Answer:
False, counterexample below
Step-by-step explanation:
Denote the unit vectors of R^3 by . Now consider
and
. We have that
. Also, the vector
is not a linear combination of
because any linear combination of these two vectors will have third coordinate zero, but v_3 has third coordinate 1 so they can't be equal.
However, the set is not linearly independent, because
is a non-trivial linear combination of these vectors that equals zero.
The (i,j) minor of a matrix A is the matrix Aij obtained by deleting row i and column j from A it is true.
The (i,j) minor of a matrix A is the matrix Aij obtained by deleting row i and column j from A. A determinant of an n×n matrix can be defined as a sum of multiples of determinants of (n−1)×(n−1) sub matrices.
This is done by deleting the row and column which the elements belong and then finding the determinant by considering the remaining elements. Then find the co factor of the elements. It is done by multiplying the minor of the element with -1i+j. If Mij is the minor, then co factor,
+
.
Each element in a square matrix has its own minor. The minor is the value of the determinant of the matrix that results from crossing out the row and column of the element .
Learn more about the minor of the matrix here:
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