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CHAPTER 1 Section 1.6 إعداد د. هيله العودان (وضعت باذن معدة العرض)

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A BASIC THEOREM Theorem 1.6.1. Every system of linear equations has either no solutions, exactly one solution, or infinitely many solutions.

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SOLVING LINEAR SYSTEMS BY MATRIX INVERSION Theorem 1.6.2. If A is an invertible n × n matrix, then for each n × 1 matrix b, the system of equations Ax = b has exactly one solution, namely, x= Aˉ¹ b.

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Example 1 Consider the system of linear equations x 1 + 2 x 2 + 3 x 3 = 5 2 x 1 + 5 x 2 + 3 x 3 = 3 x 1 + 8 x 3 = 17 In matrix form this system can be written as A x= b, where A= x = b = 1 2 3 2 5 3 1 0 8 x1x2x3x1x2x3 5 3 17

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In Example 4 of the preceding section we showed that A is invertible and Aˉ¹= By Theorem 1.6.2 the solution of the system is x = Aˉ¹ b== or x 1 =1, x 2 =-1, x 3 =2. -40 16 9 13 -5 -3 5 -2 -1 -40 16 9 13 -5 -3 5 -2 -1 5 3 17 1 2

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SOLVING MULTIPLE LINEAR SYSTEMS WITH A COMMON COEFFICENT MATRIX A x = b 1, A x = b 2, A x = b 3,..., A x = b k x 1 = A ˉ¹ b 1, x 2 = A ˉ¹ b 2, x 3 = A ˉ¹ b 3,..., x k = A ˉ¹ b k [ A ¦ b 1 ¦ b 2 ¦∙∙∙ ¦ b k ] (1) In which the coefficient matrix A is “augmented” by all k of the matrices b 1, b 2,...., b k. By reducing (1) to reduced row-echelon from we can solve all k systems at once by Gauss-Jordan elimination. This method has the added advantage that it applies even when A is not invertible.

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Example 2 Solve the systems (a) x 1 +2 x 2 +3 x 3 =4 (b) x 1 +2 x 2 +3 x 3 =1 2 x 1 +5 x 2 +3 x 3 =5 2 x 1 +5 x 2 +3 x 3 =6 x 1 +8 x 3 =9 x 1 +8 x 3 =-6

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Solution The two systems have the same coefficient matrix. If we augment this coefficient matrix with the columns of constants on the right sides of these systems, we obtain Reducing this matrix to reduced row-echelon form yields(verify) It follows from the last two column that the solution of system (a) is x 1 =1, x 2 =0, x 3 =1 and of system (b) is x 1 =2, x 2 =1, x 3 =-1. 1 2 3 ¦ 4 ¦ 1 2 5 3 ¦ 5 ¦ 6 1 0 8 ¦ 9 ¦ -6 1 0 0 ¦ 1 ¦ 2 0 1 0 ¦ 0 ¦ 1 0 0 1 ¦ 1 ¦ -1

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PROPERTIES OF INVERTIBLE MATRICES Theorem 1.6.3. Let A be a square matrix. (a)If B is a square matrix satisfying BA=I, then B=Aˉ ¹. (b)If B is a square matrix satisfying AB=I, then B=Aˉ ¹.

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Theorem 1.6.4. If A is an n × n matrix then, the following are equivalent. (a) A is invertible. (b) Ax=0 has only the trivial solution. (c) The reduced row-echelon from of A is I n. (d) A is expressible as product of elementary matrices. (e) Ax= b is consistent for every n×1 matrix b. (f) Ax= b has exactly one solution for every n×1 matrix b.

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Example 3 what conditions must b1, b2, and b3 satisfy in order for the system of equations x 1 + x 2 +2 x 3 = b 1 x 1 + x 3 = b 2 2 x 1 + x 2 +3 x 3 = b 3 to be consistent?

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Solution. The augmented matrix is Which can be reduced to row-echelon form as follows. 1 1 2 b 1 1 0 1 b 2 2 1 3 b 3 1 1 2 b 1 1 0 1 b 2 - b 1 2 1 3 b 3 -2 b 1 1 1 2 b 1 0 1 1 b 1 - b 2 0 -1 -1 b 3 -2 b 1 -1 times the first row was added to the second and -2 times the first row was added to the third. The second row was Multiplied by -1.

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It is now evident form the third row in the matrix that the system has a solution if and only if b 1, b 2, and b 3 satisfy the condition b 3 - b 2 - b 1 =0or b 3 = b 1 + b 2 To express this condition another way, Ax=b is consistent if and only if b is a matrix of the form b =where b 1 and b 2 are arbitrary. 1 1 1 b 2 0 1 1 b 1 - b 2 0 0 0 b 3 -b 2 - b 1 The second row was Added to the third. b1b2b1+b2b1b2b1+b2

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DIAGONAL MATRICES A square matrix in which all of the entries off the main diagonal are zero is called a diagonal matrix ; some examples are A general n × n diagonal matrix D can be written as D= 2 0 0 -5 1 0 0 0 1 0 0 0 1 6 0 0 0 0 -4 0 0 0 0 0 0 0 8 d 1 0 · · · 0 0 d 2 · · · 0 · · · 0 0 · · · d n

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A diagonal matrix is invertible if and only if all of its diagonal entries are nonzero; in this case the inverse of (1) is Dˉ¹= 1/d 1 0 · · · 0 0 1/ d 2 · · · 0 · · · 0 0 · · · 1/ d n

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The reader should verify that DD ˉ¹=D ˉ¹D=I Powers of diagonal matrices are easy to computer; we leave it for the reader to verify that if D is the diagonal matrix (1) and k is a positive integer, then D = d 1 0 · · · 0 0 d 2 · · · 0 · · · 0 0 · · · d n k k k k

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Example 1 If A= Then Aˉ¹=A=A= 5-5 1 0 0 0 -3 0 0 0 2 1 0 0 0 - 0 0 1 0 0 0 -243 0 0 0 32 1 0 0 0 - 0 0 1313 1 243 1 32 1212

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