Suppose x1, x2, x3 are real numbers such that x1 x2 x3 ≠ 0. Let Δ=| x1+a1 b1 a1 b2 a1 b3 a2 b1 x2+a2 b2 a2 b3 a3 b1 a3 b2 x3+a3 b3 | Then (Δ/x1 x2 x3)-1 equals:

Question

Suppose x1, x2, x3 are real numbers such that x1 x2 x3 ≠ 0. Let Δ=| x1+a1 b1 a1 b2 a1 b3 a2 b1 x2+a2 b2 a2 b3 a3 b1 a3 b2 x3+a3 b3 | Then (Δ/x1 x2 x3)-1 equals: (A) (a1 b1/x1)+(a2 b2/x2)+(a3 b3/x3) (B) -1 (C) (a1 a2 a3+b1 b2 b3/x1 x2 x3) (D) 0

Tardigrade Admin · Accepted Answer

Using the sum property, write Δ=x1 Δ1+b1 Δ2 where Δ1 =|1 a1 b2 a1 b3 0 x2+a2 b2 a2 b3 0 a3 b2 x3+a3 b3 | =(x2+a2 b2)(x3+a3 b3)-a3 b2 a2 b3 =x2 x3+x2 a3 b3+x3 a2 b2 Δ2 =|a1 a1 b2 a1 b3 a2 x2+a2 b2 a2 b3 a3 a3 b2 x3+a3 b3 | Using C2 arrow C2-b2 C1, C3 arrow C3-b3 C1, we get Δ2=|a1 0 0 a2 x2 0 a3 0 x3 |=a1 x2 x3 Thus, Δ =x1[x2 x3+x2 a3 b3+x3 a2 b2]+a1 b1 x2 x3 ⇒ (Δ/x1 x2 x3)-1 =(a1 b1/x1)+(a2 b2/x2)+(a3 b3/x3)

Q. Suppose $x_{1}, x_{2}, x_{3}$ are real numbers such that $x_{1} x_{2} x_{3} \neq = 0$ .
Let $Δ = ∣ ∣ x_{1} + a_{1} b_{1} a_{2} b_{1} a_{3} b_{1} a_{1} b_{2} x_{2} + a_{2} b_{2} a_{3} b_{2} a_{1} b_{3} a_{2} b_{3} x_{3} + a_{3} b_{3} ∣ ∣$
Then $\frac{Δ}{x _{1} x _{2} x _{3}} - 1$ equals:

$\frac{a _{1} b _{1}}{x _{1}} + \frac{a _{2} b _{2}}{x _{2}} + \frac{a _{3} b _{3}}{x _{3}}$

-1

$\frac{a _{1} a _{2} a _{3} + b _{1} b _{2} b _{3}}{x _{1} x _{2} x _{3}}$

0

Q. Suppose x1​,x2​,x3​ are real numbers such that x1​x2​x3​=0. Let Δ=∣∣​x1​+a1​b1​a2​b1​a3​b1​​a1​b2​x2​+a2​b2​a3​b2​​a1​b3​a2​b3​x3​+a3​b3​​∣∣​ Then x1​x2​x3​Δ​−1 equals:

x1​a1​b1​​+x2​a2​b2​​+x3​a3​b3​​

-1

x1​x2​x3​a1​a2​a3​+b1​b2​b3​​

0

$\frac{a _{1} b _{1}}{x _{1}} + \frac{a _{2} b _{2}}{x _{2}} + \frac{a _{3} b _{3}}{x _{3}}$

$\frac{a _{1} a _{2} a _{3} + b _{1} b _{2} b _{3}}{x _{1} x _{2} x _{3}}$