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My lecture notes for Section 4-6 of Barnett Finite Mathematics.

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- 1. Matrix EquationsMatrix Equations and Systems of Linear Equations Application Math 1300 Finite Mathematics Section 4.6 Matrix Equations and Systems of Linear Equations Jason Aubrey Department of Mathematics University of Missouri university-logo Jason Aubrey Math 1300 Finite Mathematics
- 2. Matrix Equations Matrix Equations and Systems of Linear Equations ApplicationSolving a Matrix Equation Suppose we have an n × n matrix A and an n × 1 column matrices B and X . If A is invertible, then we can solve the equation AX = B as follows AX = B −1 A (AX ) = A−1 B multiply both sides by A−1 on the left (A−1 A)X = A−1 B matrix multiplication is associative −1 IX = A B A−1 A = I X = A−1 B university-logo Jason Aubrey Math 1300 Finite Mathematics
- 3. Matrix Equations Matrix Equations and Systems of Linear Equations ApplicationUsing Inverses to Solve Systems of Equations Example: Use matrix inverse methods to solve the system: x1 − x2 + x3 = 1 2x2 − x3 = 1 2x1 + 3x2 +0x3 = 1 First, we write this as a matrix equation: 1 −1 1 x1 1 0 2 −1 x2 = 1 2 3 0 x3 1 A X B university-logo Jason Aubrey Math 1300 Finite Mathematics
- 4. Matrix Equations Matrix Equations and Systems of Linear Equations ApplicationSo we have a matrix equation AX = BTo solve this equation, we ﬁnd A−1 : 1 −1 1 1 0 0 1 0 0 3 3 −1 0 2 −1 0 1 0 → · · · → 0 1 0 −2 −2 1 2 3 0 0 0 1 Row Operations 0 0 1 −4 −5 2Therefore, 3 3 −1 A−1 = −2 −2 1 −4 −5 2 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 5. Matrix Equations Matrix Equations and Systems of Linear Equations ApplicationWe know that X = A−1 B, so x1 3 3 −1 1 5 x2 = −2 −2 1 1 = −3 x3 −4 −5 2 1 −7 X A−1 BSo, we conclude that x1 = 5 x2 = −3 x3 = −7 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 6. Matrix Equations Matrix Equations and Systems of Linear Equations ApplicationUsing Inverse Methods to Solve Systems of EquationsIf the number of equations in a system equals the number ofvariables and the coefﬁecient matrix has an inverse, then thesystem will always have a unique solution that can be found byusing the inverse of the coefﬁcient matrix to solve thecorresponding matrix equation. Matrix Equation Solution AX = B X = A−1 B university-logo Jason Aubrey Math 1300 Finite Mathematics
- 7. Matrix Equations Matrix Equations and Systems of Linear Equations ApplicationInsight. . .There are two cases where inverse methods will not work:Case 1. If the coefﬁcient matrix is singular.Case 2. If the number of variables is not the same as thenumber of equations.In either case, use Guass-Jordan elimination. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 8. Matrix Equations Matrix Equations and Systems of Linear Equations ApplicationExample: An investment advisor currently has two types ofinvestments available for clients: a conservative investment A thatpays 10% per year and investment B of higher risk that pays 20% peryear. Clients may divide their investments between the two to achieveany total return desired between 10% and 20%. However, the higherthe desired return, the higher the risk. How should each client listedin the table invest to achieve the desired return? 1 2 3 k Total Investment $20,000 $50,000 $10,000 k1 Annual Return Desired $2,400 $7,500 $1,300 k2The answer to this problem involves six quantities, two for each client.We will solve this problem for an arbitrary client k with unspeciﬁedamounts k1 for total investment and k2 for annual return. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 9. Matrix Equations Matrix Equations and Systems of Linear Equations ApplicationLet x1 = amount invested in A by a given client x2 = amount invested in B by a given clientThen we have the following mathematical model: x1 + x2 = k1 total invested 0.1x1 + 0.2x2 = k2 Total annual return desiredWrite as a matrix equation: 1 1 x1 k = 1 0.1 0.2 x2 k2 A X B university-logo Jason Aubrey Math 1300 Finite Mathematics
- 10. Matrix Equations Matrix Equations and Systems of Linear Equations ApplicationIf A−1 exists, then X = A−1 B. So, we ﬁnd A−1 : 1 1 1 0 1 0 2 −10 → ··· → 0.1 0.2 0 1 0 1 −1 10 Row OperationsThus 2 −10 A−1 = −1 10 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 11. Matrix Equations Matrix Equations and Systems of Linear Equations ApplicationTherefore, x1 2 −10 k1 = x2 −1 10 k2 X A−1 BTo solve each client’s problem, we replace k1 and k2 withappropriate values from the table and multiply by A−1 : x1 2 −10 20, 000 16, 000 = = x2 −1 10 2, 400 4, 000 Client 1Solution: x1 = $16, 000 in investment A,x2 = $4, 000 in investment B university-logo Jason Aubrey Math 1300 Finite Mathematics
- 12. Matrix Equations Matrix Equations and Systems of Linear Equations Application x1 2 −10 50, 000 25, 000 = = x2 −1 10 7, 500 25, 000 Client 2Solution: x1 = $25, 000 in investment A,x2 = $25, 000 in investment B x1 2 −10 10, 000 7, 000 = = x2 −1 10 1, 300 3, 000 Client 2Solution: x1 = $7, 000 in investment A,x2 = $3, 000 in investment B university-logo Jason Aubrey Math 1300 Finite Mathematics

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