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

$\textbf{Spot the shortcut before grinding through all three unknowns.}$
We need $y_2$, and $y_2$ only uses row 2 of $\mathbf{P}^{-1}$, which is $(f, 9, 11)$. Rows 1 and 3 hold $e$ and $g$, which we never need for this particular answer. So instead of solving for $e$, $f$ and $g$ together, solve for $f$ alone.

$\textbf{Get f from Px=b.}$
Since $\mathbf{P}\mathbf{x}=\mathbf{b}$ gives $\mathbf{x}=\mathbf{P}^{-1}\mathbf{b}$, the second row of that matrix equation is:
\[ f(-16) + 9(-3) + 11(7) = x_2 = 2 \]
\[ -16f - 27 + 77 = 2 \]
\[ -16f + 50 = 2 \]
\[ -16f = -48 \]
\[ f = 3 \]

$\textbf{Use Py=d the same way, but only row 2.}$
$\mathbf{P}\mathbf{y}=\mathbf{d}$ gives $\mathbf{y}=\mathbf{P}^{-1}\mathbf{d}$, and its second row is:
\[ y_2 = f(-16) + 9(0) + 11(7) \]
Put in $f=3$:
\[ y_2 = (3)(-16) + 0 + 77 = -48 + 77 = 29 \]

$\textbf{Final answer.}$
Without ever solving for $e$ or $g$, we get $y_2 = 29$, which is option (A). This is quicker than finding the whole matrix $\mathbf{P}^{-1}$ first, though both routes agree. \[ \boxed{29} \]