|A = [||2||0||2||1||]|
For each of the following matrix B, find a matrix X such that AX = B.
|B = [||1||]|
Answer X must be a vertical vector of dimension 4. The equation AX = B is the usual system of 3 linear equations in 4 variables. It is in fact this system. By the earlier computation, the augmented matrix [A B] has the following row echelon form.
From this we find one solution
|X = [||-1||]|
|B = [||1||1||]|
|b1 = [||1||], b2 = [||1||]|
be the two columns of B. Let x, y be the two columns of X. By AX = A[x y] = [Ax Ay], AX = B is equivalent to two separate systems Ax = b1, Ay = b2. The first system has been studied in the problem above and a solution has been obtained. The second system has an obvious solution
|y = [||0||]|
Combining the solutions together, we get a solution
|X = [||1||0||]|
of AX = B.
|B = [||1||0||0||]|
Answer The columns of B are the standard basis vectors e1, e2, e3. By a similar argument as before, the columns of X are the solutions of the systems Ax = ei. Combining the augmented matrices of the three systems, we get
|[A e1 e2 e3] = [A I] = [||2||0||2||1||1||0||0||]|
Row operations (e.g., as in this exercise) change the matrix into
By taking [col 1-5], we get a solution (1/3, -1/3, 0, 1/3) of Ax = e1. By taking [col 1-4] and [col 6], we get a solution (1/3, 2/3, 0, -2/3) of Ax = e2. By taking [col 1-4] and [col 7], we get a solution (-1/3, 1/3, 0, 2/3) of Ax = e3. Putting the three columns together, we get a solution
|X = [||1/3||1/3||-1/3||]|
of AX = B.
Remark From this sequence of exercises, you can see the relation between solving equations like AX = B and our usual systems Ax = b of linear equations. Apply your new insight to this exercise and this exercise. Moreover, study the existence/uniqueness/general solution problem for AX = B, just as we have done for systems of linear equations.