How to find the basic solution system of linear algebra 1 0 0 X1 0 1 -1 * X2 = 0 0 0 0 X3 Why is the fundamental analysis to the t power of [0,1,1]?

How to find the basic solution system of linear algebra 1 0 0 X1 0 1 -1 * X2 = 0 0 0 0 X3 Why is the fundamental analysis to the t power of [0,1,1]?

Faint ~ that's not to the power of T. t means transpose. The x you seek is a column vector, and the [0,1,1] you write is a row vector, so add a T
If you expand this formula, X1 = 0, x2-x3 = 0, so X3 is a free quantity. If you give it a value (generally 1, if you don't want to use 1, no one else will stop you), you can find X2, so the basic analysis is [0,1,1] t

In linear algebra, we know the adjoint matrix A * = diag (1,1,1,8) of matrix A, and ABA ^ - 1 = Ba ^ - 1 + 3E
The answer is a * = | a ^ (- 1), | a * = | a ^ ^ 4 | a ^ (- 1) | = | a ^ ^ 3, I want to know why | a * | = | a ^ ^ 4 | a ^ (- 1) |, other steps are not needed

According to | Ka | = k ^ n ||

Given the adjoint matrix A ^ *, and ABA ^ - 1 = Ba ^ - 1 + 3E, find B
A^*= 1 0 0 0
0 1 0 0
1 0 1 0
0 -3 0 8

If we multiply a by right, we get AB = B + 3a, so (A-E) B = 3A, and multiply (A-E) ^ - 1 by left, we get b = 3 (A-E) ^ - 1A. From a *, we can get a = 2EA * ^ - 1 = 2000 200 - 200 3 / 40 1 / 4, so (A-E) ^ - 1 = 1000 1 000 2 01 0 - 4 / 3