Practice: Columns first
Three worked examples, then five problems. Try each problem on paper first, check it in the lab if you like, then open the solution.
Worked examples
Example 1 · Find the combination
Write b = (7, 4) as a combination of v = (1, 2) and w = (3, 1).
We need c and d with c·(1, 2) + d·(3, 1) = (7, 4). Component by component:
In matrix form this is Ax = b with the columns v and w: 1321 x = (7, 4), x = (1, 2).
Example 2 · Ax by rows and by columns
Compute Ax for A = 2−113 and x = (3, 2).
Same answer. The column way shows that (4, 9) lies in the column space, because it is built from the columns.
Example 3 · Build a matrix from where (1, 0) and (0, 1) land
Find the matrix that sends (1, 0) to (1, 1) and (0, 1) to (−1, 1). What does it do to areas?
The landing spots are the columns: A = 1−111.
Both arrows have length √2 and are at 90° to each other, so A turns the plane by 45° and stretches it by √2. Areas double, which matches det A = 2.
Problems
Is (5, 1) a combination of (1, 1) and (2, −1)? If so, find the numbers c and d.
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Solve c + 2d = 5 and c − d = 1. Subtracting gives 3d = 4, so d = 4/3 and c = 7/3.
Check: 7/3 + 8/3 = 5 and 7/3 − 4/3 = 1 ✓. Yes, because the two vectors point in different directions, every vector in the plane is a combination of them.
Compute Ax by columns for A = 1234 and x = (−1, 1).
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Ax = −1·(1, 3) + 1·(2, 4) = (1, 1).
Check by rows: (−1 + 2, −3 + 4) = (1, 1) ✓
For which number k are the columns of 1k26 dependent? What is the column space then?
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The columns are dependent when det = 1·6 − 2k = 0, so k = 3.
Then column 2 = (3, 6) = 3·(1, 2). The column space is the line through (1, 2). Ax = b can only be solved when b is a multiple of (1, 2).
Find the 2 by 2 matrix that mirrors the plane across the x-axis. What is its determinant, and what does the sign mean?
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(1, 0) stays put and (0, 1) flips to (0, −1), so A = 100−1.
det A = −1. Areas keep their size (|det| = 1), and the minus sign says the plane was flipped over.
R = 0−110 turns the plane by 90°. Compute R² and R⁴ and explain them without multiplying.
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R² = −100−1, a turn by 180°: every vector points the opposite way.
R⁴ = I, the identity. Four quarter turns bring every vector back to where it started.
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