When discussing what the value is, a system of linear equations has a unique solution, no solution,

Write the augmented matrix of the equation as.

2 2 2+4 Line 1 minus line 2*, line 3 minus line 2*2, swap lines 1 and 2.

0 2-2 4 2-4 Multiply line 2 by 2, line 2 minus line 3 * (1+) swap lines 2 and 3.

Obviously, the determinant of the coefficient matrix is (2-2)*6--2).

If the determinant of the coefficient matrix is not 0, i.e. it is not equal to 1, 2 or -3, then the rank of the augmented matrix must be 3 and the system of equations has a unique solution.

Whereas if is equal to 1, 2, or -3, then the system of equations may have no solution or infinite solution, and when =1, the augmentation matrix is .

0 0 4 4 Divide line 2 by -3, line 1 minus line 2*2, line 3 minus line 2*4

So the system of equations has an infinite solution, and the general solution is c*(1,-1,0) t +(2,0,1) t, and c is a constant.

When =2, the augmentation matrix is.

Obviously, the rank of the coefficient matrix is smaller than the rank of the augmentation matrix, and the system of equations has no solution.

When = -3, the augmentation matrix is.

Obviously, the rank of the coefficient matrix is smaller than the rank of the augmentation matrix, and the system of equations has no solution.

So in summary, when it is not equal to 1, 2 or -3, the system of equations has a unique solution, and when =2 or -3, the system of equations has no solution.

And when =1, the system of equations has an infinite solution, and the general solution is c*(1,-1,0) t +(2,0,1) t, and c is a constant.

Original question

x1 + x2 - x3 = 1 x1 = 1 + x3 - x22x1 +3x2 + ax3 = 3

x1 + ax2 + 3x3 = 2

Subtract x1 to get it.

x2 + a + 2 )x3 = 1 x2 = 1 - a + 2 )x3

a -1 )x2 + 4x3 = 1

Subtract x2 to get.

6-a -a 2)x3 = 2 - a( a-2 )(a + 3) x3 = a - 2 When a = -3, the left side is always 0, so there is no solution a ≠ 2 and a ≠ 3 has a unique solution a = 2 when there is an infinite number of solutions.

x2 = 1 - a + 2 )x3

x1 = 1 + x3 - x2 = (a + 3) x3, i.e. x1 = (a + 3) x3

x2 = 1 - a + 2 )x3

x3 = x3

Linear equations are also called one-time equations. Refers to an equation in which all unknowns are one-time. Its general form is ax+by+.cz+d=0。The essence of a linear equation is to multiply both sides of the equation by any identical non-zero number, and the essence of the equation is not affected.

It can be solved by using matrices, and if it is a junior high school student, it can be solved by using Gaussian elimination;

The system of equations is ax=b

To the matrix [a|b] to perform row transformations.

When r[a]=r[a|b] ≠0, there is a unique solution.

When r[a]=r[a|b], there is no solution.

When r[a]=r[a|b]=0, there are infinitely many solutions.

It can be solved by using matrices, and if it is a junior high school student, it can be solved by using Gaussian elimination; The matrix method is as follows.

Classic questions, ready-made conclusions:

Calculate the determinant of the coefficient matrix first.

When ≠1 and ≠-2, there is a unique solution known by Crammer's rule.

When =1, the sock posture width matrix is .

The general solution is: (1,0,0).'+c1(-1,1,0)'+c2(-1,0,1)'

When =-2, the augmentation matrix is.

r3+r1+r2

At this point, there is no solution to the system of equations.

Note: This method can only be used when the number of equations in the system of equations and the number of unknown quantities are easy to grasp],6,

j. Simplified.

0 -1 0 --0 0 (1)--2 -1) then at =0, r(a)=1 is not equal to r(a)=2 without solution =1, r(a)=1 is not equal to r(a)=2 without solution When it is not equal to 0 and not equal to 1, r(a)=r(a)=3 has a unique solution (by the way, how do you enter it?). ）

For the augmented matrix, the simplest line.

When a-2=0 and b-a+1≠0, i.e., a=2, b≠1, the system of equations has no solution.

When a-2≠0, i.e., a=2, the system of equations has a solution, and there is a unique solution.

Introduction Taishang Laojun's boy.

Using the coefficient matrix determinant, it is not 0 and has a unique solution.

The coefficient matrix determinant is 0 (solution = 1 or -2), which is discussed below:

When =1, the rank of the coefficient matrix is equal to the rank of the augmentation matrix, and there is a solution.

When =-2, the rank of the coefficient matrix is not equal to the rank of the augmentation matrix, and there is no solution.

Write the augmentation matrix for this system of equations and solve it using the elementary row transformation.

4 5 -5 -1 Line 2 minus line 3 multiplied by 4, line 3 minus line 1 2, line 1 divided by 2

If the system of equations has an infinite number of solutions or no solutions, then the determinant of the coefficient matrix is equal to 0, so (-1-5 4)*(3) -1+5 4)(5-2 )=0

Solution = -4 5 or 1

So when it is not equal to -4, 5 and 1, the system of equations has a unique solution.

If = -4 5, the augmentation matrix can be reduced to .

Obviously, the rank of the coefficient matrix is smaller than the rank of the augmentation matrix, and the system of equations has no solution.

If = 1, the augmentation matrix can be reduced to .

0 3 -3 -3 Divide the 2nd line by -9 4

0 3 -3 -3 Line 1 minus line 2 1 2, line 3 minus line 2 3

Then the general solution of the system of equations is c*(0,1,1) t + 1,0,1) t where c is a constant.

In summary. When it is not equal to -4 5 and 1, the system of equations has a unique solution, and when = -4 5, the system of equations has no solution.

At 1, the general solution of the system of equations is c*(0,1,1) t + 1,0,1) t, where c is a constant.

Solution: Augmentation matrix =

r1<->r3

r2-r1, r3-λr1

r3+r20 0 (1- )2+ )3( -1) When ≠1 and ≠-2, r(a)=r(a,b)=3, the system of equations has a unique solution.

When =-2, r(a)=2 and r(a,b)=3, the system of equations has no solution.

When =1, r(a)=r(a,b)=1<3, the system of equations has an infinite number of solutions.

The general interpretation is: (2,0,0).'+c1(-1,1,0)'+c2(-1,0,1)'