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**Unformatted text preview: **Chapter 3 Section 3.1 1. (a) A = 2 3- 3 1 ¶ = ⇒ A- 1 = 1 11 1- 3 3 2 ¶ . Therefore, x = A- 1 b = 1 11 1- 3 3 2 ¶ 7- 5 ¶ = 2 1 ¶ . That is, x 1 = 2 ,x 2 = 1. (b) The lines are intersecting, as shown below. –10 –5 5 10 x2 –2 –1 1 2 3 4 5 x1 2. (a) A = 1- 2 2 3 ¶ = ⇒ A- 1 = 1 7 3 2- 2 1 ¶ . Therefore, x = A- 1 b = 1 7 3 2- 2 1 ¶ 10 6 ¶ = 6- 2 ¶ . That is, x 1 = 6, x 2 =- 2. (b) Therefore, the lines are intersecting, as shown below. 1 –4 –2 2 x2 2 4 6 8 x1 3. (a) A = 1 3 2- 1 ¶ = ⇒ A- 1 = 1 7 1 3 2- 1 ¶ . Therefore, x = A- 1 b = 1 7 1 3 2- 1 ¶ ¶ = ¶ . That is, x 1 = 0, x 2 = 0. (b) Therefore, the lines are intersecting, as shown below. –2 2 4 6 8 x2 –1 1 2 3 4 x1 4. (a) A =- 1 2 2- 4 ¶ det A = 0. Therefore, A is singular which implies either there is no solution or an infinite number of solutions. Multiplying the first equation by 2, we have- 2 x 1 + 4 x 2 = 8. Our second equation is 2 x 1- 4 x 2 =- 6. Adding the two equations, we have 0 = 2 which cannot occur. Therefore, we have no solution. 2 (b) Therefore, the lines are parallel, as shown below. 1.5 2 2.5 3 3.5 4 x2 –1 1 2 3 4 x1 5. (a) A = 2- 3 1 2 ¶ = ⇒ A- 1 = 1 7 2 3- 1 2 ¶ . Therefore, x = A- 1 b = 1 7 2 3- 1 2 ¶ 4- 5 ¶ =- 1- 2 ¶ . That is, x 1 =- 1, x 2 =- 2. (b) Therefore, the lines are intersecting, as shown below. –4 –3 –2 –1 x2 –3 –2 –1 1 2 3 x1 6. (a) A = 3- 2- 6 4 ¶ det( A ) = 0. Therefore, there are no solutions or an infinite number of solutions. Multi- plying the first equation by- 2, we see the equations represent the same line. Therefore, there are an infinite number of solutions. In particular, any pair ( x 1 ,x 2 ) such that x 2 = 3 x 1 / 2 will satisfy the system of equations. 3 (b) Therefore, the lines are coincident, as shown below. –4 –2 2 4 x2 –3 –2 –1 1 2 3 x1 7. (a) A = 2- 3- 4 6 ¶ det( A ) = 0, therefore, there are no solutions or an infinite number of solutions. Mul- tiplying the first equation by- 2, we see the two equations are equations for the same line. Therefore, there are an infinite number of solutions. In particular, any numbers ( x 1 ,x 2 ) such that x 2 = (2 x 1- 6) / 3. (b) Therefore, the lines are coincident, as shown below. –2 –1 1 x2 –1 1 2 3 4 5 x1 8. (a) A = 4 1 4- 3 ¶ = ⇒ A- 1 = 1 16 3 1 4- 4 ¶ . Therefore, x = A- 1 b = 1 16 3 1 4- 4 ¶- 12 ¶ =- 3 / 4 3 ¶ . That is, x 1 =- 3 / 4, x 2 = 3. 4 (b) Therefore, the lines are intersecting, as shown below. –8 –6 –4 –2 2 4 6 x2 –1 –0.5 0.5 1 1.5 2 x1 9. (a) A = 1 4 4 1 ¶ = ⇒ A- 1 =- 1 15 1- 4- 4 1 ¶ . Therefore, x = A- 1 b =- 1 15 1- 4- 4 1 ¶ 10 10 ¶ = 2 2 ¶ . That is, x 1 = 2, x 2 = 2. (b) Therefore, the lines are intersecting, as shown below....

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