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- 1. MATRIX:Definition: A matrix is defined as an orderedrectangular array of numbers. They can be used torepresent systems of linear equations, as will beexplained below.Here are a couple of examples of different types ofmatrices: Upper LowerSymme Diago Identit Triang Triangul Zero tric nal y ular ar And a fully expanded m×n matrix A, would looklike this:... or in a more compact form:
- 2. The horizontal lines in a matrix are called rows andthe vertical lines are called columns. A matrix withm rows and n columns is called an m-by-n matrix (orm×n matrix) and m and n are called its dimensions.The places in the matrix where the numbers are, arecalled entries. The entry of a matrix A that lies in therow number i and column number j is called the i,jentry of A. This is written as A[i,j] or aij.We write to define an m × n matrix A witheach entry in the matrix called aij for all 1 ≤ i ≤ mand 1 ≤ j ≤ n. Example The matrix is a 4×3 matrix. This matrix has m=4 rows, and n=3columns.
- 3. The element A[2,3] or a23 is 7. OperationsAddition The sum of two matrices is the matrix, which (i,j)-thentry is equal to the sum of the (i,j)-th entries of twomatrices: The two matrices have the same dimensions. Here A +B = B + A is true.Subtraction If A and B are matrices of the same type then thesubtraction is found by subtracting the correspondingelements aij − bij. Here is an example of subtracting matrices. Multiplication of two matrices The multiplication of two matrices is a bit morecomplicated: So with Numbers:
- 4. Two matrices can be multiplied with each other even if they have different dimensions, as long as the number of columns in the first matrix is equal to the number of rows in the second matrix. The result of the multiplication, called the product, is another matrix with the same number of rows as the first matrix and the same number of columns as the second matrix. the multiplication of matrices is not commutative, this means, in general that the multiplication of matrices is associative, this means Special matricesThere are some matrices that are special. Square matrixA square matrix has the same number of rows as columns,so m=n.An example of a square matrix isThis matrix has 3 rows and 3 columns: m=n=3.IdentityEvery square dimension set of a matrix has a specialcounterpart called an "identity matrix". The identity
- 5. matrix has nothing but zeroes except on the maindiagonal, where there are all ones. For example:is an identity matrix. There is exactly one identity matrixfor each square dimension set. An identity matrix isspecial because when multiplying any matrix by theidentity matrix, the result is always the original matrixwith no change.Inverse matrixAn inverse matrix is a matrix that, when multiplied byanother matrix, equals the identity matrix. For example: is the inverse of .One column matrixA matrix, that has many rows, but only one column, iscalled a column vector.Transpose of Matrices DEFINITION: The transpose of a matrix is found byexchanging rows for columns i.e. Matrix A = (aij) and thetranspose of A is:
- 6. AT = (aji) where j is the column number and i is therow number of matrix A. For example, the transpose of a matrix would be: In the case of a square matrix (m = n), the transpose can be used to check if a matrix is symmetric. For a symmetric matrix A = AT.trace(A) The trace of a matrix is simply the summation of its main diagonal.AT The transpose of a matrix is switching the rows and columns. For example: a b c a d g T A =d e f A =b e h g h i c f i The adjoint of a square matrix [aij] is defined as the transpose of the matrix [Aij] where Aij are the cofactors of the elements aij. Adjoint of A is denoted by adj A. Adjoint of asquare matrix
- 7. Minors and Co-factors Minor : Minor of an element Aij of a determinant is the determinant obtained by deleting its ith row and jth column in which element aij lies. Minor of an element aij is denoted by Mij. Co-factor : Cofactor of an element Aij , denoted by Aij is defined by Aij =(-1) i+j Mij Properties of adjoint of a matrix , if A is aninvertible square matrix of order n. 1. A.(adj A) = (adj A). A = |A| In ( Note: |A(adjA)| = |A|n, if |A| ≠0, then |adjA| =|A|n-1 ) 2. adj (AB) = (adj B) . (adj A). 3. adj(adjA) = |A|n-2 .A, if a is an invertible squarematrix of order n. 4. If A is an invertible symmetric matrix, then A-1is also sym. 5. Every skew- sym. Matrix of odd order issingular
- 8. Example:Find the adjoint of the matrix.A11 = + = 4 – 5 = -1A12 = - = -4A13 = + =8A21 = - = 19A22 =+ = 14A23 = - =3
- 9. A31 = + = -8 A32 = - = -1 A33 = + =2 Inverse of a matrix If A is any square matrix of order n and there exist another square matrix B of the same order n, such that AB=BA =I , then B is called an inverse matrix of A and is denoted by A-1 =The Inverse of a Matrix
- 10. DEFINITION: Assuming we have a square matrix A, which is non-singular (i.e. det(A) does not equal zero), then there exists an n×n matrix A-1 which is called the inverse of A, such that this property holds: AA-1 = A-1A = I, where I is the identity matrix. The inverse of a 2×2 matrix Take for example an arbitrary 2×2 Matrix A whosedeterminant (ad − bc) is not equal to zero. where a,b,c,d are numbers, The inverse is: Singular Matrix : Any matrix whose determinant is zero, is singular matrix. Non-singular Matrix : Any matrix whose determinant is not zero, is non-singular matrix. Note 1: Only a square matrix can have its inverse. Note 2: From the definition, it is clear that if B is theinverse of A, then A is the inverse of B.
- 11. Note 3: Inverse of A is denoted by A-1, thus B = A-1 and AA-1 = A-1A=I.Inverse matrix An inverse matrix is a matrix that, when multipliedby another matrix, equals the identity matrix. Forexample: is the inverse of . matrix Q is orthogonal if its transpose is equal to itsinverse: which entails where I is the identity matrix. An orthogonal matrix Q is necessarily square andinvertible, with inverse Q−1 = QT. Question: Give an examples of matrices A, B and Csuch that (i) AB ≠ BA (ii) AB = O = BA, A≠ O, B ≠ O (iii) AB = O but BA ≠O (iv) AB=AC but B≠ C,A≠ O.
- 12. Answer: (i) Let A = and B = then AB= BA= (ii) Let A= and B = (iii)Let A= and B = then AB = O, BA = (iv) Let A= ,B= and C = then AB = = AC . Give an example of matrices A, B such that AB=BA A= B= . Question: If A = then A5 . Answer: A = 2I therefore A5 = (2I)5 = 32 I = 16(2I)= 16A. **Question: Given that A = and Xbe the solution set of equation Ax = A, where xЄ N-{1}. Evaluate .
- 13. Answer: A2 = A (by multiplying) and A3 =A and so on...... Thus An = A for n = 2,3,4...... Now, = = ( . ..... . )( . .... ) = 3/2. Question: (i) ≠ , why? [Not same order] (ii) Is the equation - = - Valid? [No] Question: (i) Let = and = , then - =0 [Hint = ]
- 14. (ii) If f(x) = , thenwhich is correct f(a)=0 , f(b)=0, f(0)=0 and f(1)=0 [ Hint f(0)=0 ∵ det.(skew-symm. matrix)=0]. (iii) Let f(t) = , then is equal to 0,-1,2,3. [Hint 0, = → as t→ ]. (iv) There are two values of a which makes determinant = = 86, then sum of these numbers is 4,5,-4,9. [Hint a=-4, operate R2 – 2R1]
- 15. Question: If A = , show that A2 – 6A +17I =0. Hence find A-1 . Solution: A2 = A.A = = A2 – 6A +17I = -6 + 17 = =0 A2 – 6A +17I = 0 ⇨ 17I = - A2 +6A ⇨ IA-1 = -1/17 A2 A-1 + 6/17 AA-1 = -1/17A+6/17I A-1 = 1/17 +6/17 = 1/17 . NOTE If A is symm. As well as skew-symm., then Ais a null matrix.( if A = AT then A is Symm. And if A = -AT then A is skew- symm.) A= is symmetric and B = is skew-symmetric.
- 16. NOTE: (i) If A and B are symmetric matrices, thenBA-2AB is neither symm. nor skew-symm. (ii) If A is symm. matrix then BTAB issymm. (iii) If A and B are symmetric matrices ofsame order, then AB is symm. iff AB=BA. (iv) Zero matrix is both symm. and skew-symm. (v) Sum of two skew-symm. matrices isalways skew-symm. (vi) If A is a symm., then A3 is a symm. andif A is skew-symm., then A2 is a symm. Theorem: The inverse of a square matrix if it exists, isunique. Let A be an invertible square matrix. If possible,let B and C be two inverse of A. Then AB = BA = I. AC = CA = I (by def. of inverse) Now, B = BI = B(AC) = (BA)C [ Matrix multiplication is associative] = IC = C i.e., B = C Hence the inverse of A is unique.
- 17. Theorem If A and B are two invertible matrices of thesame order, then (AB)-1 = B-1A-1. Proof: From the definition of inverse of a matrix, wehave (AB)(AB)-1 = I or A-1 (AB)(AB)-1 = A-1 I (Pre-multiplying bothsides by A-1) or (A-1A) B (AB)-1 = A-1 (Since A-1 I = A-1) or I B (AB)-1 = A-1 or B (AB)-1 = A-1 or (B-1B)(AB)-1 =B-1A-1 or I(AB)-1= B-1A-1 or (AB)-1 = B-1A-1
- 18. Properties of Inverse of Matrix In other words, a square matrix A is invertible ifand only if A is a non-singular matrix. (c) If A and B are invertible square matrices,then (AB)-1 = B-1 A-1 (d) If A and B are two non-singular square matrices of the same order, then AB and BA are also non-singular matrices of the same order. The Determinant of a MatrixDEFINITION: Determinants play an important role infinding the inverse of a matrix and also in solving systemsof linear equations. In the following we assume we have asquare matrix (m = n). The determinant of a matrix A willbe denoted by det(A) or |A|. Firstly the determinant of a2×2 and 3×3 matrix will be introduced, then the n×n casewill be shown.Determinant of a 2×2 matrix
- 19. Assuming A is an arbitrary 2×2 matrix A, where theelements are given by: then the determinant of a this matrix is as follows: Determinant of a 3×3 matrixThe determinant of a 3×3 matrix is a little more tricky andis found as follows (for this case assume A is an arbitrary3×3 matrix A, where the elements are given below).then the determinant of a this matrix is as follows: Consistent and InconsistentSolutions Consistent system : A system of equation is said to be consistent if its solution ( one or more ) exists. Inconsistent system : A system of
- 20. equation is said to be inconsistent if its solution does not exist. Working rule to check consistency: Case I When A 0 System is consistence and has unique solution. Case II When A =0 . Find Adj(A) and then find Adj(A) .B If Adj(A) .B 0 then system is inconsistence . Case III If Adj(A).B=0 Then it may have infinite solutions then it is consistence or have no solution then it is inconsistence. Solving Systems of Equations using Matrices DEFINITION: A system of linear equations is a set ofequations with n equations and n unknowns, is of the formof The unknowns are denoted by x1, x2, ..., xn and the coefficients (a and b above) are assumed to be given. In matrix form the system of equations above can be written as:
- 21. A simplified way of writing above is like this: Ax = bInverse Matrix MethodDEFINITION: The inverse matrix method uses theinverse of a matrix to help solve a system ofequations, such like the above Ax = b. By pre-multiplying both sides of this equation by A-1 gives:or alternativelySo by calculating the inverse of the matrix andmultiplying this by the vector b we can find thesolution to the system of equations directly. And fromearlier we found that the inverse is given byFrom the above it is clear that the existence of asolution depends on the value of the determinant of A.There are three cases: 1. If the det(A) does not equal zero then solutions exist using 2. If the det(A) is zero and b=0 then the solution will be not be unique or does not exist.
- 22. 3. If the det(A) is zero and b=0 then the solution can be x = 0 but as with 2. is not unique or does not exist. Looking at two equations we might have that Written in matrix form would look like and by rearranging we would get that the solutionwould look like Three simultaneous equations in x, y and z ax + by + cz = p dx + ey + fz = q gx + hy + iz = r To solve use the following By cramer’s rule
- 23. Solve the system using matrices. −x + 5y = 4 2x + 5y = −2 Always check your solutions! Answer , andTo solve the system, we need the inverse of A,which we write as A-1.Swap leading diagonal:Change signs of the other 2 elements:
- 24. Now we find the determinant of A:|A| = -5 - 10 = -15 So So the solution to the system is given by: This answer means that we have found thesolution x = -2 and y = 2/5. Is the solution correct? We check it in the original set of equations:
- 25. Substituting x = -2 and y = 2/5, we get: −(−2) + 5×(2/5) = 2 + 2 = 4 2×(−2) + 5×(2/5) = −4 + 2 = −2 So the solution to the original system ofequations is x = -2, y = 2/5.Solve the system using matrix methods.Did I mention? Its a good idea to always check yoursolutions.AnswerUse adjoint of A , we find the inverse of A to be:
- 26. So the solution to the system of equations is:Check: 22 + 2(-16) - (-16) = 6 3(22) + 5(-16) - (-16) = 2 -2(22) - (16) - 2(-16) = 4So the solution is x = 22, y = -16 and z = -16. The Determinant of a Matrix DEFINITION: Determinants play an important role in finding the inverse of a matrix and also in solving systems of linear equations. In the following we assume we have a square matrix (m = n). The determinant of a matrix A will be denoted by det(A) or |A|. Firstly the determinant of a 2×2 and 3×3 matrix will be introduced, then the n×n case will be shown.
- 27. Determinant of a 2×2 matrix Assuming A is an arbitrary 2×2 matrix A, where theelements are given by: then the determinant of a this matrix is as follows: Determinant of a 3×3 matrix The determinant of a 3×3 matrix is a little more tricky and is found as follows (for this case assume A is an arbitrary 3×3 matrix A, where the elements are given below). then the determinant of a this matrix is as follows:
- 28. Properties of Determinants:Property 1. If each element of a row (column)of a determinant is zero , thenvalue of determinant is zero.Property 2. Value of a determinant is notchanged by changing the rowsinto columns and columns into rows.Property 3. If two adjacent rows (columns)ofa determinant are interchanged , then thesign of the determinant is changed but itsnumerical value is unchanged.Property 4. If two rows (columns) areidentical, then the value of the determinant iszero.Property 5. If every element of a row(column) is multiplied by some constant k,the value of the determinant is multiplied by k.Property 6 .If each element in any row(column) consist of two terms , then thedeterminant can be expressed as the sum ofthe determinants of same order.Property 7 . The value of a determinantremain unchanged if to each element of arow (column) be add ( or subtracted)
- 29. equimultiplies of the corresponding elements of one or more rows (columns) of thedeterminant. Property 8. The value of the determinant of a diagonal matrix is equal to the product of the diagonal elements. Property 9. The value of the determinant of a skew-symmetric matrix of odd order is always zero. Property 10. The determinant of a symmetric matrix of even order is always a perfect square. Notations Let be the given determinant. Then (i)R1, R2, R3 stand for first, second and third rows of . (ii) C1, C2, C3 stand for first, second and third columns of . (iii) By R2 R2 - R3 we mean that third row is to be subtracted from 2nd row. (iv) By C1 C1 + 2C2 - 3C3, we mean that we are to add in first column, the two times of C2 and subtract three times C3.
- 30. The given determinant isy+z x yz+x z x =x+y y zBy ( R1 + R2 + R3 ) and, then, taking (x+y+z) outfrom R1,2 1 1z+x z x · (x+y+z) =x+y y zBy [ C1 - ( C2 + C3 ) ] and [ C2 - C3 ],0 0 10 z-x x · (x+y+z) =x-z y-z zNow, expanding along R1,= { 0 - 0 + 1 [ (0) - (z-x)(x-z) ] } • (x+y+z)= (x-z)(x-z)·(x+y+z)Applications of Determinants1. Area of a Triangle The area of a trianglewith vertices (x1,y1), (x2,y2), (x3,y3) is givenby
- 31. =1/2 [ x1(y2-y3)+x2(y3-y1)+x3(y1-y2)] In determinant form = 1/2 Because area is always positive so we take absolute value of determinant. Question.1 Prove that the points P (a, b+c), Q(b,c+a), R(c, a+b) are collinear. Answer : If P,Q and R are collinear then =0 By applying C2 → C2+C1 = (a+b+c) =0 (∵ C2, C3 are identical) Question.2 Find the value of k if the area of thetriangle with vertices (-2,0),(0,4) and (0,k) is 4 squareunits. Answer: Area of =½ =4
- 32. ⇨ the absolute value of ½ (-2)(4 – k) = 4 ⇨the absolute value of (k – 4) = 4 ⇨ |k – 4| = 4 ⇨ k – 4 = 4, -4 ⇨ k = 8,0. Question.3 Without expanding, show that (i) =0 Operating C1 → C1+C2+C3, we get = 0. (ii) =0 Taking out (-1) from C1,C2 and C3, we get =(-1)(-1)(-1) = -1 =-(by interchanging rows and columns) 2 =0 ⇨ =0
- 33. (iii) =0 ⇨ = = abc ( Operating C3 → C3+C1)abc(ab+bc+ac) x 0 = 0 ( two cols. Are identical) (iv) =0 ⇨ - = - - =0(Pass C3 over the first two columns.)
- 34. (v) = R.H.S. = ( applying C1 ↔ C2) = - (apply C3 → C3 – (ab+bc+ca))C1) = = (apply C2 ↔ C3and C1 ↔ C2) If a,b,c are +ve and are the pth,qth and rthterms resp. Of a G.P.,show without expanding that
- 35. (vi) =0 (put a=xyp-1,b=xyq-1 ,c=xyr-1 , apply C1 →C1-logx.C3,C1→C1+C3) (vii) = (samemethod as given below) (viii) = (Multiply by abc as R1 with a,R2 by b and R3 by cthen divide with abc ) Find the values of: (ix) (Operate C2→ C2. and value is 0
- 36. (x)(Operating C3 → C3 – cos .C2+ sin .C1 and valueis 0) Prove that : (a)= (1+a2+b2)3 ( Apply C1 → (C1 - bC3) and C2 →(C2+aC3) (b) =1(Apply R3 → sin R3 + cos R1) (C) = xyz(x+y+z)3 (Apply R1 → x R1, R2 → y R2, R3 → z R3 and takex,y,z common from C1,C2,C3 resp.)
- 37. (d) = (a-b)(b-c)(c-a)(a+b+c)(a2+b2+c2) ( Apply C1 → (C1+C2 – 2C3) (e) =2(a+b)(b+c)(c+a) (Apply C1 →(C1+C3 )and C2→(C2+C3)) (f) =(a+b+c)(a2+b2+c2) (g) = 2(apply C1→C1-C2-C3, C2→C2-C1,C3→C3-C1,C2↔C3) (h) = abc ( + ++1) = (ab+bc+ca+abc).
- 38. (Hint taking a,b,c common from each row ,apply R1→R1+R2+R3 then expand along first row). (i)= )3 Apply C1→C1-b C3, C2→C2+a C3, we get =( expand along C1, We get ( . (h) Evaluate where =C(x,1) ( binomial coefficient)
- 39. Solution: = ( takingx,y,z common fromR1,R2,R3 resp. and ½!,1/3!From C2,C3 resp.) ( by formula of C(n,r) = ) Apply C3→C3 + C2 and put a= x-1, b=y-1,c=z-1 = (a-b)(b-c)(c-a)= (x-y)(y-z)(z-x).
- 40. Question: If x,y,z are all different and if = 0 , prove that xyz = -1. Solution: =+ = + xyz =0 (1+xyz) = 0 ⇨ (x-y)(y-z)(z-x)(1+xyz) = 0 ⇨ xyz=-1 ∵ x ≠y≠ z. Elementary Transformation Elementary transformations are of the followingthree types: Interchange of any two rows (or columns) The multiplication of the elements of a row (or column) by a non-zero number.
- 41. The addition to the elements of any row (or column) the corresponding elements of any other row (or column) multiplied by any number. Any elementary operation is called a row transformation or a column transformation according as it applies to rows or columns. Definition Let Ri denotes the ith row of the matrix A = [aij] then the elementary row operations on the matrix A are defined as: 3. Ri → Ri + kRj means multiply each elementof jth row by k and add it to the correspondingelements of ith row. The corresponding column transformations are 4. In applying one or more row operations while finding A-1 by elementary row operations, we obtain all zeros in one or more, then A-1 does not exist.
- 42. Q.15 Of Ex.3.4Step 2: Transform the matrix to the reduced row echelon form - 3 1 - 3 2 3 1 0 0 1 0 0 3Row 2 2 multiply the 1st row by 2Operation 2 2 3 0 1 0 1/21: 2 2 3 0 1 0 - 3 2 0 0 1 2 - 3 2 0 0 1 2 - - 3 1 3 1 3 3 1 0 0 1 0 0 2 2Row 2 2 2 2 add -2 times the 1st rowOperation to the 2nd row -2: 2 2 3 0 1 0 0 5 0 1 0 1 - 3 2 0 0 1 - 2 3 2 0 0 1 2 - 3 1 - 3 3 1 1 0 0 3 1 0 0 2 2 2 2 2Row 2 add -3 times the 1st -Operation 0 5 0 1 0 - row to the 3rd row 13: 0 5 0 1 0 1 - - 5 - 5 3 3 2 0 0 1 0 0 1 2 2 2 2Row - 3 1 - 3 1 multiply the 2nd rowOperation 1 3 0 0 1 3 0 0 by 1/54: 2 2 2 2
- 43. 2 2 - - 0 5 0 1 0 1 1 1 0 1 0 0 - - 5 5 5 5 3 0 0 1 - - 2 5 2 2 5 3 0 0 1 2 2 2 - - 3 1 3 1 3 3 1 0 0 1 0 0 2 2 2 2 2 2 - -Row 1 1 1 add -5/2 times the 2nd 1Operation 0 1 0 0 0 1 0 0 row to the 3rd row5: 5 5 5 5 - - - - 5 5 3 5 - 1 0 0 1 0 0 1 1 2 2 2 2 2 - - 3 1 3 1 3 3 1 0 0 1 0 0 2 2 2 2 2 2 - -Row 1 1 1 multiply the 3rd row 1Operation 0 1 0 0 0 1 0 0 by -2/56: 5 5 5 5 - - - 2 1 5 - 1 2 0 0 1 0 0 1 1 5 5 2 2 5 - - 3 1 -1 -3 3 3 3 1 0 0 1 0Row add -3/2 times the 10 10 5 2 2 2Operation 2 3rd row to the 1st7: - 1 row -1 1 0 1 0 1 0 0 1 0 0 5 5 5
- 44. 5 - 2 1 2 - 0 0 1 2 1 2 5 5 0 0 1 5 5 5 5 - - 3 -1 -3 3 2 3 1 0 0 0 1 0 5 10 10 5 5 2 -Row -1 1 1 add 3/2 times the 2nd 1Operation 0 1 0 0 0 1 0 0 row to the 1st row8: 5 5 5 5 - 2 1 - 2 2 1 0 0 1 2 0 0 1 5 5 5 5 5 5Q. 16 Of Ex.3.4Step 1: Adjoin the identity matrix to the given matrix 1 3 -2 1 0 0Adjoining I3 to the given matrix, we obtain the 3x6 matrix: -3 0 -5 0 1 0 2 5 0 0 0 1Step 2: Transform the matrix to the reduced row echelon form - 1 3 1 0 0 1 3 -2 1 0 0 2Row add 3 times the 1st row -Operation - - 0 9 3 1 0 0 0 1 0 to the 2nd row 111: 3 5 2 5 0 0 0 1 2 5 0 0 0 1
- 45. 1 3 -2 1 0 0 1 3 -2 1 0 0Row - - add -2 times the 1st 0 9 3 1 0Operation 0 9 3 1 0 11 11 row to the 3rd row2: - - 2 5 0 0 0 1 0 4 0 1 1 2 1 3 -2 1 0 0 1 3 -2 1 0 0 - 1 1Row - 11 0 9 3 1 0 multiply the 2nd 0 1 0Operation 11 row by 1/9 3 93: 9 - - 0 4 0 1 1 2 - - 0 4 0 1 1 2 1 3 -2 1 0 0 1 3 -2 1 0 0 - 1 1 - 11 1 1 0 1 0Row 11 0 1 0 add 1 times the 2nd 3 9Operation 9 3 9 row to the 3rd row4: 9 - 25 1 - - 5 0 4 0 1 0 0 1 1 2 9 9 3 1 3 -2 1 0 0 1 3 -2 1 0 0 - - 1 1 1 1 11 11 0 1 0 0 1 0Row 3 9 multiply the 3rd 3 9Operation 9 9 row by 9/255: - - 25 1 1 9 5 3 0 0 1 0 0 1 9 9 25 25 3 5 1 3 -2 1 0 0 - 1 3 1 0 0 - 2Row 1 1 add 11/9 times theOperation 11 3rd row to the 2nd - 0 1 0 4 116: 3 9 row 2 9 0 1 0 25 25 0 0 1 - 1 9 5
- 46. 3 - 1 9 25 25 3 0 0 1 5 25 25 5 - 2 18 - 1 1 3 1 0 0 1 3 0 2 25 25 5 - 4 11 2 -Row 0 1 0 4 11 add 2 times the 3rd 2Operation 25 25 0 1 0 5 row to the 1st row7: 25 25 5 - 1 9 3 - 0 0 1 1 9 3 25 25 0 0 1 5 25 25 5 - 2 18 -2 -3 1 1 3 0 1 0 0 1 25 25 5 5 5 - - 4 11Row 4 11 add -3 times the 2 2 0 1 0Operation 0 1 0 2nd row to the 1st 25 258: 25 25 row 5 5 - - 1 9 1 9 3 3 0 0 1 0 0 1 25 25 25 25 5 5Step 1: Adjoin the identity matrix to the given matrix 2 6 -2 1 0 0Adjoining I3 to the given matrix, we obtain the 3x6 matrix: 1 4 0 0 1 0 2 -5 1 0 0 1Step 2: Transform the matrix to the reduced row echelon form
- 47. 1 - - 2 6 1 0 0 1 3 0 0 2 1Row 2 multiply the 1st rowOperation 1 4 0 0 1 0 by 1/2 1 4 0 0 1 01: - 2 1 0 0 1 - 5 2 1 0 0 1 5 1 - 1 3 0 0 1 1 - 2 1 3 0 0 1Row 2 - add -1 times the 1stOperation 1 1 4 0 0 1 0 row to the 2nd row 0 1 1 1 02: - 2 2 1 0 0 1 5 - 2 1 0 0 1 5 1 1 - - 1 3 0 0 1 3 0 0 1 1 2 2Row - - add -2 times the 1stOperation 1 1 0 1 1 1 0 row to the 3rd row 0 1 1 1 03: 2 2 - - - 2 1 0 0 1 0 3 0 1 5 11 1 1 1 - 1 3 -1 0 0 1 3 0 0 1 2 2 -1Row - add 11 times the 0 1 1 1 0Operation 1 2nd row to the 3rd 0 1 1 1 0 24: row 2 - 13 - - 0 0 14 11 1 0 3 0 1 11 1 2Row 1 1 multiply the 3rd -Operation 1 3 -1 0 0 1 3 0 0 row by 1/14 15: 2 2
- 48. -1 -1 0 1 1 1 0 0 1 1 1 0 2 2 - - 11 1 13 13 0 0 14 11 1 0 0 1 14 14 2 28 1 1 - - 1 3 0 0 1 3 0 0 1 1 2 2 -1 -1 3 -1Row add -1 times the 0 1 1 1 0 0 1 0Operation 3rd row to the 2 28 14 146: 2nd row - - 11 1 11 1 13 13 0 0 1 0 0 1 14 14 14 14 28 28 1 1 11 1 - 1 3 0 0 1 3 0 1 2 28 14 14 -1 3 -1 -1 3 -1Row add 1 times the 0 1 0 0 1 0Operation 3rd row to the 28 14 14 28 14 147: 1st row - - 11 1 11 1 13 13 0 0 1 0 0 1 14 14 14 14 28 28 1 11 1 1 1 2 1 3 0 1 0 0 28 14 14 7 7 7 -1 3 -1 -1 3 -1Row add -3 times the 0 1 0 0 1 0Operation 2nd row to the 28 14 14 28 14 148: 1st row - - 11 1 11 1 13 13 0 0 1 0 0 1 14 14 14 14 28 28Q. 14 OF EX. 4.6
- 49. - - 1 2 7 1 2 7 1 1Row - add -3 times the 1st row to the - -Operation 3 4 -5 0 7 5 2nd row 11 261: - - 2 3 12 2 3 12 1 1 - 1 2 7 - 1 1 2 7 1Row - - add -2 times the 1st row to theOperation 0 7 - - 11 26 3rd row 0 72: 11 26 - 2 3 12 0 1 -1 -2 1 1 -1 2 7 1 -1 2 7Row -11 -26Operation 0 7 -11 -26 multiply the 2nd row by 1/7 0 13: 7 7 0 1 -1 -2 0 1 -1 -2 - 1 2 7 - 1 1 2 7 1 - -Row - - 11 26 add -1 times the 2nd row to the 0 1Operation 11 26 0 1 3rd row4: 7 7 7 7 4 12 0 1 -1 -2 0 0 7 7 1 -1 2 7 1 -1 2 7 -11 -26Row 0 1 -11 -26Operation 7 7 multiply the 3rd row by 7/4 0 15: 7 7 4 12 0 0 0 0 1 3 7 7Row - add 11/7 times the 3rd row to the -Operation 1 2 7 1 2 7 1 2nd row 16:
- 50. - - 0 1 0 1 11 26 0 1 0 0 1 3 7 7 0 0 1 3 1 -1 2 7 1 -1 0 1RowOperation 0 1 0 1 add -2 times the 3rd row to the 1st row 0 1 0 17: 0 0 1 3 0 0 1 3 -Row 1 0 1 1 0 0 2 1Operation add 1 times the 2nd row to 0 1 0 18: 0 1 0 1 the 1st row 0 0 1 3 0 0 1 3ANSWER IS X=2,Y=1, Z=3. Mathematics reference Rules for matrices Basic properties of matrices. A, B, a and n are scalars. Basic. -A == (-1) A A - B == A + (-B)
- 51. 1A=A0A=OA+O=O+A=AIA=AI=AA-A=OAddition and scalar product.A+B=B+A(A + B) + C = A + (B + C)r (A + B) = r A + r B(r + s) A = r A + s A(r s) A = r (s A)Matrix product.A0 == IA2 == A AAn = A An - 1(A B) C = A (B C)A (B + C) = A B + A C(A + B) C = A C + B CTranspose and inverse.IT = I
- 52. (AT)T = A (A + B)T = AT + BT (r A)T = r AT (A B)T = BT AT I-1 = I A A-1 = A-1 A = I (A B)-1 = B-1 A-1 (A-1)T = (AT)-1 Trace. tr (A + B) = tr A + tr B tr (r A) = r tr A tr (A B) = tr (B A) Determinant and adjoint. det O = 0 det I = 1 det A = det AT det (A B) = (det A) (det B)DefinitionsMatch the following terms with their definitions.
- 53. ____address>_______________1. Diagonal from theupper left corner entry to the bottom right corner entry____ determinant of 2x2 matrix>____________2. Arectangular array of numbers enclosed in brackets____ dimensions>____________3. Variation in size of amatrix____ main diagonal>__________4. Any matrix that hasthe same number of rows as it does columns____ matrix>________________5. Matrix in which allof the entries are zero____ scalar>________________6. The difference of theproducts of the diagonals____ square matrix>__________7. Number locatedoutside of a single matrix which is multiplied by eachentry of the matrix____ zero matrix>____________8. Describes whereeach value, or entry, of a matrix lives

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