Demonstrate that a matrix that has a null row is not invertible

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Homework Statement


Demonstrate that a matrix that has a null row is not invertible.

2. The attempt at a solution I know that saying that a matrix A is invertible is equivalent to say that A is row-equivalent and column-equivalent to the identity matrix.
And also that there exist elemental matrices E_1, E_2,...,E_r and F_1,...,F_s such that the identity matrix is equal to E_1...E_r A F_1...F_s.
So I could show that if A has at least a null row, then one of these properties cannot be true.
But still I don't know how to proceed. I understand clearly that if A has a null row, it's obvious that it cannot be equivalent to the identity matrix... but to prove it formally, I don't know how I can start. A little help is welcome.
 
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Suppose it did have an inverse. What does its product with its inverse supposed to look like? How does it actually look?
 
Think of A as the coefficient matrix such that Ax= b and let xn be the component of x corresponding to the 0 row. If the corresponding component of b is not 0, there is NO value of xn that makes that true. If the corresponding component of b is 0, then any value of xn makes that true. In either case there is NOT single solution to Ax= b. If A-1 exists, then A-1b would be 'the' solution to Ax= b.
 
Hurkyl said:
Suppose it did have an inverse. What does its product with its inverse supposed to look like? How does it actually look?

It is supposed to look like the identity matrix, however if the matrix has a null row it means that the product of this matrix with its inverse has a null column which means it's not the identity matrix, so that in fact any matrix having a null row doesn't have an inverse, hence is not invertible.
Is it a good proof? Or too informal?
 
fluidistic said:
Is it a good proof? Or too informal?
I can't answer that one; it would depend on what your teacher expects.
 
Hurkyl said:
I can't answer that one; it would depend on what your teacher expects.
Ok I understand. I hope it's enough.
And thank you very much HallsOfIvy, I got what you mean... nice way to do the proof.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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