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4 years ago

Use a truth table to show that P Qand (~PV Q) A (~QV P) are equivalen

Mathematics

1 answer:

kati45 [8]4 years ago

8 0

Answer: The given logical equivalence is proved below.

Step-by-step explanation: We are given to use truth tables to show the following logical equivalence :

P ⇔ Q ≡ (∼P ∨ Q)∧(∼Q ∨ P)

We know that

two compound propositions are said to be logically equivalent if they have same corresponding truth values in the truth table.

The truth table is as follows :

P Q ∼P ∼Q P⇔ Q ∼P ∨ Q ∼Q ∨ P (∼P ∨ Q)∧(∼Q ∨ P)

T T F F T T T T

T F F T F F T F

F T T F F T F F

F F T T T T T T

Since the corresponding truth vales for P ⇔ Q and (∼P ∨ Q)∧(∼Q ∨ P) are same, so the given propositions are logically equivalent.

Thus, P ⇔ Q ≡ (∼P ∨ Q)∧(∼Q ∨ P).

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Help please!!!!!!!!!

In-s [12.5K]

ANSWER

EXPLANATION

For a matrix A of order n×n, the cofactor $C_{ij}$ of element $a_{ij}$ is defined to be

$C_{ij} = (-1)^{i+j} M_{ij}$

$M_{ij}$ is the minor of element $a_{ij}$ equal to the determinant of the matrix we get by taking matrix A and deleting row i and column j.

Here, we have

$C_{11} = (-1)^{1+1} M_{11} = M_{11}$

M₁₁ is the determinant of the matrix that is matrix A with row 1 and column 1 removed. The bold entries are the row and the column we delete.

$\begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \end{aligned}$

Since the determinant of a 2×2 matrix is

$\det\left( \begin{bmatrix} a & b \\ c& d \end{bmatrix} \right) = ad-bc$

it follows that

so $C_{11} = M_{11} = 24$

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6/12<br> is that equivalent to<br> 1/12

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4 years ago

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kari74 [83]

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