solve the system of equations using cramer's rule.?

Question

1) 5x+7y= -3
2x+3y= -1

2) 2x+3y=7
8x+12y=2

3) 2x-y+3z=9
x +2z=3
3x+2y+z=10

Answer 1

a) ∆=15-14=1 ; ∆(x)= -9 + 7 = -2 ; ∆(y)= -5+6 =1

--> x=∆(x)/∆ = -2 ; y=∆(y)/∆ =1

b) ∆=24-24=0 System not compatible

c) ∆=-7 ; ∆(x)= -35 ; ∆(y)= 14 ; ∆(z)= 7

--> x=∆(x)/∆ = 5 ; y=∆(y)/∆ = -2 ; z=∆(z)/∆ = -1

Saludos.

Answer 2

I assume you know how to find the determinants of matrices

1)
you have the matrix
A
=
[ 5 7 ]
[ 2 3 ]
so
detA = 5*3 - 7*2 = 1
replace the first column vector (5, 2) with (-3, -1) gives the matrix
A1
=
[-3 7 ]
[-1 3 ]
so
detA1 = -3*3 - 7*(-1) = -2
similarly replace the second column vector (7, 3) with (-3,-1) you get
detA2 = 1
then
x = detA1/detA = -2/1 = -2
&
y = detA2/detA = 1/1 = 1

2) the determinant of the matrix is zero so the system is inconsistent

3)The third one follows the same process as the first, it is just a matter of swapping the vectors around and calculating a few determinants
A good hint for the 3rd one is to use the fact that the second equation causes there to be a zero term in the matrix, so expanding the determinants about a vector containing this zero will save you a lot of work. It is to difficult for me to expand the determinant of a 3x3 matrix on here because of the spacing problem on this website
I got
detA = -7, detA1 = -35, detA2 = 14, detA3 = 7
so the answer is
x = 5, y = -2, z = -1

,.,.,.,

Answer 3

I´ll show you the last

2 -1 3
D= 1 0 3 =6-9 +1-12=-14
3 2 1

9 -1 3
x= 3 0 3 /(-14) = =69/14
10 2 1

2 9 3
1 3 3
y=3 10 1 = 6+30+81 -27 -60-9 = -75/14

2 -1 9
z=1 0 3 = 11/14
3 2 10

Thats enough

Answer 4

Cramers Rule includes determinants interior the numerator and denominators. The numerators variety for each variable, yet no longer the denominator. The denominator is | one million one million -one million| | 5 -one million one million| = one million(3 - 2) - one million(-15-one million) -one million(10+one million) increasing for the period of | one million 2 -3| row one million = one million + sixteen - eleven = 6 The numerator for x, which I write as X, is | -one million one million -one million| |-eleven -one million one million| in simple terms the previous numbers, yet column one | 0 2 -3| makes use of the constants from the unique difficulty! = -one million(3 - 2) - one million(33-0) -one million(-22+0) increasing for the period of row one million = -one million -33 + 22 = -12 So x = -12/6 = -2. further, Y is | one million -one million -one million| | 5 -eleven one million| in simple terms the previous numbers, yet column 2 | one million 0 -3| makes use of the constants from the unique difficulty! = one million(33-0)+one million(-15-one million)-one million(0+eleven) = 33 - sixteen -eleven = 6 So y= 6/6 = one million. finally, Z is | one million one million -one million| | 5 -one million -eleven| in simple terms the previous numbers, yet column threee | one million 2 0| makes use of the constants from the unique difficulty! = one million(0+22)-one million(0+eleven)-one million(10+one million) = 22 - eleven - eleven = 0 So z= 0/6 = 0.

Answer 5

I failed the second half of algebra 2 in high school so i doubt i would be that much help. i knew how to do it but not anymore.