Matrix multiplier
Let A be a 2 × 3 matrix and B be a 3 × 2 matrix. If A × B and B × A both defined and determinant of the matrix A B is 4, then evaluate the determinant of the matrix B A .
The answer is 0.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
3 solutions
Thanks sir for the solution .But would u explain further
Log in to reply
@Arka Dutta Where do you get these questions from???
Log in to reply
A book named Advanced problems in mathematics for JEE of Balaji pub.
Log in to reply
@Arka Dutta – Ooooh!!! Nice book!!! Btw, do you have any other books like this for Physics and Chemistry??? Coz I am also preparing for JEE.....
Log in to reply
@Aaghaz Mahajan – If u want very challenging problem book in physics. You can try Pathfinder for Olympiad and JEE Adv. Physics from Pearson. It's written by Sachin Singh ( HOD physics Allen Kota). And for chem. I am confused about that subject always.😥
Log in to reply
@Arka Dutta
–
Thanks a lot!!! Even I find Chemistry confusing....:P
But hey, if you wanna learn Organic Chemistry, then Solomons is a good read.....
Each of the three columns in B A will be a linear combination of the two columns of B , and so they span an at most two-dimensional space. For B A to be nonsingular, they would have to span the whole of R 3 . Thus B A is singular.
See David Vreken's solution below. The rank of a matrix can be at most the lesser of its two dimensions. The rank of the product of two matrices can be at most the lesser of the two factors' ranks. The product must be square. If a matrix has a rank less than its dimension and is square, then, it is singular (that is, can not be inverted). A singular matrix has a zero determinant. This is covered in linear algebra or matrix algebra courses fairly early in the course.
Log in to reply
David's proof, while accurate, does not give any insight into what is going on. The key fact is (the nontrivial result) that the row rank and column rank of a matrix are the same, which is what tells us that the concept of matrix rank is well-defined.
Log in to reply
Without any disagreement with you, the original issue in this thread of discussion is that the requester did not understand matrix rank. David Vreken's solution provides an illustration of what you were discussing. I guess that Arka Dutta is at the stage of solving multiple equations in multiple unknowns using Cramer's Rule. The discussion of matrix rank, although correct, was not at the level Arka Dutta was at.
Log in to reply
@A Former Brilliant Member – My reply to Arka expanded on the term of rank to an extent he seems content with - he was asking for an amplification of my proof, and not a different one.
While, with David's proof, there is evidence that backs up my argument (that the columns of B A are linear combinations of the columns of B ) that evidence is not obvious, unless you go looking for it.
Yeah it's in our course
Log in to reply
Are you studying in college right now???
Log in to reply
@Aaghaz Mahajan – Nope . Isn't rank of matrix included in jee adv. Syllabus!?
Log in to reply
@Arka Dutta – IDK.......In FIITJEE we haven't started Matrices.......Our class twelfth started just this week.......
Yeah. Surely I couldn't get completely of Mark Hennings sir's solution at first. But after going through concepts of rank and the amplification of the proof , I got that. David Vreken sir's proof is also very good. But the exam the question is asked for has limited time for each one. So it's must relying on concepts than a note down proof.
Let A = [ a d b e c f ] and B = ⎣ ⎡ g i k h j l ⎦ ⎤ .
Then irregardless of det A B ,
det ( B A )
= det ( ⎣ ⎡ g i k h j l ⎦ ⎤ [ a d b e c f ] )
= det ( ⎣ ⎡ a g + d h a i + d j a k + d l b g + e h b i + e j b k + e l c g + f h c i + f j c k + f l ⎦ ⎤ )
= ( a g + d h ) ( b c i k + b f i l + c e j k + e f j l − b c i k − c e i l − b f j k − e f j l ) − ( b g + e h ) ( a c i k + a f i l + c d j k + d f j l − a c i k − c d i l − a f j k − d f j l ) + ( c g + f h ) ( a b i k + a e i l + b d j k + d e j l − a b i k − b d i l − a e j k − d e j l )
= ( a g + d h ) ( b f i l + c e j k − c e i l − b f j k ) − ( b g + e h ) ( a f i l + c d j k − c d i l − a f j k ) + ( c g + f h ) ( a e i l + b d j k − b d i l − a e j k )
= a b f g i l + a c e g j k − a c e g i l − a b f g j k + b d f h i l + c d e h j k − c d e h i l − b d f h j k − a b f g i l − b c d g j k + b c d g i l + a b f g j k − a e f h i l − c d e h j k + c d e h i l + a e f h j k + a c e g i l + b c d g j k − b c d g i l − a c e g j k + a e f h i l + b d f h j k − b d f h i l − a e f h j k
= 0
Good solution sir.
you are so brave！
A n x m matrix always represents a linear transformation from R^m to R^n, an thus AB is a composition of two linear transformations, one from R^2 to R^3 (B) and then another from R^3 to R^2 (A). Notice that the composition is done from right to left since by associativity (AB)x = A(Bx). But this means BA is a transformation from R^3 to R^2 and then back to R^3, which can never be surjective (onto). We come then to the conclusion that det(BA) = 0. (:
Every column of B A is a linear combination of the two columns of B , and hence the (column) rank of B A is at most 2 . Thus we deduce that B A is singular.