y = x - 1
y = -x + 1 (add both the equations)
2y = 0
y = 0
when y = 0, x = 1
So that the only solution is the intersection point of the lines, (1, 0).
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i think its pretty much the same thing because matrix X1 X2 IS ACTUALLY X1 X2
4
It is linear. The highest power is 1 (x = x1, y = y1) so it is linear.
This question can only be answered if the probability distribution functions of X1, X2 and X3 are known. They are not and so the question cannot be answered.