Exercise 2.4 Assume that a system Ax = b of linear equations has at least two distinct solutions y and z.
a. Show that xk = y+k(y−z) is a solution for every k.
b. Show that xk = xm implies k = m. [Hint: See Example 2.1.7.]
c. Deduce that Ax = b has infinitely many solutions.
2.4. Let the system Ax = b of linear equations have at least two distinct solutions y and z. Then Ay = b and Az = b.
a. Let k be an arbitrary scalar. Now, if xk =y+k(y−z),then Axk= A(y+k(y−z)) = Ay +kA(y-z) = b+k(Ay-Az) = b+k(b-b) = b. Hence, xk = y+k(y−z) is a solution to Axx = b for all k.
b. If xk = xm, then y+k(y−z) = y+m(y−z) or, k(y-z) = m(y-z) so that k = m ( as y ≠ z).
c. As per part (b) above, if the equation Ax = b has two distinct solutions y and z , then xk = y+k(y−z) is also a solution for every k. This implies that the equation Ax = b has infinite solutions.
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