Question 2887: I need your help on these three problems....................
(1).
-4 5
-5
The determinant of the matrix A is 1. Find the missing element of the matrix A
a.1
b.3
c.6
d.–24
(2). Find the determinant of the matrix C.
C=
4 0 –7
3 0 6
-5 2 3
a.45
b.24
c.-90
d.-21
(3). Use Cramer’s Rule to solve y.
x-4y=4
4x-y=1
a.-8/15
b. –1
c. 17/5
d. 0
Answer by xcentaur(357) (Show Source):
You can put this solution on YOUR website! [1]
-4 5
-5 X
"The determinant of the matrix A is 1."
In matrix X=
Determinant = ab - cd
Determinant = (-4*X)-(-5*5)
= -4X+25
Since we know the determinant is one,
1=-4X+25
-4X=-24
X=-24/-4
=6
Therefore,correct answer is: c.6
[2]
Matrix C=
In matrix X of order 3=
Determinant of X is=
=x1(y2z3-y3z2)-x2(y1z3-y3z1)+x3(y1z2-z1y2)
Determinant of C is=
= 4[0*3-(2*6)]-0[3*3-(-5*6)]+(-7)[3*2-(-5*0)]
= 4[0-12]-0[9+30]-7[6-0]
= 4[-12]-0[39]-7[6]
= -48-0-42
= -90
Therefore,correct answer is: c.-90
[3]
Cramer's Rule - Given the system
p1x + q1y = c1 ... [eqn1]
p2x + q2y = c2 ... [eqn2]
Determinant detA = p1q2 - p2q1
Determinant detX = c1q2 - c2q1
Determinant detY = p1c2 - p2c1
Then x=detX/detA
Then y=detY/detA
Cramer’s Rule to solve y.
x-4y=4
4x-y=1
Consider the determinant formed by the x,y and constant term of the given equations.
detA = p1q2 - p2q1
=(1*-1)-(4*-4)
=-1-(-16)
=-1+16
=15
detY = p1c2 - p2c1
=(1*1)-(4*4)
=1-16
=-15
By Cramer's Rule,y=detY/detA
y=-15/15
=-1
Therefore,the correct answer is: b. –1
Hope this helps,
best of luck.
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