What is the Expansion by Minors in Linear Algebra?

[Bob19]

if one is presented with an n x n where n = 4.

What is the rule called which allows the i'th row and the j'th column to be removed from the matrix in order to make calculating the determinant easier ?

Secondly if one wants to square $$\left[ \begin{array}{cccc} 2 & 5 & 7 & 6 \\ 2 & 9 & 2 & 1 \\ 0 & 1 & -2 & 1 \\ 6 & 7 & 1 & -5\end{array}\right ] ^2$$

Do I square every element of matrix individually ?

/Bob

 

[HallsofIvy]

if one is presented with an n x n where n = 4.

What is the rule called which allows the i'th row and the j'th column to be removed from the matrix in order to make calculating the determinant easier ?

What you are talking about is "expansion by minors". The (n-1) x(n-1) matrix you get by removing the "i'th row and j'th column is the "minor" at that point. Choose anyone row or column, calculate the minor for each element in that row or column. The determinant is the sum of the product of the element itself times its minor times either plus or minus one, depending on whether i+j is even or odd.

Secondly if one wants to square $$\left[ \begin{array}{cccc} 2 & 5 & 7 & 6 \\ 2 & 9 & 2 & 1 \\ 0 & 1 & -2 & 1 \\ 6 & 7 & 1 & -5\end{array}\right ] ^2$$

Do I square every element of matrix individually ?

/Bob

No! squaring a matrix means multiplying the matrix by itself, not the individual elements. You won't be able to square a matrix if you don't know how to multiply two matrices. The simplest way to remember that is to think of each row of the first matrix and each column of the second as "vectors". The i, j element of the product is the dot product of the ith row of the first matrix and the jth column of the second.

 

[Gokul43201]

I think Bob may have been asking about row/column transformations to simplify the evaluation of a determinant.

 

[HallsofIvy]

I don't. He specifically said "which allows the i'th row and the j'th column to be removed from the matrix". That's the "expansion by minors", not "row reduction".