Matrix linear algebra, linear algebra matrix

The rank of the matrix is 4, and the solution process is as follows:

The first step is to swap the first and fourth rows.

The second step is to divide all the elements in the second row by 2

In the third step, add the second row to the third row, remove the two -1s from the third row, and multiply the second row by -1 and add it to the fourth row, eliminating the two 1s from the fourth row

The fourth step is to add the third line to the fourth line, and remove the -2 of the fourth line The fifth step, the matrix is already a stepped matrix, and it can be seen that the rank of the matrix is 4 if you are satisfied

The equation ba a+4e is multiplied by a* at both ends.

baa*＝aa*+4a*

Since aa* = |a|e

So |a|b = |a|e + 4a*

Calculate |a*|, according to |a*| = |a|(n-1) determines |a|

Substituting the above equation gives b

b-e)a = 4e

Explain that the inverse matrix of a is a -1 = 1 4 * b-e) a -1 is again equal to a* |a|

So 1 4 * b-e) = a* |a|b = 4a* / |a| +e

The matrix p is an elementary matrix that is obtained by interchanging the lines of the identity matrix e1,2. So matrix a left multiplication p is equivalent to swapping the 1,2 rows of a once, and multiplying the right is equivalent to swapping the 1,2 columns of a once. So after swapping so many times, the result is.

Because a is not singular, it can be expressed as the product of several elementary matrices, a = p1p2....P.S. Why.

Hehe, that's also a theorem, don't you have it in your book?

Because a is not singular, i.e., a is reversible, its equivalent standard form is e

That is, a can be transformed into e by elementary transformation

So there is an elementary matrix such that p1p2....ps a q1q2...qt = e

Invert these elementary matrices to the right of the equation, and the inverse of the elementary matrix is still the elementary matrix, so a is expressed as the product of the elementary matrix.

Example 3 a = 2e + b， b =

then b 2 =

b^n = o (n ≥ 3)

a^n = (2e+b)^n

2 ne + n2 (n-1)eb+ [n(n+1) 2]2 (n-2)eb 2 + ellipsis terms are all zero matrices).

2^ne + n2^(n-1)b + n(n+1)/2]2^(n-2)b^2 =

2^n n2^(n-1) n(n+1)2^(n-3)]

0 2^n n2^(n-1)]

0 0 2^n ]

Example 5 a^2 = 2a, a^3 = 4a, .a^n = 2^(n-1)a

a^n - 2a^(n-1) = 2^(n-1)a - 2*2^(n-2)a = o

<> Divide the funny Layou slip b into blocks and promote them.

<> such as a frank touch of socks let the excitement be noisy.

Discussed in two cases, when both a and b are reversible, c*=|c|c^(-1)

a||b|*

a^(-1) 0

0 b^(-1)

a||b|*

a*/|a| 0

0 b*/|b|

a*|b| 0

0 b*|a|

And when there are irreversible matrices in a and b, it is also easy to verify c=|c|i

i.e. is the final result.

The answer is | |b|a* 0 ||0 |a|b* |

Because c c*=|c|·e

So c*=|c|C inverse.

So c* is equal to |a||b|The inverse of xc and then the inverse of c you should know is a inverse and b inverse in the original order.

So you multiply the determinant of a by the determinant of b and get a diagonal determinant, which is |a||b|xa inverse sum |a||b|XB inverse.

And then |a|Multiply a inverse is a*, and you get the answer.

A 2 = 2a can be verified directly by multiplication, i.e. a 2-2a = o, so a n-2a (n-1) = a (n-2) {a 2-2a) = o.