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Two Determinants Product

Prove that the product of determinants of third order is again a determinant of the third order.

Now substituting the values of

The system of equations given by

Or

If

Thus we conclude that system of equation

Thus we see that the system of equations given by

Obviously

Let

Surely, both the determinants

Hence the product of two determinants of third order is again a determinant of third order.

**Proof:**Consider the system of three linear homogeneous equations.Now substituting the values of

**X**,**Y**and**Z**from**(2)**in the system of equations given by**(1)**and re-arranging, we getThe system of equations given by

**(3)**will hold only if the system of equations given by**(1)**holds. But the equations given by**(1)**will hold either ifOr

**X = 0**,**Y = 0**,**Z = 0**.If

**X = 0**,**Y = 0**,**Z = 0**, then we haveThus we conclude that system of equation

**(1)**holds for non-zero values of**x, y, z**when either**(4)**or**(5)**is true. If**(2)**holds for non-zero values of**X**,**Y**and**Z**, then we must haveThus we see that the system of equations given by

**(3)**holds if the system of equations given by**(1)**holds. But the system of equations given by**(1)**holds for non-zero values of**x, y**and**z**when either**(4)**or**(5)**holds. Hence we conclude that**(6)**holds if either**(4)**or**(5)**is true.Obviously

**Δ***must contain**Δ**and**Δ1**as factors.Let

**Δ* = k Δ. Δ**_{1}Surely, both the determinants

**Δ***and**k Δ Δ**_{1}are of six dimensions. By comparing the coefficients of the unique term we get**k = 1****∴ Δ* = Δ Δ**_{1}Hence the product of two determinants of third order is again a determinant of third order.

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