show that the matrix is non singular if and only if this condition is met

**Jskid** · February 5th 2011, 01:40 PM

Show that the matrix $A=\left[\begin{matrix}a&b\\c&d\end{matrix}\right]$ is nonsingular if and only if ad-bc≠0. If this condition holds, show that $\left[\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right]$

I think what I want to do is show that $\left[\begin{matrix}a&b\\c&d\end{matrix}\right]$ is row equivalent to $I_2$ . When I was doing this I got stuck at $\left[\begin{matrix}ca&da\\0&1\end{matrix}\right]$

**Tinyboss** · February 5th 2011, 07:48 PM

Showing that the inverse is given by that particular matrix (with ad-bc in the denominators) gives you one direction. To do the other, suppose ad-bc=0 and try multiplying the top row by d and the bottom row by b.

**HallsofIvy** · February 6th 2011, 04:51 AM

Show that $\left[\begin{matrix}a&b\\c&d\end{matrix}\right]\left[\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right]= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$ .

Since the inverse of a matrix is unique, it follows that the denominator of that fraction cannot be 0.

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