Question

Given \( \mathbf{x} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \), \( \mathbf{b} = \begin{pmatrix} -16 \\ -3 \\ 7 \end{pmatrix} \), \( \mathbf{P}^{-1} = \begin{pmatrix} e & -6 & -7 \\ f & 9 & 11 \\ -1 & -2 & g \end{pmatrix} \) and \( \mathbf{P}\mathbf{x} = \mathbf{b} \).
If \( \mathbf{P}\mathbf{y} = \mathbf{d} \) with \( \mathbf{y} = \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} \) and \( \mathbf{d} = \begin{pmatrix} -16 \\ 0 \\ 7 \end{pmatrix} \), the value of \( y_2 \) is

• A. \( 29 \)
• B. \( 27 \)
• C. \( 25 \)
• D. \( 5 \)

Hint:

Use Px=b to pin down the unknown entries of P inverse first, then apply the same P inverse to the new vector d.

Answer 1

The Correct Option is A

Step 1: Turn Px=b into a usable equation for the unknowns in P^{-1}.
We are told \( \mathbf{P}\mathbf{x} = \mathbf{b} \). Multiplying both sides on the left by \( \mathbf{P}^{-1} \) gives \( \mathbf{P}^{-1}\mathbf{P}\mathbf{x} = \mathbf{P}^{-1}\mathbf{b} \), and since \( \mathbf{P}^{-1}\mathbf{P} = I \), this simplifies to \( \mathbf{x} = \mathbf{P}^{-1}\mathbf{b} \).
So the known vectors \( \mathbf{x} \) and \( \mathbf{b} \) let us pin down the unknown entries \( e, f, g \) inside \( \mathbf{P}^{-1} \), without ever needing \( \mathbf{P} \) itself.

Step 2: Write out the three scalar equations.
Row 1: \( e(-16) + (-6)(-3) + (-7)(7) = 1 \)
Row 2: \( f(-16) + (9)(-3) + (11)(7) = 2 \)
Row 3: \( (-1)(-16) + (-2)(-3) + g(7) = 1 \)

Step 3: Solve each row for its unknown.
Row 1: \( -16e + 18 - 49 = 1 \Rightarrow -16e - 31 = 1 \Rightarrow -16e = 32 \Rightarrow e = -2 \).
Row 2: \( -16f - 27 + 77 = 2 \Rightarrow -16f + 50 = 2 \Rightarrow -16f = -48 \Rightarrow f = 3 \).
Row 3: \( 16 + 6 + 7g = 1 \Rightarrow 22 + 7g = 1 \Rightarrow 7g = -21 \Rightarrow g = -3 \).
So \( \mathbf{P}^{-1} = \begin{pmatrix} -2 & -6 & -7 \\ 3 & 9 & 11 \\ -1 & -2 & -3 \end{pmatrix} \).

Step 4: Apply the same P^{-1} to the new system.
By the same logic as Step 1, \( \mathbf{P}\mathbf{y} = \mathbf{d} \) means \( \mathbf{y} = \mathbf{P}^{-1}\mathbf{d} \), using the exact \( \mathbf{P}^{-1} \) just found in full.
\( y_2 \) comes from row 2 of \( \mathbf{P}^{-1} \):
\[ y_2 = (3)(-16) + (9)(0) + (11)(7) = -48 + 0 + 77 = 29 \]
Options (B) 27, (C) 25 and (D) 5 come from arithmetic slips, such as dropping a sign on the -48 term or picking the wrong row of \( \mathbf{P}^{-1} \).

Final Answer:
The value of y2 is 29. \[ \boxed{29} \]