The Rank of a Matrix with a Parameter
Krystian Karczyński
Founder and General Manager of eTrapez.
Graduate of Mathematics at Poznan University of Technology. Mathematics tutor with many years of experience. Creator of the first eTrapez Courses, which have gained immense popularity among students nationwide.
He lives in Szczecin, Poland. He enjoys walks in the woods, beaches and kayaking.
Let’s calculate the rank of the matrix:
Solution
There are a few ways to approach this, and probably the quickest way is to multiply the fifth column by -1 and add it to the first, second, third, and fourth columns, resulting in:
Now, let’s take the determinant of the 4th-degree matrix:
A determinant where all elements except the main diagonal are zero is equal to the product of the elements on the main diagonal (I will prove this someday 🙂 ), so:
This determinant is non-zero for all a not equal to 1. Therefore, for such a, the rank of our matrix, which we need to calculate, is 4 (because a non-zero 4th-degree minor can be extracted from it, and a larger one cannot).
What about the case where 
. We then get the rank of the matrix (substitute 1 for a):
And this rank is 1 (you can, for example, again use the fifth column on the others and eliminate the zero columns).
So, for a not equal to 1, the rank of the matrix is 4, and for a equal to 1, the rank of the matrix is 1.
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