I suggest that you check your definitions of determinant and of singular matrix.
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I worked out that you can get the determinant of a singular matrix like this.
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Alternatively, A-\ lambda _ i I is a singular matrix and its span is a plane.
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Then the singular matrix will turn into a non-singular matrix and you can take the determinant of the square matrix ! !!
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Then the singular matrix will turn into a non-singular matrix and you can take the determinant of the square matrix ! !!
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Note that this range does not include the values for \ pi, as this leads to a singular matrix, which can't be solved.
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Note that in " Car Robots ", there is not one singular Matrix, but multiple ones, each held by a high-ranking Autobot.
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Every real non-singular matrix can be uniquely factored as the product of an orthogonal matrix and a symmetric positive definite matrix, which is called a polar decomposition.
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If A is a singular matrix of rank k, then it admits an " LU " factorization if the first k leading principal minors are nonzero, although the converse is not true.
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It seems that what you mean by " Determinant of a Singular Matrix " has to be translated into : " norm of a non-square matrix ", which is something definitely not worth an exclamation mark.
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