Proving Determinant of Matrix Unchanged by Adding Columns

[Unusualskill]

Any1 can explain in an easy way?I know is something to do with the facts that row operations do not change the determinant of a matrix and also determinant of A=determinant of A transpose.But just dk how to prove in full sentence. Appreciate any1 help.

 

[homeomorphic]

I'll explain by example. Let's say u is some vector in R^2 and v is a multiple of it. Then suppose we form a 2 by 2 matrix (u, v) by taking u as the first column and v as the second. The determinant is 0 because the columns are linearly dependent. We can write it as
det(u, v) = 0 = det(v, u).

Now, suppose we have a matrix (v, w). We want to see what happens when we add a multiple of v to another column, like this: (v, w+u). We can use the fact that the determinant is linear in each variable.

det(v, w + u) = det(v, w) + det(v, u) = det(v, w) + 0 = det(v,w).

The same thing will happen if we look at n by n matrices because you're adding something that's going to be linearly dependent.

Geometrically, this is like the geometry theorem that you can shear a parallelogram without changing its area. It's always base times height, so if you don't change the height, it stays the same. Try to see how that relates to my argument by drawing the parallelogram spanned by some vectors v and w, then the one spanned by v and w+u. Remember that the meaning of the determinant is that it is the signed area of this parallelogram (or volume in higher dimensions).

 

[jbunniii]

Suppose $M$ is any $n \times n$ matrix, and $1 \leq i,j \leq n$ with $i \neq j$.

If $I$ is the $n \times n$ identity matrix and $J$ is the $n \times n$ matrix with a $1$ in row $i$, column $j$, and $0$ everywhere else, then define $A = I + J$. It's easy to check that $AM$ is the matrix formed by replacing the $i$'th row of $M$ with the sum of the $i$'th and $j$'th rows. Note that $A$ is either upper triangular (if $i > j$) or lower triangular (if $i < j$), so its determinant is the product of its diagonal elements, which are all $1$. Therefore $\det(A) = 1$. Consequently, $\det(AM) = \det(A)\det(M) = \det(M)$.