², which equals A² − 2AB + B² only if BA = AB — again false in general (same counterexample).
(iii) (A + B)(A − B) = A² − AB + BA − B², which equals A² − B² only if BA = AB — false in general. So all three familiar school identities need the extra condition AB = BA for matrices.
At = [[−1, 1, −3], [2, 0, 5], [3, 2, 3]].
A + At = [[−2, 3, 0], [3, 0, 7], [0, 7, 6]] — symmetric, as A + At always is.
A − At = [[0, 1, 6], [−1, 0, −3], [−6, 3, 0]] — skew-symmetric (diagonal zero, opposite signs across the diagonal), as A − At always is.
AAt: dot each row of A with each row of A (columns of At are rows of A): row1·row1 = 1+4+9 = 14, row1·row2 = −1+0+6 = 5, row1·row3 = 3+10+9 = 22, row2·row2 = 1+0+4 = 5, row2·row3 = −3+0+6 = 3, row3·row3 = 9+25+9 = 43. So AAt = [[14, 5, 22], [5, 5, 3], [22, 3, 43]].
AtA: dot columns of A: col1·col1 = 1+1+9 = 11, col1·col2 = −2+0−15 = −17, col1·col3 = −3+2−9 = −10, col2·col2 = 4+0+25 = 29, col2·col3 = 6+0+15 = 21, col3·col3 = 9+4+9 = 22. So AtA = [[11, −17, −10], [−17, 29, 21], [−10, 21, 22]]. Note AAt ≠ AtA here — another face of non-commutativity.
(At)t = A — transposing twice returns the original matrix.
X = A² − 5A + 4I. First A² = A·A: row1: [2·2+0·2+1·1, 2·0+0·1+1·(−1), 2·1+0·3+1·0] = [5, −1, 2]; row2: [2·2+1·2+3·1, 2·0+1·1+3·(−1), 2·1+1·3+3·0] = [9, −2, 5]; row3: [1·2+(−1)·2+0·1, 1·0+(−1)·1+0·(−1), 1·1+(−1)·3+0·0] = [0, −1, −2]. So A² = [[5, −1, 2], [9, −2, 5], [0, −1, −2]]. Then 5A = [[10, 0, 5], [10, 5, 15], [5, −5, 0]] and 4I = [[4, 0, 0], [0, 4, 0], [0, 0, 4]]. Hence X = [[5−10+4, −1−0+0, 2−5+0], [9−10+0, −2−5+4, 5−15+0], [0−5+0, −1+5+0, −2−0+4]] = [[-1, −1, −3], [−1, −3, −10], [−5, 4, 2]].
Start with A² and rewrite one factor using BA = A: A² = A·A = A·(BA) = (AB)·A = B·A = A. (Each step uses only associativity and the two given relations.) Similarly B² = B·B = B·(AB) = (BA)·B = A·B = B. Adding: A² + B² = A + B ✓. The trick is spotting that A can be replaced by BA and B by AB inside the squares.
(i) Expand along row 1: 1·|−1 −3; 3 2| − (−2)·|3 −3; −2 2| + (−4)·|3 −1; −2 3| = 1(−2 + 9) + 2(6 − 6) − 4(9 − 2) = 7 + 0 − 28 = −21.
(ii) Do R2 → R2 − R1 and R3 → R3 − R1 (determinant unchanged): rows become [a+b, a−b, a], [−b, 2b, −b], [−2b, b, b]. Take b common from rows 2 and 3: det = b² · |a+b a−b a; −1 2 −1; −2 1 1|. Expand: (a+b)(2+1) − (a−b)(−1−2) + a(−1+4) = 3(a+b) + 3(a−b) + 3a = 9a. Hence the determinant is 9ab².
(iii) Take x common from row 1, y from row 2, z from row 3: det = xyz · |2 1 1; 1 2 1; 1 1 2| = xyz · [2(4−1) − 1(2−1) + 1(1−2)] = xyz · (6 − 1 − 1) = 4xyz.
(i) C2 → C2 − C1 gives [1, 1, 1]t and C3 → C3 − C2 also gives [1, 1, 1]t — two identical columns, so the determinant is 0.
(ii) Row 3 is a combination of rows 1 and 2: −10·[5, 6, −1] + 26·[2, 2, 0] = [2, −8, 10] = R3. A row that is a combination of other rows makes the determinant 0 (equivalently, R3 + 10R1 − 26R2 = 0 row).
(iii) The matrix is skew-symmetric: its transpose is its negative (check: (1,2) entry 0 = −0, (1,3) entry b = −(−b) after transposing, etc.). For any 3×3 skew-symmetric matrix, |A| = |At| = |−A| = −|A|, so |A| = 0.
(iv) C2 → C2 + C1 makes column 2 equal to [l+m+n, l+m+n, l+m+n]t = (l+m+n)·C3 — two proportional columns, so the determinant is 0.
(v) Column 3 = [3x, 9x, 15x]t = 3x·[1, 3, 5]t = 3x·C2 — two proportional columns, so the determinant is 0.
(vi) Multiply row 1 by a, row 2 by b, row 3 by c (determinant multiplied by abc): the determinant becomes |abc a² a³; abc b² b³; abc c² c³| with the factor abc outside. Take abc common from column 1 and divide back: the left side equals |1 a² a³; 1 b² b³; 1 c² c³|. (If one of a, b, c is zero the same identity is checked directly — e.g. a = 0 gives b²c²(c − b) on both sides.)
(i) Take 5 common from row 2 and 5 from column 2 of the left determinant (25 in all). Expanding both sides confirms it: LHS = 3(25−250) − 5(5−70) = −675 + 325 = −350, and the right-hand determinant is 3(1−10) − (1−14) = −14, so 25·(−14) = −350 ✓.
(ii) R1 → R1 + R2 + R3 makes row 1 = [3a+x, 3a+x, 3a+x]; take (3a+x) out, then C2 → C2 − C1, C3 → C3 − C1 leaves [[1, 0, 0], [a, x, 0], [a, 0, x]], whose determinant is x². Hence x²(3a+x).
(iii) and (iv) together: subtract row 1 from rows 2 and 3 in both determinants. For (iii): rows become [1, x, yz], [0, y−x, z(x−y)], [0, z−x, y(x−z)] = (y−x)(z−x)·|1 x yz; 0 1 −z; 0 1 −y| = (y−x)(z−x)(z−y). For (iv): the same steps give (y−x)(z−x)·|1 x x²; 0 1 x+y; 0 1 x+z| = (y−x)(z−x)(z−y). The two determinants are equal, and (y−x)(z−x)(z−y) = (x−y)(y−z)(z−x) (each bracket flips sign twice over).
(v) C2 → C2 − C1, C3 → C3 − C1: rows become [1, 0, 0], [a+1, b−a, c−a], [(a+1)², (b−a)(b+a+2), (c−a)(c+a+2)]. Take (b−a) and (c−a) out; the remaining 2×2 determinant simplifies to (c−b), giving (b−a)(c−a)(c−b) = (a−b)(b−c)(c−a).
(vi) R1 → R1 − R2 − R3 gives [0, −2b², −2a²]. Expanding along this row: det = 2b²·(a²(c²+a²) − a²b²) − 2a²·(a²b² − b²(b²+c²)) = 2a²b²c² + 2a⁴b² − 2a²b⁴ − 2a⁴b² + 2a²b⁴ + 2a²b²c² = 4a²b²c².
(vii) C1 → C1 + C2 + C3 gives first column [a+b+c, 2(a+b+c), 2(a+b+c)]t; take (a+b+c) out, then R2 → R2 − 2R1, R3 → R3 − 2R1 leaves determinant a²+b²+c²−ab−bc−ca. Multiply: (a+b+c)(a²+b²+c²−ab−bc−ca) = a³+b³+c³−3abc.
(viii) R1 → R1 + R2 + R3 gives [a+b+c+t, a+b+c+t, a+b+c+t]; take (a+b+c+t) out, then C2 → C2 − C1, C3 → C3 − C1 leaves [[1, 0, 0], [b, t, 0], [c, 0, t]], determinant t². Hence t²(a+b+c+t).
(ix) R1 → R1 + R2 + R3 gives [a+b+c, a+b+c, a+b+c]; take (a+b+c) out, then C2 → C2 − C1, C3 → C3 − C1 leaves [[1, 0, 0], [2b, −(a+b+c), 0], [2c, 0, −(a+b+c)]], determinant (a+b+c)². Hence (a+b+c)³.
(x) Row 1 = (x+y+z)·[1, 1, 1] − row 2, so the determinant equals (x+y+z)·|1 1 1; x y z; x² y² z²| (the second determinant, with row 1 = row 2, is zero) = (x+y+z)·(x−y)(y−z)(z−x) by part (iv) applied to columns.
(xi) C2 → C2 − C1, C3 → C3 − C1: the second row becomes [b²−a², c²−a²] and the third [b³+b−a³−a, c³+c−a³−a] = (b−a)(b²+ba+a²+1) and (c−a)(c²+ca+a²+1). Taking (b−a)(c−a) out, the 2×2 determinant is (c−b)(ab+bc+ca−1), giving (a−b)(b−c)(c−a)(ab+bc+ca−1).
(xii) C1 → C1 − C3, C2 → C2 − C3 gives [[a, 0, 1], [0, b, 1], [−c, −c, 1+c]]; expanding along row 1: a·(b(1+c) + c) + 1·(bc) = ab + abc + ac + bc = abc + ab + bc + ca.
(xiii) Split column 3 into [a², b², c²]t minus [bc, ca, ab]t: the determinant splits into |1 a a²; 1 b b²; 1 c c²| minus |1 a bc; 1 b ca; 1 c ab|. The second determinant equals the first (both expand to ab² + bc² + ca² − a²b − b²c − c²a — check by the same row-subtraction as part (iv)), so their difference is 0.
(xiv) C1 → C1 − 2C2 makes column 1 = [−1, −1, −1]t; take −1 out, then R2 → R2 − R1, R3 → R3 − R1 leaves determinant (c−a) − 2(b−a) = a + c − 2b = 0 by the given condition 2b = a + c. Hence the determinant is 0.
(xv) C1 → C1 + C2 + C3 gives factor (a+b+c); the remaining factor is a²+b²+c²−ab−bc−ca (same computation as part (vii)), and using ω³ = 1 and 1 + ω + ω² = 0, a²+b²+c²−ab−bc−ca = (a+bω+cω²)(a+bω²+cω). Hence the determinant is (a+b+c)(a+bω+cω²)(a+bω²+cω).
(i) A13 = (+1)·|0 −5; −2 −2| = 0·(−2) − (−5)(−2) = −10. A23 = (−1)·|1 2; −2 −2| = −(−2 + 4) = −2. A33 = (+1)·|1 2; 0 −5| = −5. Then |A| = a13A13 + a23A23 + a33A33 = (−3)(−10) + 0·(−2) + 7·(−5) = 30 − 35 = −5 (expanding down column 3).
(ii) B31 = (+1)·|−2 5; −1 4| = −8 + 5 = −3. B32 = (−1)·|−5 5; −3 4| = −(−20 + 15) = 5. B33 = (+1)·|−5 −2; −3 −1| = 5 − 6 = −1. Then |B| = b31B31 + b32B32 + b33B33 = (−2)(−3) + (−1)(5) + 2(−1) = 6 − 5 − 2 = −1 (expanding along row 3).
(i) Expand along row 1: 2·|−4 −3; 1 0| − 1·|−1 −3; x 0| + x·|−1 −4; x 1| = 2(0+3) − (0+3x) + x(−1+4x) = 6 − 3x − x + 4x² = 4x² − 4x + 6. Set equal to 5: 4x² − 4x + 1 = 0, i.e. (2x − 1)² = 0, so x = 1/2 (a repeated root).
(ii) Expand along row 1: 1·|x+1 2; −3 x| − (x−1)·|−1 2; 2 x| + 3·|−1 x+1; 2 −3| = (x²+x+6) + (x−1)(x+4) + 3(1−2x) = 2x² − 2x + 5. Set equal to 9: 2x² − 2x − 4 = 0, i.e. x² − x − 2 = (x−2)(x+1) = 0, so x = 2 or x = −1.
(iii) Expand along row 1: 1·(x²−12) − 1·(2x−6) + 1·(12−3x) = x² − 5x + 6 = (x−2)(x−3). Set equal to 0: x = 2 or x = 3.
(i) AAt is 2×2: [[9+4+1, −6+2−3], [−6+2−3, 4+1+9]] = [[14, −7], [−7, 14]], so |AAt| = 196 − 49 = 147. AtA is 3×3: [[13, −4, 9], [−4, 5, 1], [9, 1, 10]], and its determinant is 13(50−1) + 4(−40−9) + 9(−4−45) = 637 − 196 − 441 = 0. (It must be zero: A has only 2 rows, so AtA cannot have full rank 3.)
(ii) AAt is 3×3: [[10, 8, 6], [8, 8, 8], [6, 8, 10]], determinant 10(80−64) − 8(80−48) + 6(64−48) = 160 − 256 + 96 = 0 (A has only 2 columns). AtA is 2×2: [[14, 10], [10, 14]], so |AtA| = 196 − 100 = 96. The zero and non-zero swap between the two parts — exactly the rank reason above.
Multiplying the whole matrix by k multiplies every entry of every row by k. A determinant takes one factor out per row (property (v) of section 4.4.3): pulling k out of row 1 gives k|A′|, out of row 2 gives k²|A″|, out of row 3 gives k³|A|. Hence |kA| = k³|A|. (In general, for order n, |kA| = kⁿ|A|.)
Singular means determinant zero.
For A: |A| = 4(λ·1 − 6·3) − 2(7·1 − 6·2) + 3(7·3 − λ·2) = 4(λ − 18) − 2(−5) + 3(21 − 2λ) = 4λ − 72 + 10 + 63 − 6λ = 1 − 2λ. Set = 0: λ = 1/2.
For B: |B| = −2·|−2 1; λ 0| − 4·|1 1; 2 0| + 5·|1 −2; 2 λ| = −2(0 − λ) − 4(0 − 2) + 5(λ + 4) = 2λ + 8 + 5λ + 20 = 7λ + 28. Set = 0: λ = −4.
|A| = 1·|0 4; 4 0| − 2·|−5 4; 5 0| + 1·|−5 0; 5 4| = −16 + 40 − 20 = 4 ≠ 0, so the inverse exists. Cofactors: A11 = −16, A12 = 20, A13 = −20, A21 = 4, A22 = −5, A23 = 6, A31 = 8, A32 = −9, A33 = 10. Transposing the cofactor matrix gives adj A = [[−16, 4, 8], [20, −5, −9], [−20, 6, 10]]. Hence A−1 = (1/4) adj A = [[-4, 1, 2], [5, −5/4, −9/4], [−5, 3/2, 5/2]]. Check A−1A: row 1 gives [−4−5+10, −8+0+8, −4+4+0] = [1, 0, 0]; row 2 gives [5+25/4−45/4, 10+0−9, 5−5+0] = [0, 1, 0]; row 3 gives [−5−15/2+25/2, −10+0+10, −5+6+0] = [0, 0, 1]. So A−1A = I3 ✓.
(i) AB = [[1+3+0, 1+2+2], [0+9+0, 0+6+1]] = [[4, 5], [9, 7]], so (AB)t = [[4, 9], [5, 7]]. Bt = [[1, −3, 0], [1, −2, 1]] and At = [[1, 0], [−1, −3], [2, 1]], so BtAt = [[1+3+0, 0+9+0], [1+2+2, 0+6+1]] = [[4, 9], [5, 7]]. Both sides agree ✓. Note the order reverses: Bt first, then At.
(ii) AB = [[1−4, −3+2], [1−8, −3+4], [2−2, −6+1]] = [[−3, −1], [−7, 1], [0, −5]], so (AB)t = [[−3, −7, 0], [−1, 1, −5]]. Bt = [[1, −2], [−3, 1]] and At = [[1, 1, 2], [2, 4, 1]], so BtAt = [[1−4, 1−8, 2−2], [−3+2, −3+4, −6+1]] = [[−3, −7, 0], [−1, 1, −5]]. Both sides agree ✓.
Method for each: form [A⋮I] and apply row operations until the left side is I; the right side is then A−1. (The row-operation chains are long; the key steps and the verified results are given, and each inverse was checked by multiplying back to I.)
(i) |A| = −12 ≠ 0. Reducing [A⋮I]: R1 ↔ R3 and scaling lead, after the usual elimination, to A−1 = [[1, 17/4, 1/2], [0, −1/2, 0], [1/3, 11/6, 1/3]]. Check: A·A−1 = I (e.g. row 1 of A times column 1 of the inverse: 2·1 + 6·0 + (−3)(1/3) = 1 ✓).
(ii) |A| = 10 ≠ 0. Reducing [A⋮I] gives A−1 = [[-2/5, −2/5, 7/5], [4/5, 3/10, −4/5], [1/5, 1/5, −1/5]], and A·A−1 = I ✓.
(iii) |A| = −3 ≠ 0. Reducing [A⋮I] gives A−1 = [[-13/3, 8/3, 26/3], [2/3, −1/3, −4/3], [2/3, −1/3, −1/3]], and A·A−1 = I ✓.
(i) R2 + 2R1 → [0, −8, 7, 1]; R3 − 3R1 → [0, 4, −5, −5]; R3 + (1/2)R2 → [0, 0, −3/2, −9/2]. Three non-zero rows remain, so the rank is 3.
(ii) R2 − 2R1 = [0, 0, 0] (row 2 is twice row 1 — dependent). The remaining rows [1, −2, 3], [−1, 0, 2], [0, 1, −1] reduce to three non-zero rows (e.g. [1, −2, 3], [0, 1, −1], [0, 0, 3]), so the rank is 3.
(iii) Reducing to echelon form leaves four non-zero rows (no row becomes zero; the four rows are independent), so the rank is 4 — full row rank for a 4×5 matrix.
(i) |A| = 2(−1−4) − (1−6) − (2+3) = −10. Replace each column by [1, 3, 4]t in turn: Dx = −10, Dy = 0, Dz = −10. So x = (−10)/(−10) = 1, y = 0/(−10) = 0, z = (−10)/(−10) = 1. Solution (1, 0, 1) ✓ (check: 1 − 0 + 2 = 3 ✓).
(ii) |A| = 1(−2−3) − 2(8+2) − 3(12−2) = −55. D1 = −68, D2 = −59, D3 = −62. So x1 = 68/55, x2 = 59/55, x3 = 62/55. ⚠️ The fractions look untidy but they check: 68/55 + 118/55 − 186/55 = 0 ✓, and 272/55 − 59/55 + 62/55 = 5 ✓.
(iii) |A| = 2(−2+4) + (−1−2) + (−2−2) = −3. D1 = −8, D2 = −6, D3 = 7. So x1 = 8/3, x2 = 2, x3 = −7/3 (check: 16/3 − 2 − 7/3 = 1 ✓).
(i) A = [[1, −2, 1], [3, 1, −2], [0, 1, −1]], |A| = −2. Cofactor matrix = [[1, 3, 3], [−1, −1, −1], [3, 5, 7]], so adj A = [[1, −1, 3], [3, −1, 5], [3, −1, 7]] and A−1 = (−1/2) adj A. Then X = A−1[−1, 4, 1]t = (−1/2)[−2, −2, 0]t = (1, 1, 0) ✓ (1 − 2 + 0 = −1 ✓).
(ii) A = [[2, 1, 3], [1, 3, −2], [−3, −1, 2]], |A| = 36. adj A = [[4, −5, −11], [4, 13, 7], [8, −1, 5]]. Then X = (1/36) adj A·[3, 0, 4]t = (1/36)[−32, 40, 44]t = (−8/9, 10/9, 11/9) ✓ (check first equation: −16/9 + 10/9 + 33/9 = 3 ✓).
(iii) A = [[1, 1, 0], [2, 0, −1], [0, 2, −3]], |A| = 8. adj A = [[2, 3, −1], [6, −3, 1], [4, −2, −2]]. Then X = (1/8) adj A·[2, 1, −1]t = (1/8)[8, 8, 8]t = (1, 1, 1) ✓.
(i) Augmented matrix [[1, 2, −2, −1], [2, 3, 1, 1], [5, 4, −3, 1]]. R2 − 2R1 = [0, −1, 5, 3]; R3 − 5R1 = [0, −6, 7, 6]; R3 − 6R2 = [0, 0, −23, −12]. Echelon form: [[1, 2, −2, −1], [0, −1, 5, 3], [0, 0, −23, −12]]. Back substitution: x3 = 12/23, x2 = −9/23, x1 = 19/23. Reduced echelon form: [[1, 0, 0, 19/23], [0, 1, 0, −9/23], [0, 0, 1, 12/23]].
(ii) Augmented matrix [[1, 2, 1, 2], [2, 1, 2, 3], [2, 3, −1, 7]]. R2 − 2R1 = [0, −3, 0, −1]; R3 − 2R1 = [0, −1, −3, 3]. Scale and eliminate to echelon form [[1, 2, 1, 2], [0, 1, 0, 1/3], [0, 0, 1, −10/9]]. Back substitution: z = −10/9, y = 1/3, x = 22/9. Reduced echelon form: [[1, 0, 0, 22/9], [0, 1, 0, 1/3], [0, 0, 1, −10/9]].
(iii) Augmented matrix [[1, 4, 1, 2], [2, 1, −2, 9], [3, 1, −1, 12]]. R2 − 2R1 = [0, −7, −4, 5]; R3 − 3R1 = [0, −11, −4, 6]; R3 − (11/7)R2 = [0, 0, 16/7, −13/7]. Echelon form (with leading 1s): [[1, 4, 1, 2], [0, 1, 4/7, −5/7], [0, 0, 1, −13/16]]. Back substitution: x3 = −13/16, x2 = −1/4, x1 = 61/16. Reduced echelon form: [[1, 0, 0, 61/16], [0, 1, 0, −1/4], [0, 0, 1, −13/16]].
(i) Matrix [[1, 4, −2], [2, 1, 5], [5, 2, 8]] → R2 − 2R1 = [0, −7, 9], R3 − 5R1 = [0, −18, 18], R3 − (18/7)R2 = [0, 0, −36/7]. Rank 3 = number of variables, so by Rule I only the trivial solution (0, 0, 0).
(ii) Matrix [[1, 4, 2], [2, 1, −3], [3, 2, −4]] → R2 − 2R1 = [0, −7, −7], R3 − 3R1 = [0, −10, −10], and the third row becomes zero after R3 − (10/7)R2. Rank 2 < 3, so by Rule II there are non-trivial solutions: −7x2 − 7x3 = 0 gives x2 = −x3, and x1 + 4x2 + 2x3 = 0 gives x1 = 2x3. Solution: (2t, −t, t) for any real t.
(iii) Matrix [[1, 2, −1], [1, −1, 5], [2, 1, 4]] → R2 − R1 = [0, −3, 6], R3 − 2R1 = [0, −3, 6], and the third row becomes zero. Rank 2 < 3: −3x2 + 6x3 = 0 gives x2 = 2x3, and x1 + 2x2 − x3 = 0 gives x1 = −3x3. Solution: (−3t, 2t, t) for any real t.
Reflection over the y-axis uses [[−1, 0], [0, 1]]: (x, y) → (−x, y). So A′ = [[−1, 0], [0, 1]]·[4, 1]t = (−4, 1); B′ = (2, 5); C′ = (0, −3) (on the axis, so unchanged). The reflected triangle has vertices (−4, 1), (2, 5), (0, −3).
PA = A′ means −5·(x, y, z) = (30, 20, −5). Divide by −5: x = 30/(−5) = −6, y = 20/(−5) = −4, z = (−5)/(−5) = 1. So A = (−6, −4, 1). (Equivalently, A = P−1A′ with P−1 = (−1/5)I.)
The transformation sends (x, y) to (x′, y′) = (x + 2y, 3x + 4y). To find the image equation, invert it: the inverse matrix is (−1/2)[[4, −2], [−3, 1]] = [[−2, 1], [3/2, −1/2]], so x = −2x′ + y′ and y = (3x′ − y′)/2. Substitute into y = x²: (3x′ − y′)/2 = (−2x′ + y′)² = 4x′² − 4x′y′ + y′². Multiply by 2: 3x′ − y′ = 8x′² − 8x′y′ + 2y′². Dropping the primes, the image curve is 8x² − 8xy + 2y² − 3x + y = 0.
With A–Z = 1–26 and space = 27 (the convention the book states in Q11), the message “KEEP IT UP” is K=11, E=5, E=5, P=16, space=27, I=9, T=20, space=27, U=21, P=16 — 10 characters. ⚠️ The book does not state a padding rule, but the coding matrix is 3×3, so the characters must be grouped in triples; the last triple is padded here with two spaces (27, 27) to complete it — flagged as an assumption, not a book rule. Groups: [11, 5, 5], [16, 27, 9], [20, 27, 21], [16, 27, 27]. Multiply each on the left by A: A·[11, 5, 5]t = [16, 22, 15]t; A·[16, 27, 9]t = [25, 14, 45]t; A·[20, 27, 21]t = [41, 34, 69]t; A·[16, 27, 27]t = [43, 32, 81]t. The coded message is [16, 22, 15], [25, 14, 45], [41, 34, 69], [43, 32, 81].
To decode, multiply by A−1. |A| = −1. Cofactors give adj A = [[−1, −2, 1], [1, 3, −2], [1, 1, −1]], so A−1 = (1/−1) adj A = [[1, 2, −1], [−1, −3, 2], [−1, −1, 1]]. Decode each block: A−1·[11, 20, 43]t = [11+40−43, −11−60+86, −11−20+43]t = [8, 15, 12]t = H, O, L; A−1·[25, 10, 41]t = [25+20−41, −25−30+82, −25−10+41]t = [4, 27, 6]t = D, space, F; A−1·[22, 14, 41]t = [22+28−41, −22−42+82, −22−14+41]t = [9, 18, 5]t = I, R, E. Reading together: “HOLD FIRE” ✓.
اس باب میں کیا ہے؟ Matrices کی پوری بنیاد: matrix کیا ہے اور اسکی قسمیں, operations (جمع, تفریق, scalar اور matrix ضرب), ان کی خصوصیات, determinants (minor, cofactor, expansion, khususiyaat), adjoint اور inverse, row operations, echelon forms اور rank, پھر linear systems کے تین حل (reduced echelon, inversion, Cramer), homogeneous systems کے لیے Gaussian elimination, اور آخر میں اصل دنیا کے استعمال (reflection aur coding).
Revision guidance: سب سے پہلے ضرب کا rule (row × column) اور AB ≠ BA پکا کرو — آدھے سوال اسی پر ہیں. Determinant میں پہلے properties آزماؤ, پھر expansion; “without expansion” والوں میں identical یا proportional rows/columns ڈھونڈو. Inverse میں transpose مت بھولو, اور systems میں آخری row کو غور سے پڑھو: 0 = 0 مطلب آذادی, 0 = 4 مطلب کوئی حل نہیں. Exercise: 📝 Official Exercise Questions — with Answers
اس باب میں میٹرکس کی تعریف اور اقسام، عوامل، خصوصیات، دترمیننٹ، ایڈجوائنٹ اور معکوس، قطار عملیں، ایشلون صورتیں اور رینک، خطی نظاموں کے حل، اور عملی استعمالات (عکس اور کوڈنگ) شامل ہیں۔
نظرثانی کی رہنمائی: ضرب کا قاعدہ اور AB ≠ BA اچھی طرح یاد کریں، دترمیننٹ میں پہلے خصوصیات آزمائیں، معکوس میں ٹرانسپوز نہ بھولیں، اور نظاموں میں آخری قطار سے مطابقت کا فیصلہ کریں۔
Chapter-wise tests, fresh random papers اور instant score کے لیے PunjabTest.com کا میں test section دیکھو:
PunjabTest.com — Complete Guide Book • FSc Part 1 Mathematics, Chapter 4 (New Book, Session 2026–27) • English + Urdu