矩阵和乘法的逆 [英] Inverse of matrix and multiplication

问题描述

我是新的矩阵世界，对于这个基本问题，我无法想像：

I am new to world of matrix, sorry for this basic question I could not figure out:

我有四个矩阵（一个未知）。

I have four matrix (one unknown).

矩阵X

x <- c(44.412, 0.238, -0.027, 93.128, 0.238, 0.427, -0.193, 0.673, 0.027, 
     -0.193, 0.094, -0.428, 93.128, 0.673, -0.428, 224.099)

X <- matrix(x, ncol = 4 )

矩阵B：需要解决，1 X 4（列x nrows）与b1，b2，b3，b4值

Matrix B : need to be solved , 1 X 4 (column x nrows), with b1, b2, b3, b4 values

Matrix G

g <- c(33.575, 0.080, -0.006, 68.123, 0.080, 0.238, -0.033, 0.468, -0.006, 
-0.033, 0.084, -0.764, 68.123, 0.468, -0.764, 205.144)

G <- matrix(g, ncol = 4)

Matrix A

a <- c(1, 1, 1, 1) # one this case but can be any value 
A <- matrix(a, ncol = 1)

解决方案：

B = inv(X) G A  # inv(X) is inverse of the X matrix multiplied by G and A 

我不知道如何正确地解决这个问题，特别是矩阵。感谢你的帮助。

I did not know how to solve this properly, particularly inverse of the matrix. Appreciate your help.

推荐答案

我猜猜，Nick和Ben都是老师，比做别人的作业要大得多但是，完整解决方案的道路真的非常明显，在下一步中没有什么意义：

I'm guessing that Nick and Ben are both teachers and have even greater scruples than I do about doing other peoples' homework, but the path to a complete solution was really so glaringly obvious that it didn't make a lot of sense not to tae the next step:

B = solve(X) %*% G %*% A 
> B
             [,1]
[1,] -2.622000509
[2,]  7.566857261
[3,] 17.691911600
[4,]  2.318762273

可以通过提供一个身份矩阵作为第二个参数来调用QR方法：

The QR method of inversion can be invoked by supplying an identity matrix as the second argument:

> qr.solve(G, diag(1,4))
                [,1]             [,2]          [,3]             [,4]
[1,]  0.098084556856 -0.0087200426695 -0.3027373205 -0.0336789016478
[2,] -0.008720042669  4.4473233701790  1.7395207242 -0.0007717410073
[3,] -0.302737320546  1.7395207241703 13.9161591761  0.1483895429511
[4,] -0.033678901648 -0.0007717410073  0.1483895430  0.0166129089935

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