1.4.3 (ii) Cramer’s Rule

This rule can be applied only when the coefficient matrix is a square matrix and non-singular. It is explained by considering the following system of equations: a x a x a x a x a x a x a x a x a x = b , where the coefficient matrix is non-singular. Then ¹ . Let us put D = .

Then, we have x D = x a x a x a x a x a x a x a x a x a x a x a x a x = ∆ Since D ¹ , we get x = ∆ ∆ Similarly, we get x = where ∆ ∆ = ∆ ∆ ∆= ∆= b a b a b a Thus, we have the Cramer’s rule x = ∆ ∆ = ∆ ∆ = ∆ ∆ where D = b a b a ∆= b a b a b a b a ∆= ∆= Note Replacing the first column elements a of D with b b b respectively, we get D . Replacing the second column elements a of D with b b b respectively, we get D . Replacing the third column elements a of D with b b b respectively, we get D . If ∆= , Cramer’s rule cannot be applied.

Example . Solve, by Cramer’s rule, the system of equations First we evaluate the determinants ∆= ≠ ∆= , = , ∆= ∆= = . By Cramer’s rule, we get x = ∆ ∆ = = ∆ ∆ = −= − So, the solution is ( Example . In a T20 match, a team needed just runs to win with ball left to go in the last over.

The last ball was bowled and the batsman at the crease hit it high up. The ball traversed along a path in a vertical plane and the equation of the path is y ax bx with respect to a xy -coordinate system in the vertical plane and the ball traversed through the points ( , ),( ),( , can you conclude that the team won the match? Justify your answer. (All distances are measured in metres and the meeting point of the plane of the path with the farthest boundary line is ( , ).) The path y ax bx passes through the points ( , ),( ),( .

So, we get the system of equations . To apply Cramer’s rule, we find