Question

. Let A be an n × n matrix and suppose that two of the rows of A are equal. Use cofactor expansion to prove that det A = 0. Hint: Start small! Prove that this is true for 2×2 matrices, then use the 2 × 2 result to do the 3 × 3 case, etc. More generally, assume it works for (n − 1) × (n − 1) matrices, then prove the n × n case.1

Answer 1

Proof by mathematical induction

Explanation:

A. Case n=2:

Let A be a matrix 2x2 with two equal rows:

`A=\left[\begin{array}{cc}x_(1)&x_(2)\\x_(3)&x_(4)\end{array}\right] =\left[\begin{array}{cc}a&b\\a&b\end{array}\right]`

By definition of Determinant and uniqueness of it:

`Det(A)=x_(1)x_(4)-x_(2)x_(3)=ab-ab=0`

B. Case n=3:

Let A be a matrix 3x3 with two equal rows. without loss of generality, assume the first 2 rows are equal:

`\left[\begin{array}{ccc}a&b&c\\a&b&c\\d&e&f\end{array}\right]`

Then, Using the cofactor expansion, and the fact that no matter what row or column you choose the cofactor expansion gives an unique determinant:

`Det(A)=d(\left[\begin{array}{cc}b&c\\b&c\end{array}\right])-e(\left[\begin{array}{cc}a&c\\a&c\end{array}\right] )+f(\left[\begin{array}{cc}a&b\\a&b\end{array}\right])`

Notice choosing at first the row that does not equal to none of the others, lead the cofactors to be 2x2 matrices with two equal rows (this will always happen)

`Det(A)=d(bc-bc)-e(ac-ac)+f(ab-ab)=0`

That proves what we want because you can always choose to expand starting with the row that´s different if there´s any

C. Hypothesis of Induction:

If A is a matrix (n-1)x(n-1) and it has 2 equal rows, then Det(A)=0

D. Case n:

Let A be a matrix nxn that has two equal rows, then (as we´ve shown earlier) we can always choose a row different of the ones that are equal and then, start expanding from it.

Notice that, if A nxn has two equal rows, and we start from one different row from the others, the matrices (n-1)x(n-1) that results from the cofactor expansion, have two equal rows as well (as we seen in the Case n=3).

That happen because we start expanding from a different row of those equal, we are taking out a column and a row that contains the cofactor. The row that is removed, does not have any issue because we choose it as above (different of the others).

The column that is taken away, has one value of every row in that column but if two rows are equal then the two values that are in the removed column are equal and the elements left in those rows are also equal. So the matrices that results from a cofactor expansion described as before, they have two equal rows as well

Because our Hypothesis tell us whenever we have an (n-1)x(n-1) with two equal rows its determinant is 0, we conclude that the determinant for A must be 0