Proving the Invertibility of Non-Singular Matrices

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A matrix is non-singular if its determinant is non-zero, which implies that it has an inverse. The discussion highlights the need to prove that a matrix is non-singular if and only if its determinant is non-zero. To find the inverse of a matrix, one can use the adjoint method, which requires the determinant to be non-zero. Additionally, if the determinant is non-zero, it ensures that the inverse exists since 1/det is defined. Therefore, demonstrating both conditions establishes the relationship between non-singularity and invertibility.
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a matrix is non singular only if its det does not equal zero. Calculate its inverse.

How do I go about proving this? I can only think of a counter example where matrix is singular given identical rows or columns or multiples of each other, which will generate a det of 0.

What do you think?
 
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Prove what? That a matrix is non-singular if and only if its determinant is non-zero? What is your definition of "non-singular". I suspect, from that addtional "Calculate its inverse" that "non-singular" is defined as "has an inverse" (or, more precisely, that "singular" is defined as "does not have an inverse" and "non-singular" is the reverse of that. Okay, how would you find the inverse of a matrix? Does the determinant come into that?
 
By non singular I mean a inverse exists. I believe the inverse is the adjoint/ det of the matrix. So the det can't be 0.
 
That's half way. You also need to show that if the determinant is non-zero then the matrix is invertible. Since if the determinant is non-zero, 1/det exists, all you need to do is show that "adjoint" always exists.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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