inverse of a matrix + determinant

JamesGoh

To determine if a matrix is invertible or not, can we determine this by seeing if the determinant of the matrix is zero or non-zero ?

If it's zero, then the matrix doesn't exist because the inverse of the determinant would be an infinite number ?

AlephZero

Quote by JamesGoh

To determine if a matrix is invertible or not, can we determine this by seeing if the determinant of the matrix is zero or non-zero ?

Yes.

If it's zero, then the matrix doesn't exist because the inverse of the determinant would be an infinite number ?

I think "matrix" was a typo for "inverse matrix".

You need to be careful about how you use the word "infinite" in mathematics, but your basic idea is right.

HallsofIvy

A reason for that is that the inverse of an invertible matrix is the matrix in which the "ij" term is the minor of the "ji" term of the original matrix divided by the determinant. Given an "ji" term, it has a minor so the only problem is that we cannot divide by 0.

Bacle

Another way of looking at it (for nxn-matrices):

Det(AB)=DetADetB;

In our case, say we have an inverse A' for A, then:

AA'=I , so that,

Det(AA')=DetADetA'=DetI=1,

So you need two numbers whose product equals one, and that rules out DetA=0.

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JamesGoh	inverse of a matrix + determinant Tweet To determine if a matrix is invertible or not, can we determine this by seeing if the determinant of the matrix is zero or non-zero ? If it's zero, then the matrix doesn't exist because the inverse of the determinant would be an infinite number ?

HallsofIvy Recognitions: Gold Member Science Advisor Staff Emeritus	A reason for that is that the inverse of an invertible matrix is the matrix in which the "ij" term is the minor of the "ji" term of the original matrix divided by the determinant. Given an "ji" term, it has a minor so the only problem is that we cannot divide by 0.

Bacle	inverse of a matrix + determinant Another way of looking at it (for nxn-matrices): Det(AB)=DetADetB; In our case, say we have an inverse A' for A, then: AA'=I , so that, Det(AA')=DetADetA'=DetI=1, So you need two numbers whose product equals one, and that rules out DetA=0.