Let a be a matrix of order n and satisfy that the square of a = e, and prove that R (A-E) + R (a + e) = n

Let a be a matrix of order n and satisfy that the square of a = e, and prove that R (A-E) + R (a + e) = n

First you have to know R (a + b)

Let n-order matrix a satisfy a square = a, and E be n-order identity matrix. It is proved that R (a) + R (A-E) = n

N-order matrix a satisfies a square = a
===>r(A)≤n
When R (a) = n, = = = > A = e = = > R (A-E) = 0 = = = > R (a) + R (A-E) = n
When R (a) a is an identity matrix with at least one row of all zeros
===>r(A)+r(A-E)=n.
===>N-order matrix a satisfies a square = a, R (a) + R (A-E) = n

The problem of linear algebraic matrix
If a is a matrix of order m * n, what does R (a) = n mean
Of course, I know that the rank of a is n. but for a matrix of order 3 * 2, R (a) = n = 2, doesn't that mean that the determinant of the matrix is 0?

If and only if M = n, DET (a) can be defined. Generally, the rank r (a) of a matrix can be defined from different angles, and its meaning is equivalent. For example, R (a) = row rank of a matrix, that is, the number of vectors in the maximal linearly independent group of row vectors; R (a) = column rank of a matrix, that is, the number of vectors in the maximal linearly independent group of column vectors

In a matrix, the first line is 120 and the other two lines are zero. Why are the basic solution systems - 210 and 001

The corresponding equation is X1 + 2x2 = 0
X 2.x 3 is 1,0 and 0,1 respectively

1. Write an M function to calculate the product of two matrices (two matrices as input parameters, matrix multiplication operator cannot be used)

Function C = Ji (a, b) [M n] = size (a); [M n] = size (b); if = merror ('the number of columns of the former is different from the number of rows of the latter, which cannot be multiplied '); else for P = 1: m for Q = 1: nfor t = 1: Nd (P, Q, t) = a (P, t) * B (T, q); end C (P, q) = sum (d (P, Q,:); end

Why is the number of operations needed to multiply two n-order matrices to the third power of N?
Great Xia, help a! Ha ha

The result of n-order matrix multiplication is still n-order matrix, so there are n square elements in the result, and each element is obtained by multiplication of 2n numbers, that is, n times operation. So it is the third power of n

How to calculate if a is a second-order matrix to the power a of E?

e^A=E+A+A^2/2!+A^3/3!+...+A^n/n!+...
First find all the powers of a until a certain power is 0, then add them up according to the above formula

It is known that the inverse matrix of the third-order square matrix A is 1 1 2 1 1 3. Find the inverse matrix of the adjoint matrix A *

Because a (- 1) = a * / | a |
A*=A(-1)|A|
[A*](-1)=[A(-1)|A|](-1)
Since | a | is a numerical value, the left side = [a (- 1)] (- 1) / | a | = A / | a |
Because the inverse matrix of a matrix is not fully explained in your problem, there is no way to give specific results

Let a be a square matrix of third order, and | a | = 2, a * be the adjoint matrix of a, and | 3A * | =?

A*=|A|A^(-1)=2A^(-1)
From | a | = 2 to | a ^ (- 1) | = 1 / 2
|3A*|=|6A^(-1)|=6³|A^(-1)|=6³×1/2=108
A ^ (- 1) denotes the inverse matrix of A

Try to use the elementary transformation of matrix to find the inverse matrix of the down stream square matrix 321A = 315 3223

(A,E) =
3 2 1 1 0 0
3 1 5 0 1 0
3 2 3 0 0 1
r2-r1,r3-r1
3 2 1 1 0 0
0 -1 4 -1 1 0
0 0 2 -1 0 1
r2-2r3,r1-(1/2)r3
3 2 0 3/2 0 -1/2
0 -1 0 1 1 -2
0 0 2 -1 0 1
r1+2r2
3 0 0 7/2 2 -9/2
0 -1 0 1 1 -2
0 0 2 -1 0 1
r1*(1/3),r2*(-1),r3*(1/2)
1 0 0 7/6 2/3 -3/2
0 1 0 -1 -1 2
0 0 1 -1/2 0 1/2
If you have any questions, please let me know