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Determinants of Square Matrices Computing (Hyper-)Volumes and Testing Invertibility Introduction Defining Determinants for n × n Matrices Computing Determinants Determinants of Square Matrices Computing (Hyper-)Volumes and Testing Invertibility A. Havens Department of Mathematics University of Massachusetts, Amherst February 28- March 3, 2018 A. Havens Determinants of Square Matrices Introduction Defining Determinants for n × n Matrices Computing Determinants Outline 1 Introduction Determinants for 2 × 2 Matrices and Areas Determinants for 3 × 3 Matrices and Volumes 2 Defining Determinants for n × n Matrices Treating Columns as Inputs: Multilinear Functions A Theorem and a Definition Properties from the Definition 3 Computing Determinants A Recursive Definition of Determinants Row Reduction and Elementary Matrices Volumes and Cramer's Rule A. Havens Determinants of Square Matrices Introduction Defining Determinants for n × n Matrices Computing Determinants Determinants for 2 × 2 Matrices and Areas The 2 × 2 Determinant Recall that for a 2 × 2 matrix ñ a a ô A = 11 12 a21 a22 the quantity ∆ = a11a22 − a12a21 determines if A is invertible: A−1 exists () ∆ 6= 0 : Definition The quantity det(A) := ∆(A) = a11a22 − a12a21 is called the determinant of the 2 × 2 matrix A. For the following discussion we consider an arbitrary real 2 × 2 matrix A with determinant det A = ∆. A. Havens Determinants of Square Matrices Introduction Defining Determinants for n × n Matrices Computing Determinants Determinants for 2 × 2 Matrices and Areas The 2 × 2 Determinant as Area Geometrically we can interpret the meaning of ∆ as follows. If î ó A = a1 a2 then ∆ is the signed area of the parallelogram with vertices whose positions are given by 0, a1, a2, and a1 + a2. We call this the parallelogram spanned by the two vectors a1 and a2. The sign will be determined by the spatial relationship of a1 to a2. We will explain this below via a notion of orientations. We can prove the relation to area using elementary trigonometry. A. Havens Determinants of Square Matrices Introduction Defining Determinants for n × n Matrices Computing Determinants Determinants for 2 × 2 Matrices and Areas Some Elementary Geometry Assume for the moment that 0 ≤ α < β ≤ π=2 so the vectors lie in the first quadrant, and consider the diagram. A. Havens Determinants of Square Matrices A = b · h = ka1kka2k sin(β − α) Ä ä = ka1kka2k sin(β) cos(α) − cos(β) sin(α) = ka1kka2k sin(β) cos(α) − ka1kka2k cos(β) sin(α) Ä äÄ ä Ä äÄ ä = ka1k sin(β) ka2k cos(α) − ka2k cos(β) ka1k sin(α) = a11a22 − a12a21 = ∆ : It is left as an exercise to check that one still gets ±∆ as the area if the vectors are not in the first quadrant. When will the determinant be the negative of the area? Introduction Defining Determinants for n × n Matrices Computing Determinants Determinants for 2 × 2 Matrices and Areas Some Trigonometry The length of the base of the parallelogram can be taken to be b = ka1k, and the length of the altitude is h = ka2k sin(β − α). The area A of the parallelogram is related to the determinant ∆ via the sine angle subtraction identity: A. Havens Determinants of Square Matrices = ka1kka2k sin(β − α) Ä ä = ka1kka2k sin(β) cos(α) − cos(β) sin(α) = ka1kka2k sin(β) cos(α) − ka1kka2k cos(β) sin(α) Ä äÄ ä Ä äÄ ä = ka1k sin(β) ka2k cos(α) − ka2k cos(β) ka1k sin(α) = a11a22 − a12a21 = ∆ : It is left as an exercise to check that one still gets ±∆ as the area if the vectors are not in the first quadrant. When will the determinant be the negative of the area? Introduction Defining Determinants for n × n Matrices Computing Determinants Determinants for 2 × 2 Matrices and Areas Some Trigonometry The length of the base of the parallelogram can be taken to be b = ka1k, and the length of the altitude is h = ka2k sin(β − α). The area A of the parallelogram is related to the determinant ∆ via the sine angle subtraction identity: A = b · h A. Havens Determinants of Square Matrices Ä ä = ka1kka2k sin(β) cos(α) − cos(β) sin(α) = ka1kka2k sin(β) cos(α) − ka1kka2k cos(β) sin(α) Ä äÄ ä Ä äÄ ä = ka1k sin(β) ka2k cos(α) − ka2k cos(β) ka1k sin(α) = a11a22 − a12a21 = ∆ : It is left as an exercise to check that one still gets ±∆ as the area if the vectors are not in the first quadrant. When will the determinant be the negative of the area? Introduction Defining Determinants for n × n Matrices Computing Determinants Determinants for 2 × 2 Matrices and Areas Some Trigonometry The length of the base of the parallelogram can be taken to be b = ka1k, and the length of the altitude is h = ka2k sin(β − α). The area A of the parallelogram is related to the determinant ∆ via the sine angle subtraction identity: A = b · h = ka1kka2k sin(β − α) A. Havens Determinants of Square Matrices = ka1kka2k sin(β) cos(α) − ka1kka2k cos(β) sin(α) Ä äÄ ä Ä äÄ ä = ka1k sin(β) ka2k cos(α) − ka2k cos(β) ka1k sin(α) = a11a22 − a12a21 = ∆ : It is left as an exercise to check that one still gets ±∆ as the area if the vectors are not in the first quadrant. When will the determinant be the negative of the area? Introduction Defining Determinants for n × n Matrices Computing Determinants Determinants for 2 × 2 Matrices and Areas Some Trigonometry The length of the base of the parallelogram can be taken to be b = ka1k, and the length of the altitude is h = ka2k sin(β − α). The area A of the parallelogram is related to the determinant ∆ via the sine angle subtraction identity: A = b · h = ka1kka2k sin(β − α) Ä ä = ka1kka2k sin(β) cos(α) − cos(β) sin(α) A. Havens Determinants of Square Matrices Ä äÄ ä Ä äÄ ä = ka1k sin(β) ka2k cos(α) − ka2k cos(β) ka1k sin(α) = a11a22 − a12a21 = ∆ : It is left as an exercise to check that one still gets ±∆ as the area if the vectors are not in the first quadrant. When will the determinant be the negative of the area? Introduction Defining Determinants for n × n Matrices Computing Determinants Determinants for 2 × 2 Matrices and Areas Some Trigonometry The length of the base of the parallelogram can be taken to be b = ka1k, and the length of the altitude is h = ka2k sin(β − α). The area A of the parallelogram is related to the determinant ∆ via the sine angle subtraction identity: A = b · h = ka1kka2k sin(β − α) Ä ä = ka1kka2k sin(β) cos(α) − cos(β) sin(α) = ka1kka2k sin(β) cos(α) − ka1kka2k cos(β) sin(α) A. Havens Determinants of Square Matrices = a11a22 − a12a21 = ∆ : It is left as an exercise to check that one still gets ±∆ as the area if the vectors are not in the first quadrant. When will the determinant be the negative of the area? Introduction Defining Determinants for n × n Matrices Computing Determinants Determinants for 2 × 2 Matrices and Areas Some Trigonometry The length of the base of the parallelogram can be taken to be b = ka1k, and the length of the altitude is h = ka2k sin(β − α). The area A of the parallelogram is related to the determinant ∆ via the sine angle subtraction identity: A = b · h = ka1kka2k sin(β − α) Ä ä = ka1kka2k sin(β) cos(α) − cos(β) sin(α) = ka1kka2k sin(β) cos(α) − ka1kka2k cos(β) sin(α) Ä äÄ ä Ä äÄ ä = ka1k sin(β) ka2k cos(α) − ka2k cos(β) ka1k sin(α) A. Havens Determinants of Square Matrices It is left as an exercise to check that one still gets ±∆ as the area if the vectors are not in the first quadrant. When will the determinant be the negative of the area? Introduction Defining Determinants for n × n Matrices Computing Determinants Determinants for 2 × 2 Matrices and Areas Some Trigonometry The length of the base of the parallelogram can be taken to be b = ka1k, and the length of the altitude is h = ka2k sin(β − α). The area A of the parallelogram is related to the determinant ∆ via the sine angle subtraction identity: A = b · h = ka1kka2k sin(β − α) Ä ä = ka1kka2k sin(β) cos(α) − cos(β) sin(α) = ka1kka2k sin(β) cos(α) − ka1kka2k cos(β) sin(α) Ä äÄ ä Ä äÄ ä = ka1k sin(β) ka2k cos(α) − ka2k cos(β) ka1k sin(α) = a11a22 − a12a21 = ∆ : A. Havens Determinants of Square Matrices Introduction Defining Determinants for n × n Matrices Computing Determinants Determinants for 2 × 2 Matrices and Areas Some Trigonometry The length of the base of the parallelogram can be taken to be b = ka1k, and the length of the altitude is h = ka2k sin(β − α). The area A of the parallelogram is related to the determinant ∆ via the sine angle subtraction identity: A = b · h = ka1kka2k sin(β − α) Ä ä = ka1kka2k sin(β) cos(α) − cos(β) sin(α) = ka1kka2k sin(β) cos(α) − ka1kka2k cos(β) sin(α) Ä äÄ ä Ä äÄ ä = ka1k sin(β) ka2k cos(α) − ka2k cos(β) ka1k sin(α) = a11a22 − a12a21 = ∆ : It is left as an exercise to check that one still gets ±∆ as the area if the vectors are not in the first quadrant. When will the determinant be the negative of the area? A. Havens Determinants of Square Matrices Introduction Defining Determinants for n × n Matrices Computing Determinants Determinants for 2 × 2 Matrices and Areas Orientations in the Plane Note that if instead 0 ≤ β < α ≤ π=2, the sign is reversed (as sin(β − α) < 0), whence, if a1 would rotate clockwise through the parallelogram towards a2 instead of counterclockwise, ∆ would return the negative of the area.
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