Question

Let \(A=\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}\). If \(u_1\) and \(u_2\) are column matrices such that \(Au_1 = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}\) and \(Au_2 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}\), then \(u_1 - u_2\) is

• A. \(\begin{bmatrix} -1 \\ 2 \\ -1 \end{bmatrix}\)
• B. \(\begin{bmatrix} -1 \\ 1 \\ -1 \end{bmatrix}\)
• C. \(\begin{bmatrix} 1 \\ -2 \\ -1 \end{bmatrix}\)
• D. \(\begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}\)

Hint:

If \(Au_1\) and \(Au_2\) are known, then \(A(u_1-u_2)=Au_1-Au_2\) because matrix multiplication spreads over subtraction. So we first find the right hand side, then solve a simple system.

Answer 1

The Correct Option is A

Concept:
If \(Au_1\) and \(Au_2\) are known, then \(A(u_1-u_2)=Au_1-Au_2\) because matrix multiplication spreads over subtraction. So we first find the right hand side, then solve a simple system.

Step 1:
Subtract the two given results: \[Au_1 - Au_2 = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} - \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}.\] So \(A(u_1-u_2) = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}\).

Step 2:
Let \(u_1-u_2 = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\). Multiplying A by this vector gives three equations: \[x+y=1,\quad y+z=1,\quad z=-1.\]

Step 3:
From the third equation \(z=-1\). Put this in the second: \(y-1=1\), so \(y=2\). Put \(y=2\) in the first: \(x+2=1\), so \(x=-1\).

Step 4:
Therefore \(u_1-u_2 = \begin{bmatrix} -1 \\ 2 \\ -1 \end{bmatrix}\).

Answer: Option (1) — \(\begin{bmatrix} -1 \\ 2 \\ -1 \end{bmatrix}\).