Question 1167704
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The solution will be much easier if you represent the original matrices as the vectors, placing the second row after the first


    A = (2, 6, 1, -3),  B = (3, 2, 1, 0).


Now use the formula for the cosine of angle between the vectors in n-dimensional space {{{R^n}}}


    cos(A,B) = {{{(A*B)/(abs(A)*abs(B))}}}.



In your case, n= 4,  the scalar product (A*B) = 2*3 + 6*2 + 1*1 + (-3)*0 = 19,


the length of the vector  |A| = {{{sqrt(2^2 + 6^2 + 1^2 + (-3)^2)}}} = {{{sqrt(50)}}};


the length of the vector  |B| = {{{sqrt(3^2 + 2^2 + 1^2 + 0^2)}}} = {{{sqrt(14)}}};


    cos(A,B) = {{{19/(sqrt(50)*sqrt(14))}}} = 0.718  (approximately).



Having the cosine, find the angle between the vectors as  arccos(0.718).


Use your calculator.
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Solved and explained in all details.