Question

If \( \begin{pmatrix} 2x+y & x+y \\ p-q & p+q \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \), then \( (x, y, p, q) \) equals:

• A. \( 0, 1, 0, 0 \)
• B. \( 0, -1, 0, 0 \)
• C. \( 1, 0, 0, 0 \)
• D. \( 0, 1, 0, 1 \)
• E. \( 1, 0, 1, 0 \)

Hint:

If \( p+q=0 \) and \( p-q=0 \), the only solution is \( p=0, q=0 \). Recognizing this pattern saves time on calculations.

Answer 1

The Correct Option is A

Concept: Matrix equality implies that each corresponding element must be identical. This creates two independent systems of linear equations: one for \( x \) and \( y \), and another for \( p \) and \( q \).

Step 1: Solving for \( x \) and \( y \).
From the first row: \[ 1) \quad 2x + y = 1 \] \[ 2) \quad x + y = 1 \] Subtracting equation (2) from (1): \[ (2x + y) - (x + y) = 1 - 1 \implies x = 0 \] Substitute \( x = 0 \) into (2): \[ 0 + y = 1 \implies y = 1 \]

Step 2: Solving for \( p \) and \( q \).
From the second row: \[ 3) \quad p - q = 0 \implies p = q \] \[ 4) \quad p + q = 0 \] Substituting \( p = q \) into (4): \[ q + q = 0 \implies 2q = 0 \implies q = 0, \text{ hence } p = 0 \] The ordered quadruple is \( (0, 1, 0, 0) \).