Question

For an n x n matrix A, explain how to find each value: (a) The minor Mij of the entry aij: Take the determinant of the (n - 1) x (n - 1) matrix that is left after deleting the ith row and jth column. (b) The cofactor Cij of the entry aij: Cij = (-1)^(i+j) * Mij. (c) The determinant of A: IAI = a11C11 + a12C12 + ... + a1nC1n.

          For an n x n matrix A, explain how to find each value:
(a) The minor Mij of the entry aij: Take the determinant of the (n - 1) x (n - 1) matrix that is left after deleting the ith row and jth column.
(b) The cofactor Cij of the entry aij: Cij = (-1)^(i+j) * Mij.
(c) The determinant of A: IAI = a11C11 + a12C12 + ... + a1nC1n.
        
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Added by Margarita R.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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For an n x n matrix A, explain how to find each value: (a) The minor Mij of the entry aij: Take the determinant of the (n - 1) x (n - 1) matrix that is left after deleting the ith row and jth column. (b) The cofactor Cij of the entry aij: Cij = (-1)^(i+j) * Mij. (c) The determinant of A: IAI = a11C11 + a12C12 + ... + a1nC1n.
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For an n x n matrix A, explain how to find each value: (a) The minor Mij of the entry aij: Take the determinant of the (n - 1) x (n - 1) matrix that is left after deleting the ith row and jth column. (b) The cofactor Cij of the entry aij: Cij = (-1)^(i+j) * Mij. (c) The determinant of A: IAI = a11C11 + a12C12 + ... + a1nC1n.

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Theorem 2.2.2 An n % n matrix A is singular if and only if det(A) = 0 Proof The matrix A can be reduced to row echelon form with a finite number of row operations. Thus, U = E_{k}E_{k-1} ⋯ E_{1}A where U is in row echelon form and the E_{i}’s are all elementary matrices. It follows that det(U) = det(E_{k}E_{k-1} ⋯ E_{1}A) = det(E_{k}) det(E_{k-1}) ⋯ det(E_{1}) det(A) Since the determinants of the E_{i}’s are all nonzero, it follows that det(A) = 0 if and only if det(U) = 0. If A is singular, then U has a row consisting entirely of zeros, and hence det(U) = 0. If A is nonsingular, then U is triangular with 1’s along the diagonal and hence det(U) = 1. From the proof of Theorem 2.2.2, we can obtain a method for computing det(A). We reduce A to row echelon form. U = E_{k}E_{k-1} ⋯ E_{1}A If the last row of U consists entirely of zeros, A is singular and det(A) = 0. Otherwise, A is nonsingular and det(A) = [det(E_{k}) det(E_{k-1}) ⋯ det(E_{1})]^{-1} Actually, if A is nonsingular, it is simpler to reduce A to triangular form. This can be done using only row operations I and III. Thus, T = E_{m}E_{m-1} ⋯ E_{1}A and hence, det(A) = ± det(T) = ±t_{11}t_{22} ⋯ t_{nn} where the t_{ii}’s are the diagonal entries of T. The sign will be positive if row operation I has been used an even number of times and negative otherwise.

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Transcript

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00:01 First of all i'll tell how to find minor mij of a matrix so suppose we have a matrix a11 a1 2 a 1 3 and so on and here like a2 1 2 a 31 and so on it will be a 2 2 a 3 2 like this to find the minor suppose we want to find the minor of this value for this this we will exclude this column i will exclude this is j and this will be i here so we'll exclude i a through and jeth column and the remaining matrix that we are that we come up with we will find the determinant of that matrix so that will be the minor of this m i .j now for the co -factor matrix so the four factor we do c .i .j will be equal to minus 1 to the power i plus j mij.
01:16 When i plus j is even, the co -factor is equal to minor and when i plus j is on the co -factor is negative...
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