forall u, v, ((2,3,5,7), (13,17,19,23)) * ((64,28,-18), (-64,-27,18), (15,5,-5), (0,0,1)) * ((1), (u), (v)) = ((11), (29)) ?

1 Answer

Please refer to a Proof in the Explanation.

Explanation:

Let, [A]_(2xx4)=[(2,3,5,7),(13,17,19,23)],

[B]_(4xx3)=[(64,28,-18),(-64,-27,18),(15,5,-5),(0,0,1)],

[C]_(3xx1)=[(1),(u),(v)].

Then, [A]_(2xx4)*[B]_(4xx3)*[C]_(3xx1) is defined, and, is a

(2xx1) Matrix. .

Now, [B]_(4xx3)*[C]_(3xx1) is a 4xx1 Matrix, given by,,

=[(64+28u-18v),(-64-27u+18v),(15+5u-5v),(v)].

Next, [A]_(2xx4)*{[B]_(4xx3)*[C]_(3xx1)} is a 2xx1 Matrix

and, [A]_(2xx4)*{[B]_(4xx3)*[C]_(3xx1)},

=[(2,3,5,7),(13,17,19,23)]*[(64+28u-18v),(-64-27u+18v),(15+5u-5v),(v)].

Now,

2(64+28u-18v)+3(-64-27u+18v)+5(15+5u-5v)+7(v),

=(128-192+75)+u(56-81+25)+v(-36+54-25+7),

=11, and,

13(64+28u-18v)+17(-64-27u+18v)+19(15+5u-5v)+23(v),

=(832-1088+285)+u(364-459+95)+v(-234+306-95+23),

=29.

:.[(2,3,5,7),(13,17,19,23)]*[(64+28u-18v),(-64-27u+18v),(15+5u-5v),(v)]=[(11),(29)].

From here on, consider the variables exchange

alpha = 3 + u - v

beta = u

ABC = K

ADE = K

D = ((10, 18, 10), (-10, -18, -9), (0, 5, 0), (3, -1, 1)), E = ((1), (alpha), (beta))

DE = BC

E = MC

DMC = BC => DM = B => M = ((1, 0, 0), (3, 1, -1), (0, 1, 0))