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SECTION 11.3 Systems of Linear Equations: Determinants 725You
are asked to prove this result for a 3 by 3 determinant using row 2
in
Problem 63.
EXAMPLE 8 Demonstrating Theorem (14)
I ! ~ I 6 - 8 = -2I k 2kl 1 1 2 16 = 6k - 8k = -2k = k( -2) = k
4 6
T H E O R E M If the entries of any row (or any column) of a
determinant are multiplied bya nonzero number k and the result is
added to the corresponding entries ofanother row (or column), the
value of the determinant remains unchanged.
(15).JIn Problem 65, you are asked to prove this result for a 3
by 3 determinant using
rows 1 and 2.
EXAMPLE 9 Demonstrating Theorem (15)
I ~ ~ I= -14 1 3 4 1 ~ 1 - 7 1 = -5 2 5 2 14rMultiply row 2 by
-2 and add to row 1.
11.3 AssessYour
Understanding~~------------------------------------------------------Concepts
and Vocabulary1. Cramer's Rule uses
equations.2. D =: : 1 =--
to solve a system of linear 3. True or False A 3 by 3
determinant can never equal O.4. True or False The value of a
determinant remains un-
changed if any two rows or any two columns are
inter-changed.
Skill BuildingIn Problems 5-14, find the value of each
determinant.5. I ~~ I 6. I ~ ~ I 7. I - ~ : 1 8. I : - ~ I 9. I - ~
- ~ I
~ I3 4 2 1 3 -2 4 -1 2 3 -9 41 - 4 ..II. 1 -1 5 12. 6 1 -5 13. 6
0 14.10. -5 -1 1 4 01 2 -2 8 2 3 1 -3 4 8 -3 1
In Problems 15-42, solve each system of equations using Cramer's
Rule if it is applicable. If Cramer's Rule is not applicable, say
so.' > 0 . _ {x + Y = 8 {x + 2y = 5 {5X - y = 13 { x + 3y = 5~,
i:-. 16. 17. 18.x - y = 4 x - y = 3 2x + 3y = 12 2x - 3y = -8
{ 3x =2419. x + 2y = 0 {4X + 5v = -320. "-2y = -4 {3X - 6y =
2421. 5x + 4y = 12 {2X + 4y = 1622. 3x - 5y = -9
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726 CHAPTER 11 Systems of Equations and Inequalities
23. { 3 X - 2y =46x - 4y =0{ 2x - 3y = -127. lOx + lOy = 5
31. { 3x - 5y = 315x + 5y = 21
{
X + 2y - z = -335. 2x - 4Y + z = -7
-2x + 2y - 3z = 4
{
X + 2y - z = 039. 2x - 4y + Z = 0
-2x + 2y - 3z = 0
{-x + 2y = 524. 4x - 8y = 6
2 8 { 3 X - 2y = 0 5x + lOy = 4
{2X - y = -1
32. 1 3x + -y =-2 2
{
X + 4y - 3z = -836. 3x - y + 3z = 12
x + Y + 6z = 1
{
X + 4y - 3z = 040. 3x - y + 3z = 0
x + Y + 6z = 0
{2X - 4y =-225. 3x + 2y = 3
{2X + 3y = 6
29. 1x - Y = "
{x+ Y- z= 6
.D. 3x - 2y + Z = -5x + 3y - 2z =14
{
X - 2y + 3z =137. 3x + y - 2z =0
2x - 4y + 6z = 2
{
X - 2y + 3z =041. 3x + y - 2z = 0
2x - 4y + 6z =0
{3x + 3y = 8 3
26. 4x + 2y = 3 "
{.!.x+ y=-2
30. 2x - 2y = 8
{
X - y + Z = -434. 2x - 3y + 4z = -15
5x + y - 2z = 12
{
X - y + 2z = 538. 3x + 2y = 4
-2x + 2y - 4z = -10
{
X - y + 2z = 042. 3x + 2y = 0
-2x + 2y - 4z = 0
In Problems 43-48, solve for x. 1 1I ~~ I I ~~ I x43. = 5 44. =
-2 45. 4 3 2 =2-1 2 53 2 4 x 2 3 x 1 2
46. 1 x 5 =0 47. 1 x 0 =7 48. 1 x 3 =-4x0 -2 6 1 -2 0 1 2
In Problems 49-56, use properties of determinants to find the
value of each determinant if it is known thatx y zu v w =41 2 3
1 2 3 x y z x y z 1 2 349. u v w 50. u v w 51. -3 -6 -9 52. x -
u y - v z - w
x y z 2 4 6 u v w u v w1 2 3 x y z - x 2 3 x + 3 y + 6 z + 9
53. x-3 y-6 z - 9 54. u v w-u 55. 2x 2y 2z 56. 3u -1 3v-2 3w-32u
2v 2w 1 2 2 u - 1 v - 2 w - 3 1 2 3
Applications and Extensions57. Geometry: Equation of a Line An
equation of the line con-
taining the two points (X I, Y I) and (X l , Y2 ) may be
expressedas the determinant
x YX I y, = 0X l Y l
Prove this result by expanding the determinant and compar-ing
the result to the two-point form of the equation of a line.
58. Geometry: Collinear Points Using the result obtained
inProblem 57, show that three distinct points ( x " Y I), (X l, Y2
),and (x), Y 3) are collinear (lie on the same line) if and only
if
X I y, 1X l Y l 1 = 0x) Y 3 1
59. Geometry: Area (If a Triangle A triangle has vertices(x " Y
I), (xc, Y l), and (x) , Y 3)' Show that the area of thetriangle is
given by the absolute value of D, where
1 x , X2 X3D = - Y l Y 2 Y 3 ' Use this formula to find the area
of a
2 1 1 1triangle with vertices (2,3), (5, 2), and (6,5).
X2 x 160. Show that I Y 1 = (y - z)(x - y)(x - z).
Z2 Z 161. Complete the proof of Cramer's Rule for two equations
con-
taining two variables.[Hint: In system (5), page 719, if a = 0,
then b * 0 and c * 0,since D = =bc * O. Now show that equation (6)
provides a
