Get a high score for a patient solution to a system of homogeneous linear equations!

1.It is known that (1,0,1,0) t is the basic solution system of ax=0.

So ax=0 contains a linearly independent solution vector.

Since a is a matrix of order 4 and r(a) = 3 = 4-1, so r(a*) = 1

Relation to r(a) and r(a*): <

2.Because r(a)=3 so a*a = |a|e = 0, so a column vector.

are all solutions of a*x=0.

and r(a*) =1, so the base solution system of a*x=0 contains 4-r(a*) = 4-1=3 solution vectors.

The rank of a is 3, and the column vectors are all solutions of a*x=0, so the column vector group of a contains the basic solution system of a*x=0.

3.Since a1+a3=0, a1 and a3 can be expressed linearly with each other.

And because r(a)=3, a has 3 linearly independent column vectors.

Therefore, as long as a1, a2, a3, and a4 do not contain a1, a3 are all extremely irrelevant groups of the column vector group of a.

So only (d) matches in the selection, so a1, a2, a4 are removed

First question:

Because the title says that (1,0,1,0) t is a fundamental solution system for the system of equations ax=0.

For a vector, as long as it is not a zero vector, it is linear independent.

So ax=0 contains a linearly independent solution vector.

If r(a)=n, then r(a*)=n

If r(a)=n-1, then r(a*)=1

If r(a) the second question:

Because r(a)=3, i.e., a is not a full-rank matrix.

So|a|=0

So a*a=|a|e=0

A column vector that can be a from a*a=0 must infer the a*x=0 solution, but not necessarily the basic solution system.

Here we can deduce the basic solution system because r(a*)=1, and the basic solution system contains 3 solution vectors.

r(a)=3, i.e., the column vector group of a contains exactly 3 linearly independent vectors.

Then the column vector group of a contains the basic solution system of a*x=0.

One last question:

Because r(a)=3, i.e., the column vector group of a is of rank 3

A1, A2, A4 can be expressed linearly by A1, A2, A3, A4, A1, A2, A3, A4 can also be expressed linearly by A1, A2, A4.

i.e. a1, a2, a4 are equivalent to a1, a2, a3, a4.

So r(a1,a2,a4)=r(a1,a2,a3,a4)=3, so a1,a2,a4 is linearly irrelevant.

In the same way, a2, a3, and a4 are linearly independent.

It's not that a1, a2, a4 have been removed, it's just that there is no such option.

The first question: The relation between r(a) and r(a*) is the total order of r(a) + r(a*) = a.

ax=0 contains a linearly independent solution vector, which should be seen from the conditional basis solution system in the problem.

The second question Since r(a)=3 and a is a matrix of order 4, the modulo of a is 0, i.e., |a|=0

a|e is the form of the basic solution, because e is the form of the simplest solution, and it cannot be simplified.

The third question a1+a3=0 is not a basic solution system if it contains both of them, so the answer can exclude options a and c;

and r(a*)=1, so the basic solution system is 3 solution vectors, and d is directly selected

It's been 13 years since I graduated from university, and I don't remember a lot of things in my head, so I tried to do it first.

To solve this homogeneous linear equation, we can convert it to matrix form and apply Gaussian elimination to the matrix. First, write the coefficients as a matrix:

a = 1 1 -1 -1 |

Our goal is to convert matrix A into a stepped matrix. First, use the first line to eliminate the first Zheng element of the following two lines:

r2 = r2 - r1

r3 = r3 - 2 * r1

Get: a = 1 1 -1 -1 |

We find that r2 and r3 are the same, and we remove one of them:

r3 = r3 - r2

Get: a = 1 1 -1 -1 |

Now we have a simplified stepped matrix. We can solve for variables by algebra:

Solve x2:from the second equation

x2 = 3x3 - 4x4

Substitute the expression of x2 into the first equation to solve x1:

x1 = x2 + x3 + x4

x1 = 3x3 + 4x4 - x3 - x4x1 = 2x3 + 3x4

So, we get the general solution of the system of homogeneous linear equations:

x1 = 2x3 + 3x4

x2 = 3x3 - 4x4

x3 = x3

x4 = x4

where x3 and x4 are arbitrary real numbers. We can express the general solution in terms of vectors:

x = x3 * 2, -3, 1, 0) +x4 * 3, -4, 0, 1)

This is the general solution of a system of homogeneous linear equations.

A simple calculation is sufficient, and the answer to the head of the limb celery calendar is shown in the first life picture.

It is the determinant of the number matrix, and the 2nd and 3rd columns are added to the 1st column, and then the -1 times of the first row are added to the 2nd and 3rd rows respectively, then the early tomato becomes the upper triangle determinant, and the |a| =2)(λ1)^2

When 2, and 1, |a|≠0, the system of equations has a unique solution;

When 1, the augmentation matrix (a, b) = elementary row transformation is.

r(a, b) =2, r(a) =1, there is no solution in the Fang Hail Changcheng group.

When 2, the augmentation matrix (a, b) = elementary row transformation is.

The elementary row is transformed to.

The elementary row is transformed to.

r(a, b) =r(a) =2 < 3, systems of equations have infinitely many solutions.

The equation is systemized as.

x1 = 2+x3

x2 = x3

take x3 = 0, get the special solution (-2, 0, 0) t;

The export group is. x1 = x3

x2 = x3

Take x3 = 1 to get the base solution system (1, 1, 1) t of ax = 0

The general solution of ax = b is x = k (1, 1, 1) t + 2, 0, 0) t.

There are original questions on the linear algebra books.

A simple calculation is sufficient, and the answer to the head of the limb celery calendar is shown in the first life picture.

1. The augmented moment Li Nai wheel array B is transformed into a row ladder. If r(a)2 and r(a)=r(b), then b is further reduced to the simplest form of the row.

3. Let r(a)=r(b)=r; The unknowns corresponding to the non-0 first of r non-zero rows in the simplest form of the row are represented by the remaining n-r unknowns (free unknown mu empty numbers), and the free unknowns are equal to c1, c2, c3 ,.. respectivelycn-r, you can write a general solution with n-r parameters.

2) Elementary envy macrocode transformations for the following augmented matrices:

1 2 1 4 and add 1, -1, and 3 times in the fourth row.

One, two, three brothers, got.

1 2 1 4, divide the second row by 2 and add the -3, -4, and -1 times of the second row to the first row.

One, three, four lines, got.

0 4 1 4, add -2 times from the first row to the third row.

If the third equation is not true, there is no solution to the problem.

It feels like there should be no solution, and the rank of the corresponding augmentation matrix is larger than that of the unknown coefficient matrix.

x1+2x2-3x3=0，2x1 ＋5x2－3x3=0，x1 ＋4x2－3x3=0

Solution: coefficient matrix a =

r2-2r1,r3-r1

r3-2r2

So r(a)=3, the system of equations has only zero solutions.

Solution: coefficient matrix =

r1-3r3,r2-2r3

r2*(1/12),r1-16r2,r3+5r20 0 0 0 0

r1<->r3

So a1=(9,-3,4,0,0) t,a2=(3,7,0,4,0) t,a3=(1,5,0,0,-4) t is the fundamental solution.

The general solution of the system of equations is c1a1+c2a1+c3a3, and c1,c2,c3 are arbitrary constants.

The coefficient matrix is formed into a stepped matrix.

r(a)=2, so the base solution system has 5-2=3 free variables, so that x3, x4, x5 are free variables.

(x3, x4, x5) = (1, 0, 0), (010) (001) to obtain (x1x2) = (9 4, -3 4), (3 4, 7 4), (1 4, -5 4).

So the basic solution is.

t1=(9/4，-3/4,1,0,0）^t，t2=（3/4,7/4,0,1,0）^t，t3=（-1/4，-5/4,0,0,1）^t

The general solution is (x1,x2,x3,x4,x5) t=c1t1+c2t2+c3t3,c1,c2,c3 is any constant.

See the following picture:

A simple calculation is sufficient, and the answer to the head of the limb celery calendar is shown in the first life picture.