Simultaneous Equations Involving Matrices - 2
This is a follow-up problem. The previous version can be found here .
Consider the system of equations:
A x + B y = m C x + D y = n
Here,
A = ⎣ ⎡ 3 0 0 0 3 0 0 0 3 ⎦ ⎤ ; B = ⎣ ⎡ 2 4 7 3 5 9 ⎦ ⎤ ; C = B T ; D = [ 1 2 2 4 ]
m = ⎣ ⎡ 1 0 1 1 1 2 ⎦ ⎤ ; n = [ 1 3 1 4 ]
x ∈ I R 3 × 1 ; y ∈ I R 2 × 1
The goal is to find the column matrices x and y . Let the solution be stacked into a larger column matrix
S = [ x y ] T ; S ∈ I R 5 × 1
Let all the sum of the elements of S be P . Find ⌊ 1 0 ∣ P ∣ ⌋
Note:
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⌊ . ⌋ denotes the floor function.
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∣ . ∣ denotes the absolute value function.
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The superscript T denotes the transpose operation carried out on a matrix.
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It may be easy to rewrite the system as a set of 5 linear equations involving 5 unknowns. However, I encourage the solvers to use linear algebra. Try and evaluate a closed-form algebraic expression of the solution and use a calculator in the final few steps. The recommended approach becomes useful while dealing with systems involving a very large number of variables. Otherwise, rewriting the equation set as a system of linear equations and solving them is a mammoth task.
The answer is 102.
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1 solution
Very nice solution, thank you!
Two further questions:
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this solution used the most basic type of algebraic manipulation to solve the equations. For scalars, there are other methods (eg Cramer's rule) - can these be extended to matrices?
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is there a nice linear algebra shortcut to get straight to the sum of all entries of S ? (Essentially this is dotting S with a vector with 1 s in every position)
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Thanks for the suggestions. What I have in mind is the following:
The sum of all elements of x and y can be found in the following equation
P = [ 1 1 1 1 1 ] [ A B T B D ] − 1 [ m n ]
Of course the use of a computer can easily enable such a calculation. In order to be able to do this by hand, one needs to invert the block matrix comprising of A , B and D , which would be a rather tedious task. There is a way to factorise the block matrix. I will post another comment following this once I find the means to do so.
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The following link demonstrates how to factorise and invert a block matrix. I would make use of the technique described in the link.
https://en.wikipedia.org/wiki/Schur_complement
We can manipulate this in the same we way would an equation in scalars, as long as we're careful about the interpretation of multiplication and division; we need to remember that matrix multiplication is not (in general) commutative (so U V is not necessarily equal to V U ), and in place of "division", we will multiply by inverse matrices (so we can only do this with non-singular square matrices).
Left-multiply the first equation by A − 1 to get
x + A − 1 B y = A − 1 m
Now left-multiply by C to get
C x + C A − 1 B y = C A − 1 m
(these two steps could of course be done at the same time, but I've separated them for clarity). Now subtract the second given equation to get
C A − 1 B y − D y = C A − 1 m − n
or
( C A − 1 B − D ) y = C A − 1 m − n
We can now solve for y :
y = ( C A − 1 B − D ) − 1 ( C A − 1 m − n )
and use this value to find x :
x = A − 1 m − A − 1 B y
Plugging the numbers in, we find S = [ 3 . 8 6 , 1 . 3 6 , 0 . 8 0 , 1 4 . 3 5 , − 1 0 . 1 0 ] T (all the values have been rounded to 2 decimal places), leading to the answer ⌊ 1 0 ∣ P ∣ ⌋ = 1 0 2 .