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Matrix Multiplication

Matrix Multiplication. =. A 32. A 22. The sign of (-1) i+j is given by. of the (i,j)th entry. Cofactor. (-1) i+j (determinant obtained by deleting the i th row and j th column of the matrix). A ij. A 11. A 12. A 13. next. The sign of (-1) i+j is given by. –. +.

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Matrix Multiplication

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  1. MatrixMultiplication =

  2. A32 A22 The sign of (-1)i+j is given by of the (i,j)th entry Cofactor (-1)i+j (determinant obtained by deleting the ith row and jth column of the matrix) Aij A11 A12 A13 next

  3. The sign of (-1)i+j is given by – + = a31A31+a32 A32+a33A33 3rd row first column = a11 A11+a21 A21+a31A31 second column = a12 A12+a22 A22+a32A32 third column = a13 A13+a23 A23+a33A33 Determinant +a21a32a13 +a31a12a23 +a11a22a33 direct expansion –a11a32a23 –a31a22a13 –a21a12a33 expandedalong = a11A11+a12 A12+a13A13 1st row = a21A21+a22 A22+a23A23 2nd row

  4. a11A21+a12A22+a13A23 detA 0 0 a21A11+a22A12+a23A13 0 detA 0 0 0 detA Inverse Proof Explanation Explanation =I. Similarly, it can be proved that A-1 hence found. END

  5. +a21a32a13 +a31a12a23 +a11a22a33 –a11a32a23 –a31a22a13 –a21a12a33 =a11(a22a33-a32a23) – a12(a21a33-a31a13) a13(a21a32-a31a22) + – + = a11A11 + a12 A12 a13A13 (expanded along first row) + – + The sign of (-1)i+j is given by Proof back

  6. a11 a12 a13 a11 a12 a13 Note that a21 A21+a22 A22 +a23A23= x y z xy z What about along 2nd row a11 A21+a12A22+a13A23? =0 along 1st row Proof2 a11 A11+a12 A12+a13A13= det AExpand along first row 0 a21 A21+a22 A22+a23A23= det AExpand along second row a31 A31+a32 A32+a33A33 = det AExpand along third row back to inverse a21A11+a22A12+a23A13?

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