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Related problems of matrix in linear algebra~ Let a be a third-order matrix, and AJ be the j-th column of a (J = 1,2,3), Matrix B = (a3,3a2-a3,2a1 + 5A2), (the numbers after a are subscripts) If | a | = - 2, then | B | = (?)
|B|=|A3,3A2-A3,2A1+5A2|
=|A3,3A2,2A1+5A2|
=3|A3,A2,2A1|
=6|A3,A2,A1|
=-6|A1,A2,A3|
=12
The way to solve the problem is: any row or column of the matrix plus any multiple of the remaining rows or columns, its value will not change. But if any row is exchanged once, or any column is exchanged once, its value will be multiplied by - 1
(1) Given n-order matrix a satisfies a ^ 3 = 3A (A-I), find (A-I) ^ - 1; (2) n-order square matrix A, B satisfies a + B = AB, find (A-I) ^ - 1
(1) The book does this: A ^ 3-3a ^ 2 + 3A = O, that is (A-I) (- A ^ 2 + 2a-i) = I, the final answer is - A ^ 2 + 2a-i
This is what I do. After multiplying a ^ - 1 on both sides of the equation, I reduce it to a ^ 2-3a + 2I = - I, and get (A-I) (a-2i) = - I, then the answer is 2i-a
Did I do it wrong,
(2) The book is like this: (A-I) (B-I) = ab-a-b + I = I, then the answer is B-I
This is what I do: a = (A-I) B, multiply a ^ - 1 on both sides to the left, and get (A-I) BA ^ - 1 = I, then the answer is ba ^ - 1
It's different. Am I wrong about both questions
1
This a is not necessarily reversible. If it is not, a ^ (- 1) does not exist
two
The same mistake as the first one
Given that matrix A is a 3-matrix, and a = 1 / 2, then - 2A = what?
Take - 2 and get the cubic power of - 2 = 8, then use - 8 * | a | = - 4
Finding inverse matrix 1 2 1'2 1 5'1 2 3
7/6,2/3,-3/2
1/6,-1/3,1/2
-1/2,0.1/2
Use the block matrix to find the inverse matrix of the following matrix 14000013000001000035000008
Use the block matrix to find the inverse matrices of the following matrices: 140000, 130000, 00110, 00350, 00008
Diagonally partitioned matrix
A 0
0 B
Its inverse matrix is
A^(-1) 0
0 B^(-1)
So here it is
1 4 0 0 0
1 3 0 0 0
0 0 1 1 0
0 0 3 5 0
0 0 0 0 8
and
1 4
1 3
The inverse matrix of is easily obtained as
-3 4
1 -1
Similarly,
1 1 0
3 5 0
0 0 8
You can also find the inverse in blocks,
Its inverse matrix is
5/2 -1/2 0
-3/2 1/2 0
0 0 1/8
So the inverse matrix of the original matrix is:
-3 4 0 0 0
1 -1 0 0 0
0 0 5/2 -1/2 0
0 0 -3/2 1/2 0
0 0 0 0 1/8
Matrix 2 2 3 1 2 3 50 1
How to turn matrix into standard
2 2 3
1 2 3
5 0 1
r1-r2
1 0 0
1 2 3
5 0 1
r2-r1,r3-5r1
1 0 0
0 2 3
0 0 1
r2-3r3
1 0 0
0 2 0
0 0 1
r2*(1/2)
1 0 0
0 1 0
0 0 1
Let a be a third-order matrix and a = 2, then (a *) - 1 = ()
If a * is an adjoint matrix, because the elements of matrix A are transposed by the matrix obtained by replacing them with the algebraic cofactors in determinant a, this matrix is called the adjoint matrix of A. the left and right multiplication results of adjoint matrices of a and a are all diagonal matrices whose elements on the main diagonal are all determinants of A
Let a, B, | a | = 2, | B | = 1 / 2, find | a + B|
Let a and B be matrices of third order, and | a | = 4, | B | = 2, | a ^ - 1 + B | = 2, then | a + B ^ - 1 | =?
The answer seems to be 4, which is wrong,
Let a be a third-order matrix, and | a | = 1 / 2 (see Fig
A*=|A|A^(-1)=1/2 * A^(-1)
So the original formula = | 3A ^ (- 1) - A ^ (- 1) | = | 2A ^ (- 1) | = 2 ^ 3 | a ^ (- 1) | = 8 * 2 = 16
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