High Linear Algebra Problems, Linear Algebra, High Numbers Problems...

The inverse of a = a adjoint matrix iai

So, (3a) inverse-2 times a adjoint matrix = inverse of a 3-2a adjoint matrix = adjoint matrix of 2a 3-2a adjoint matrix = adjoint matrix of 4a 3

IA adjoint matrix I=IAI (n-1).

So the determinant is (-4 3) 3*iai 2=-16 27

A inverse|=23a) inverse=1 3(ainverse).

a adjoint matrix = |a|A inverse = 1 2a inverse.

2x a adjoint matrix = 2*|a|a inverse = 2 * (1 2a inverse) = a inverse.

3a) Inverse-2 times a adjoint matrix|=|1 3 (a inverse) - a inverse |=|-2 3(a-inverse)|=[(-2/3)^3]|A inverse|=-8/27*2=-16/27

a||3a) Inverse-2 times a adjoint matrix|

a[(3a)inverse-2 times a adjoint matrix]|

1/3)i-a|

3a) Inverse-2 times a adjoint matrix|=|(1/3)i-a|/|a|=2|(1/3)i-a|

The rest won't.

3a) Inverse-2 times a adjoint matrix|

3a) inverse -2 times (a) inverse *|a||

1 3 (a inverse) - a inverse |

-2 3(a-inverse)|

8/27|a|Inverse.

Upstairs is not right, I just finished graduate school last year, I don't know yet.

Give me the mailbox, I'll send it to you, this can't have a formula, I can't call you.

Summary. To truly understand it, don't classify (1) for systems of non-homogeneous linear equations ax = b has a solution r(a)=r(a,b) has a unique solution r(a)=r(a,b)=n (the number of unknowns, or the number of columns of a) has an infinite number of solutions r(a)=r(a,b) 1A is the square matrix, and the determinant can be found. When |a|≠0, r(a)=n, the system of equations has a solution and the solution is unique;|a|=0 is indefinite, depending on rank 2

It doesn't make sense to have more rows than columns3When there are more columns than rows, if there are solutions to the system of equations, there must be an infinite number of solutions (look at the rank) (2) Aligning the system of sublinear equations is simple, ax=0 always has a solution (zero solution), just focus on whether there is only a zero solution. r(a)=n has only zero solution r(a).

To truly understand it, don't classify it like this: (1) For systems of non-homogeneous linear equations ax = b has a solution r(a)=r(a,b) has a unique solution r(a)=r(a,b)=n (the number of unknowns, or the number of columns of a) has an infinite number of solutions r(a)=r(a,b) a is a square matrix, and the determinant can be found. When |a|≠0, r(a)=n, the system of equations has a unique solution;|a|=0 is indefinite, depending on rank 2There is no more line than column or depletion meaning3

When there are more columns than rows, if there is a solution to the equation there must be an infinite number of solutions (look at the rank) (2) Aligning the system of sublinear equations is simple, ax=0 always has a solution (zero solution), just pay attention to whether there is only a zero solution. r(a)=n has only zero solution r(a).

Second-order differential equations have 2 linearly independent solutions, right? If y2=2y1, the two solutions are linearly related, that is, there is another solution that is not expressed, so it is not a general solution.

Y1 and Y2 must be linearly independent when they are expressed as general solutions, otherwise they can be written as Y2=KY1, that is, Y=C1Y1+C2Y2=(C1+KC2)Y1=Cy1, and the two solutions are linearly related, that is, there is another solution that is not expressed.

Here, c1, c2, k, and c can all be expressed as arbitrary constants, which means that y=c1y1+c2y2 has no meaning.

The combination of two arbitrary constants y=c1y1+c2y2 is not necessarily a general solution of a second-order homogeneous linear differential equation, because it is possible to merge two arbitrary constants into one (as for the arbitrary constants in the general solution of differential equations, the textbook explains that the number cannot be reduced after the merger). When the two solutions y1 are linearly related to y2, i.e., y2 y1 = constant k, then the two constants in y=c1y1+c2y2 can be combined into one, i.e., y=cy1(c=c1+kc2), so it cannot be a general solution of a second-order homogeneous linear equation. Only when the two functions y1 are linearly independent of y2, y=c1y1+c2y2 contains two arbitrary constants and becomes the general solution.

To truly understand, don't categorize it that way.

1) For a system of non-homogeneous linear equations ax = b

There is a solution <=> r(a)=r(a,b).

There is a unique solution <=> r(a)=r(a,b)=n (the number of unknown quantities, or the number of columns of a).

There are infinite solutions <=> r(a)=r(a,b) that apply to the 3 cases you have divided above:

1.A is the square matrix, and the determinant can be found. When |a|≠0, r(a)=n, the system of equations has a solution and the solution is unique;|

a|=0 varies from time to time, depending on the rank.

2.It doesn't make much sense to have more rows than columns.

3.When there are more columns than rows, if there are solutions to the system of equations, there must be an infinite number of solutions (look at the rank) (2) It is simple to align the system of sublinear equations.

ax=0 always has a solution (zero solution), just focus on whether there is only a zero solution.

r(a)=n <=> only zero solution.

r(a) has a non-zero solution.

1.a is the square matrix, then r(a)=n <=> |a|≠0 <=> There are only zero solutions.

2.Pointless.

3.There must be a non-zero solution.

Dizziness, that's the most basic.

When I first studied, I used the ternary equations of junior high school to understand it, and I had no problem coping with general exams.

The four vectors are all three-dimensional column vectors, so the vector group a1, a2, a1, and a2 composed of the four vectors must be linearly related, so there are non-zero real numbers x1, x2, y1, y2, such that x1a1+x2a2-y1b1-y2b2=0, so x1a1+x2a2=y1b1+y2b2.

From the fact that a1, a2 and b1 and b2 are all linearly independent, it can be seen that the coefficients x1 and x2 cannot all be zero, and y1 and y2 are not all zero (because: if x1 and x2 are all zero, then y1b1 + y2b2 = 0, since b1 and b2 are linear and independent, y1 = y2 = 0, which is contradictory to x1, x2, y1, y2 are not all zero.

Similarly, y1 and y2 cannot all be zero). So x1a1+x2a2=y1b1+y2b2≠0.

The vector m=x1a1+x2a2=y1b1+y2b2, then m≠0, and m can be represented linearly by a1, a2, and b1, b2.

If b1 or b2 can be represented linearly, then b1 or b2 itself is m;

Otherwise, linearly independent, because a maximum of three linear independent vectors can be found in a three-dimensional linear space, then b2 must be represented linearly, then b2 is m

The second step is to split the determinant into several determinants according to the nature of the determinant to add (subtract), such as:

The third step is to add -1 times of the 3rd column to the first column, which is obtained.

Thank you, big hands, is the earth simple? Now in the process of cleaning up the line of advanced mathematics, some practical problems can also be solved.

The sum of the elements in line ith of the matrix multiplied by the algebraic remainders of line j (i,j is not equal) results in zero because.

In fact, this is equivalent to replacing the original determinant, line j, with line i, and then finding this new determinant (in this case, according to line j, it is obviously the sum of the elements of line i of the original matrix multiplied by the algebraic remainder of line j (i, j are not equal)), and this new determinant, obviously, line i is equal to line j, so the new determinant is 0

Quadratic type f=x'ax is transformed into a standard type y by an orthogonal transformation'by (b is a diagonal matrix), then there is an orthogonal matrix c such that c'ac=b。At this point, symmetric matrices A and B contracts.

Because c is an orthogonal matrix, c'The inverse matrix is the same as c, so a and b are still similar. Similarity matrices have the same eigenvalues, so the eigenvalues of a are the diagonal elements of the diagonal matrix b.

Therefore, as long as the n-element quadratic form is transformed into a standard type through orthogonal transformation, then the coefficients of those square terms in the standard type are the eigenvalues of the matrix of the quadratic type (if the number of square terms is less than n, the remaining eigenvalues are all 0).

After finding the eigenvalues of a as 1,4,0, the determinant |a|is equal to the product of eigenvalues, and the sum of the diagonal elements of a is equal to the sum of eigenvalues. This gives you the |a|=2b-b 2-1-=0,1+a+1=1+4+0, so a=3,b=1.

f(x,y,z)=x 2+ay 2+z 2+2bxy+2xz+2yz can be transformed into f=m 2+4n 2 by orthogonal transformation

then the eigenvalues are the same, so 1,4,0

a=1 b 1b a 1

So there is 1+a+1 = 1+4+0, and we get a= 3

and then by |a|= 0 = -(b-1) 2 gives b = 1

After f is converted to the standard type, the coefficient of the standard type is the eigenvalue, so the eigenvalue of the eigenmatrix A is 0 1 4

The landlord does not have a deep understanding of the quadratic standard type.