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Question 79 of 84

Q.(a) Show that p↔q≡((∼p)∨q)∧((∼q)∨p)p\leftrightarrow q\equiv((\sim p)\vee q)\wedge((\sim q)\vee p) OR

(b) Solve the system of linear equations by Cramer's Rule. 3x−4y−2z−1=0,\dfrac{3}{x}-\dfrac{4}{y}-\dfrac{2}{z}-1=0, 1x+2y+1z−2=0,\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}-2=0, 2x−5y−4z+1=0\dfrac{2}{x}-\dfrac{5}{y}-\dfrac{4}{z}+1=0
Puducherry TnboardTamil Nadu HSC (DGE) Board 2024Subjective· 5mImportance★★★★★
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(a) Builds the 4-row truth table for p↔qp\leftrightarrow q and for ((¬p)∨q)∧((¬q)∨p)((\lnot p)\vee q)\wedge((\lnot q)\vee p) and confirms they match; (b) substitutes u=1/xu=1/x etc. to linearise the system, then solves by Cramer's rule. Both alternatives answered below.

(a) Prove p↔q≡((¬p)∨q)∧((¬q)∨p)p\leftrightarrow q\equiv((\lnot p)\vee q)\wedge((\lnot q)\vee p)

1. Truth table (T = true, F = false):

ppqqp↔qp\leftrightarrow q¬p\lnot p¬p∨q\lnot p\vee q¬q\lnot q¬q∨p\lnot q\vee p(¬p∨q)∧(¬q∨p)(\lnot p\vee q)\wedge(\lnot q\vee p)
TTTFTFTT
TFFFFTTF
FTFTTFFF
FFTTTTTT

2. Compare the p↔qp\leftrightarrow q column with the final column: they agree in all 44 rows (T,F,F,T in both).

3. Conclusion. Since the truth tables are identical, p↔q≡((¬p)∨q)∧((¬q)∨p)p\leftrightarrow q\equiv((\lnot p)\vee q)\wedge((\lnot q)\vee p).

(b) Solve by Cramer's Rule

1. Substitute u=1x, v=1y, w=1zu=\dfrac1x,\ v=\dfrac1y,\ w=\dfrac1z to linearise:

3u−4v−2w=1,u+2v+w=2,2u−5v−4w=−13u-4v-2w=1,\qquad u+2v+w=2,\qquad2u-5v-4w=-1

2. Coefficient determinant.

D=∣3−4−21212−5−4∣=3[(2)(−4)−(1)(−5)]−(−4)[(1)(−4)−(1)(2)]+(−2)[(1)(−5)−(2)(2)]D=\begin{vmatrix}3&-4&-2\\1&2&1\\2&-5&-4\end{vmatrix}=3[(2)(-4)-(1)(-5)]-(-4)[(1)(-4)-(1)(2)]+(-2)[(1)(-5)-(2)(2)]

=3(−8+5)+4(−4−2)−2(−5−4)=3(−3)+4(−6)−2(−9)=−9−24+18=−15=3(-8+5)+4(-4-2)-2(-5-4)=3(-3)+4(-6)-2(-9)=-9-24+18=-15

3. DuD_u (replace column 1 with the RHS (1,2,−1)(1,2,-1)):

Du=∣1−4−2221−1−5−4∣=1(−8+5)+4(−8+1)−2(−10+2)=−3−28+16=−15D_u=\begin{vmatrix}1&-4&-2\\2&2&1\\-1&-5&-4\end{vmatrix}=1(-8+5)+4(-8+1)-2(-10+2)=-3-28+16=-15 …

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