Tuesday, 8 April 2014

linear algebra - If $B$ is an $(n times n)$ matrix, and the RREF of $A$, does $det(A) neq 0 Rightarrow det(B) neq 0$



Suppose I have a square matrix $A$ with $\det(A) \neq 0$. And suppose $B$ is the matrix $A$ but in row reduced echelon form. Does it follow that $\det(B) \neq 0$?



Now I am aware of the effects that row operations have on the determinant. However, the thing I am unsure about is multiplying a row by a scalar $\lambda$, which may be $0$ in general. Do we ever need to multiply by $0$ to get a matrix to RREF?



Answer



Check out this, it is the invertible matrix theorem. $A$ is invertible if and only if the determinant is not zero, which means $A$ is invertible in your case. By the Invertible matrix theorem, the RREF of an invertible matrix is the identity matrix. So in your case, $B$ is the identity matrix. So it follows that the determinant of $B$ is also not zero.



And for your other question, you can not multiply a row by zero when reducing to RREF. It changes the entire matrix.


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