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# 2.2 - 2.2 Inverse of a Matrix Denition square n n matrix A...

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2.2 Inverse of a Matrix Definition: square n × n matrix A is invertible if there is an n × n matrix C satisfying AC = I n and CA = I n . C is called an inverse of A If B is another inverse, then BA = AB = I n , so B = B ( AC ) = ( BA ) C = C . Therefore, the inverse is unique, and we write C = A 1 EX: Does A = bracketleftBigg 0 1 0 0 bracketrightBigg have an inverse? If so, its inverse C satisfies CA = AC = I = bracketleftBigg 1 0 0 1 bracketrightBigg , so A = A · I = A ( AC ) = ( A · A ) C = A 2 C . But A 2 = bracketleftBigg 0 1 0 0 bracketrightBiggbracketleftBigg 0 1 0 0 bracketrightBigg = bracketleftBigg 0 0 0 0 bracketrightBigg = O implies A = A 2 C = O , which is a contradiction. So A cannot have an inverse! Note: If A does not have an inverse, then we do NOT in general have these properties: · AB = O = A = O or B = O · AB = AC = B = C 1

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Definition: If n × n A is invertible, it is called a nonsingular matrix . If it is not invertible, it is called a singular matrix . A: When is 2 × 2 matrix A = bracketleftBigg a b c d bracketrightBigg invertible? Q: If ad bc negationslash = 0, and then its inverse is C = A 1 = 1 ad bc bracketleftBigg d b c a bracketrightBigg We’ll show AC = I : AC = bracketleftBigg a b c d bracketrightBiggparenleftBigg 1 ad bc bracketleftBigg d b c a bracketrightBiggparenrightBigg = 1 ad bc bracketleftBigg ad bc ba + ba dc dc bc + ad bracketrightBigg = 1 ad bc bracketleftBigg ad bc 0 0 ad bc bracketrightBigg = bracketleftBigg 1 0 0 1 bracketrightBigg = I Showing CA = I is the same, so C = A 1 EX: For A = bracketleftBigg 0 0 0 1 bracketrightBigg = bracketleftBigg a b c d bracketrightBigg , ad bc = 0 · 1 0 · 0 = 0, so A is not invertible (it is singular ).
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