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Four.I Definition of Determinant Linear Algebra Jim Hefferon
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Properties of Determinants
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Nonsingular matrices An n × n matrix T is nonsingular if and only if each of these holds: I any system T ~ x = ~ b has a solution and that solution is unique; I Gauss-Jordan reduction of T yields an identity matrix; I the rows of T form a linearly independent set; I the columns of T form a linearly independent set, a basis for R n ; I any map that T represents is an isomorphism; I an inverse matrix T - 1 exists. This chapter develops a formula to determine whether a matrix is nonsingular.
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Determining nonsingularity is trivial for 1 × 1 matrices. ( a ) is nonsingular iff a 6 = 0 Corollary Three.IV.4.11 gives the 2 × 2 formula. a b c d is nonsingular iff ad - bc 6 = 0 We can produce the 3 × 3 formula as we did the prior one, although the computation is intricate (see Exercise 10 ). a b c d e f g h i is nonsingular iff aei + bfg + cdh - hfa - idb - gec 6 = 0 With these cases in mind, we posit a family of formulas: a , ad - bc , etc. For each n the formula defines a determinant function det n × n : M n × n R such that an n × n matrix T is nonsingular if and only if det n × n ( T ) 6 = 0 .
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Remark: how we will proceed The prior slide gives a formula for the 1 × 1 , 2 × 2 , and 3 × 3 determinants. We can give a formula for the general n × n case, the permutation expansion. However, this formula has a number of disadvantages, including its complexity and that it is too slow for practical computations. Instead, we will define a determinant function as one that satisfies some conditions. These conditions let us compute the determinant via Gauss’s Method, which we know to be fast and easy. (The conditions extrapolate from the 1 × 1 , 2 × 2 , and 3 × 3 cases; see the discussion in the book.) Defining a function by giving a list of conditions it must satisfy is common in more advanced courses. But it has the downside that we must show that there is one and only one function satisfying those conditions. We will first give an algorithm that, following the conditions, goes from any square matrix input to a real number output. Later we will develop the formula for the determinant, which will show that the determinant is well-defined (that is, for any input there is exactly one associated output).
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Definition of determinant 2.1 Definition A n × n determinant is a function det : M n × n R such that 1) det ( ~ ρ 1 , . . . , k · ~ ρ i + ~ ρ j , . . . , ~ ρ n ) = det ( ~ ρ 1 , . . . , ~ ρ j , . . . , ~ ρ n ) for i 6 = j 2) det ( ~ ρ 1 , . . . , ~ ρ j , . . . , ~ ρ i , . . . , ~ ρ n ) = - det ( ~ ρ 1 , . . . , ~ ρ i , . . . , ~ ρ j , . . . , ~ ρ n ) for i 6 = j 3) det ( ~ ρ 1 , . . . , k ~ ρ i , . . . , ~ ρ n ) = k · det ( ~ ρ 1 , . . . , ~ ρ i , . . . , ~ ρ n ) for any scalar k 4) det ( I ) = 1 where I is an identity matrix (the ~ ρ ’s are the rows of the matrix). We often write | T | for det ( T ) .
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Definition of determinant 2.1 Definition A n × n determinant is a function det : M n × n R such that 1) det ( ~ ρ 1 , . . . , k · ~ ρ i + ~ ρ j , . . . , ~ ρ n ) = det ( ~ ρ 1 , . . . , ~ ρ j , . . . , ~ ρ n ) for i 6 = j 2) det ( ~ ρ 1 , . . . , ~ ρ j , . . . , ~ ρ i , . . . , ~ ρ n ) = - det ( ~ ρ 1 , . . . , ~ ρ i , . . . , ~ ρ j , . . . , ~ ρ n ) for i 6 = j 3) det ( ~ ρ 1 , . . . , k ~ ρ i , . . . , ~ ρ n ) = k · det ( ~ ρ 1 , . . . , ~ ρ i , . . . , ~ ρ n ) for any scalar k 4) det ( I ) = 1 where I is an identity matrix (the ~ ρ ’s are the rows of the matrix). We often write
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