Question

If \(A = \left[ \begin{array}{ccc}3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7\end{array} \right]\); \(B = \left[ \begin{array}{ccc}-1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4\end{array} \right]\) then, \((2A+3B)^{-1} =\) _______

• A. \(\left[ \begin{array}{ccc}\frac{1}{3} & 0 & 0 \\ 0 & -\frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{26}\end{array} \right]\)
• B. \(\left[ \begin{array}{ccc}\frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{26}\end{array} \right]\)
• C. \(\left[ \begin{array}{ccc}\frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & -\frac{1}{26}\end{array} \right]\)
• D. \(\left[ \begin{array}{ccc}-\frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & -\frac{1}{26}\end{array} \right]\)

Hint:

First rewrite the implication using and/or, then swap and with or.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The dual of a statement is obtained by interchanging \(\wedge\) and \(\vee\) (and t and c), but only after the statement uses only \(\wedge, \vee, \sim\). So we first remove the implication.

Step 2: Rewrite:
\[ (p\wedge\sim q)\rightarrow(q\wedge\sim p) \equiv \sim(p\wedge\sim q)\vee(q\wedge\sim p) \equiv (\sim p\vee q)\vee(q\wedge\sim p) \]

Step 3: Take the dual:
Swap the connectives: \((\sim p\wedge q)\wedge(q\vee\sim p)\). By absorption, this simplifies to \(\sim p\wedge q\).

Step 4: Compare the options:
Option (B): \((p\rightarrow q)\wedge\sim(q\rightarrow p) = (\sim p\vee q)\wedge(q\wedge\sim p) = q\wedge\sim p\). This is the same as \(\sim p\wedge q\).
Option (A) simplifies to \(p\wedge\sim q\), option (C) to \(p\), and option (D) to a tautology, none equal to \(\sim p\wedge q\).

Final Answer:
The dual is equivalent to option (B). \[ \boxed{(p\rightarrow q)\wedge\sim(q\rightarrow p)} \]