Solving linear equations with Cramer's rule x1+x2-2x3=-3 2x1+x2-x3=1 x1-x2+3x3=8

Solving linear equations with Cramer's rule x1+x2-2x3=-3 2x1+x2-x3=1 x1-x2+3x3=8

D =
1 1 -2
2 1 -1
1 -1 3
= 1
D1 =
-3 1 -2
1 1 -1
8 -1 3
= 1
D2 =
1 -3 -2
2 1 -1
1 8 3
= 2
D3 =
1 1 -3
2 1 1
1 -1 8
= 3
So X1 = D1 / D = 1
x2 = D2/D = 2
x3 = D3/D = 3

How to use Cramer's law to solve this linear system of equations?
I have a system of linear equations. I haven't been able to do it for a long time,
X1＝1,X2＝2,X3＝3,X4＝-1
Now we just can't find D, D1, D2, D3, D4 (it's better to have a calculation process.)
X1+X2+X3+X4=5
X1+2X2－X3+4X4=－2
2X1－3X2－X3－5X4=2
3x1 + x2 + 2x3 + 11x4 = 0

Take a look at the picture

Is there any condition to solve linear equations with Cramer's law

① Number of unknowns = number of equations
② Coefficient determinant D ≠ 0

Finding the domain of definition and value of y = (1 / 3) ^ (x + 2 / x-1)

Denominator X-1 is not equal to 0
So the domain (- ∞, 1) ∪ (1, + ∞)
(x+2)/(x-1)
=(x-1+3)/(x-1)
=(x-1)/(x-1)+3/(x-1)
=1+3/(x-1)
3/(x-1)≠0
So 1 + 1 / (x-1) ≠ 1
y≠(1/3)^1=1/3
And the exponential function is greater than 0
So the range (0,1 / 3) ∪ (1 / 3, + ∞)

The range of the function y = sin (2x-6 π), X ∈ [6 π, 2 / 3 π] is

X ∈ [6 / π, 2 / 3 π]
2x∈【π/3,4π/3】
π ∈ [π / 6,7 π / 6]
When π = 7 π / 6, the minimum value of Y is - 1 / 2
When π = π / 2, the maximum value of Y is 1
So: the range of Y is [- 1 / 2,1]

Find the range of function y = sin (2x - π / 6) - 1 in [0, π / 4]

x∈[0,π/4]
Then 2x - π / 6 ∈ [- π / 6, π / 3]
Using the image of sine function,
Then when 2x - π / 6 = - π / 6, y has a minimum value (- 1 / 2) - 1 = - 3 / 2
When 2x - π / 6 = π / 3, y has a minimum value (√ 3 / 2) - 1 = (√ 3-1) / 2
The value range is [- 3 / 2, (√ 3-1) / 2]

Y = 2x ^ 2-1, the range of (1,7) twin function, a total of several

The solution of y = 2x ^ 2-1 = 1 is x = 1 or x = - 1
The solution of y = 2x ^ 2-1 = 7 is x = 2 or x = - 2
therefore
There are four kinds of C (1,2) C (1,2) = when the domain contains two elements
When the domain contains three elements, there are four kinds: 2C (1,2) C (2,2) = 4
When the domain contains four elements, there is C (2,2) C (2,2) = one
So there are nine

The range of function y = root (- x ^ 2 + 2x + 7) is

Y = root (- x ^ 2 + 2x + 7)
=Radical [- (x-1) & sup2; + 8]
∵0

The range of the square of function y = 2x - 6x + 3 (- 1 ≤ x ≤ 1)?

The opening of the image is upward, and the symmetry axis is 3 / 2
When x is less than 3 / 2, the function decreases
So, when x = - 1, the function has the largest value, 11
When x = 1, the minimum value of the function is - 1
So the range is [- 1,11]

When x

(1) The domain of the function is r,
Let u = 6 + x-2x2, then y=（
one
two
）u．
∵ quadratic function u = 6 + x-2x2 = - 2 (x)-
one
four
）2+
forty-nine
eight
,
The range of function is {y | y ≥（
one
two
）
forty-nine
eight
}．
The axis of symmetry of quadratic function u = 6 + x-2x2 is X=
one
four
,
In[
one
four
U = 6 + x-2x2 is a decreasing function,
In (- ∞,
one
four
]It's an increasing function and a function y=（
one
two
）U is a decreasing function,
∴y=（
one
two
）6 + x-2x2 in[
one
four
It is an increasing function,
In (- ∞,
one
four
]It's a decreasing function