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The condition a matrix has a inverse

This is the first question we ask about a square matrix: Is A invertible? We don’t mean that we immediately calculate $A^{-1}$. In most problems we never compute it ! Here are six “notes” about A-l.

1. The inverse exists if and only if elimination produces n pivots (row exchanges are allowed). Elimination solves $Ax = b$ without explicitly using the matrix $A^{-1}$.

2. The matrix A cannot have two different inverses. Suppose $BA = I$ and also $AC= I$. Then $B = C$

3. If A is invertible, the one and only solution to $Ax = b$ is $x = A^{-1} b$.

4. (Important) Suppose there is a nonzero vector x such that Ax = 0. Then A cannot have an inverse. No matrix can bring O back to x. If A is invertible, then Ax = 0 can only have the zero solution x = A- 10 = 0.

5. A 2 by 2 matrix is invertible if and only if ad - be is not zero: \(2 \text { by } 2 \text { Inverse: }\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]^{-1}=\frac{1}{a d-b c}\left[\begin{array}{rr} d & -b \\ -c & a \end{array}\right] \text {. }\) This number $ad - be$ is the determinant of $A$. A matrix is invertible if its determinant is not zero. The test for n pivots is usually decided before the determinant appears.

6. A diagonal matrix has an inverse provided no diagonal entries are zero : \(\text { If } \quad A=\left[\begin{array}{ccc} d_1 & & \\ & \ddots & \\ & & d_n \end{array}\right] \text { then } A^{-1}=\left[\begin{array}{ccc} 1 / d_1 & & \\ & \ddots & \\ & & 1 / d_n \end{array}\right] \text {. }\)

Gauss-Jordan

row exchanges are certainly allowed in Gauss-Jordan and doesn’t change the sigh.

3rf

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