Question

if a1 a2 b1 and b2 take values in the set 1,-1,0 ​

Answer 1

Step-by-step explanation:

Theorem 4.1: A set containing a single non zero vector is linearly ... cn = 0. So, c1α1 + c2α2 + ......ck-1αk-1 + ckαk + 0αk+1 + ...... + 0αk = 0 ... Solution: Let α1 = (a1,a2) and α2 = (b1, b2).

Answer 2

The Actual Question is :

If a1, a2, b1 and b2 take values in the set {1, –1, 0}, then the probability that the equation a1a2 = b1b2 is satisfied is p , q (p & q are co primes).

Answer:

Considering, a1, b1, a2, b2 are taken in the set values inside the set {1,-1,0}

then the probability of the equation a1a2 = b1b2 will then be satisfied in the form of p/q, where p and q are co primes

and the resultant will be q - 2p :

a1a2 will have 3² = 9 in which equal probability outcomes will be there {1,- 1, 0, -1, 1, 0, 0, 0, 0}

so, count of 1 is two

count of -1 is two

count of 0 is five

b1b2 also has these similar outcomes.

whereas, a1a2 and b1b2 will be independent of each other.

So, the probability that both a1a2 and b1b2 is equal to $(\frac{2}{9})^2$ = $\frac{4}{81}$

the probability for 0 and -1 is equal to $(\frac{2}{9}) ^{2}$ = $\frac{4}{81\\}$ and $(\frac{5}{9}) ^2$ = $\frac{25}{81}$

therefore, the total probability will be  $\frac{4}{81}$ + $\frac{4}{81}$ + $\frac{25}{81}$ = $\frac{33}{81}$

where, 33 and 81 are both co primes

so the required probability will be q - 2p = 81 - 2x33 = 15.

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