Answer :

Let

Given that Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_{i}→ R_{i} + kR_{j} or C_{i}→ C_{i} + kC_{j}.

Applying R_{1}→ R_{1} – R_{2}, we get

Applying R_{2}→ R_{2} – R_{3}, we get

Expanding the determinant along R_{1}, we have

Δ = (p – a)[(q – b)(r) – (b)(c – r)] – (b – q)[–a(c – r)]

⇒ Δ = r(p – a)(q – b) – b(p – a)(c – r) + a(b – q)(c – r)

∴ Δ = r(p – a)(q – b) + b(p – a)(r – c) + a(q – b)(r – c)

We have Δ = 0

⇒ r(p – a)(q – b) + b(p – a)(r – c) + a(q – b)(r – c) = 0

On dividing the equation with (p – a)(q – b)(r – c), we get

Thus,

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Solve the matrix equations:

RD Sharma - Volume 1

Using properties of determinants, prove the following:

Mathematics - Board Papers

Using properties of determinants, prove the following:

Mathematics - Board Papers

Using properties of determinants, prove the following:

Mathematics - Board Papers

Solve the matrix equations:

RD Sharma - Volume 1