Guest

I came across a surprising result. If A is any matrix of order (mxn) and B is any matrix of order (nxm), such that m>n. Then determinant of AB is always equal to zero. Can we prove this?

I came across a surprising result.
 
If A is any matrix of order (mxn) and B is any matrix of order (nxm), such that m>n.
 
Then determinant of AB is always equal to zero.
 
Can we prove this?

Grade:12

1 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
7 years ago
Hello student,
Please find the answer to your qustion below
Let us take A is 2X1 matrix and B is 1X2 matrix and take some general alphabets to prove this
Let A=\begin{bmatrix} i\\ j \end{bmatrix}and B=[k l]
So AXB=\begin{bmatrix} ik &il \\ jk &jl \end{bmatrix}
So |AB|=[ijkl-ijkl]=0

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free