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Properties of Determinants

If p≠ 0,q≠ ...

Question

If $p \neq 0, q \neq 0 a n d ∣ ∣ ∣ ∣ \begin{matrix} p & q & p α + q q & r & q α + r p α + q & q α + r & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0$ , then using properties of determinants prove that either p,q,r are in G.P. or $α$ is a root of the equation $p x^{2} + 2 q x + r = 0.$

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Solution

The given determinant is: $∣ ∣ ∣ ∣ \begin{matrix} p & q & p α + q q & r & q α + r p α + q & q α + r & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0$

Applying Row operations:
$R 1 \to α \times R 1 + R 2$
$∣ ∣ ∣ ∣ ∣ \begin{matrix} p α + q & q α + r & [α (p α + q) + (q α + r)] q & r & q α + r p α + q & q α + r & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$

$R 1 \to R 1 - R 3$
$∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & α (p α + q) + (q α + r) q & r & q α + r p α + q & q α + r & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$

Expanding the determinant:
$[α (p α + q) + (q α + r)] [q (q α + r) - r (p α + q)] = 0$
$[p α^{2} + 2 q α + r] [q^{2} α - p r α] = 0$

$∴,$ $p α^{2} + 2 q α + r = 0 \to 1$
$q^{2} - p r α = 0 \to 2$

Consider equation $1$ :
$α$ is the root of the equation $p x^{2} + 2 q x + r = 0$

Consider equation $2$ :
$q^{2} = p r ⟹ p, q, r$ are in $G . P$

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Similar questions

Q. Using properties of determinants, prove that

∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(a + b)^{2}}{c} & c & c a & \frac{(b + c)^{2}}{a} & a b & b & \frac{(c + a)^{2}}{b} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 2 (a + b + c)^{3}

OR

If

p \neq 0, q \neq 0

and

∣ ∣ ∣ ∣ \begin{matrix} p & q & p α + q q & r & p α + r p α + q & q α + r & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0

, then, using properties of determinants.
Prove that at least one of the following statements is true:

(a) p, q, r are in G.P.

(b)

α

is a root of the equation

p x^{2} + 2 q x + r = 0

Q. If

p, q, r

are in

G . P .

and the equations,

p x^{2} + 2 q x + r = 0

and

d x^{2} + 8 e x + f = 0

have a common root, then show that

\frac{d}{p}, \frac{e}{q}, \frac{f}{r}

are in

A . P .

If p, q and r are in GP and the equations $p x^{2} + 2 q x + r = 0$ and $d x^{2} + 2 e x + f = 0$ have a common root, show that $\frac{d}{p}, \frac{e}{q}$ and $\frac{f}{r}$ are in AP.

Q. Let

α

and

β

be the roots of equation

p x^{2} + q x + r = 0, p \neq 0

. If

p, q, r

are in A. P. and

\frac{1}{α} + \frac{1}{β} = 4

, then the value of

| α - β |

Q. Let

α, β

be the roots of the equation

a x^{2} + 2 b x + c = 0

and

γ, δ

be the roots of the equation

p x^{2} + 2 q x + r = 0

.
If

α, β, γ, δ

are in G.P., then-

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If p≠0,q≠0and∣∣ ∣∣pqp α+qqrq α+rp α+qq α+r0∣∣ ∣∣=0, then using properties of determinants prove that either p,q,r are in G.P. or α is a root of the equation px2+2qx+r=0.

If $p \neq 0, q \neq 0 a n d ∣ ∣ ∣ ∣ \begin{matrix} p & q & p α + q q & r & q α + r p α + q & q α + r & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0$ , then using properties of determinants prove that either p,q,r are in G.P. or $α$ is a root of the equation $p x^{2} + 2 q x + r = 0.$