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What is the meaning of inverse matrices?
If, for an n*n matrix, A, there exists a matrix B such that AB = I, where I is the n*n identity matrix, then the matrix B is said to be the inverse of A. In that case, BA = I (in general, with matrices, AB ≠BA) I is an n*n matrix consisting of 1 on the principal diagonal and 0s elsewhere.
What are the next two consecutive odd integers if n is an odd integer?
If n is an odd integer then the next two consecutive odd integers are n+2 and n+4.
What is the formula for matrices?
A matrix is an array of elements in m rows and n columns.
If n plus 4 represents an odd integer the next larger number odd integer is represented by?
Every integer is either even (divisible by 2) or odd (not divisible by 2). Since an even number plus 1 is odd and an odd number plus one is even, because 1 does not divide 2. We know (n + 4) is odd. The next integer is (n + 4 + 1) = (n + 5), because an odd number plus 1 is even, (n + 5) is even. The integer after (n + 5) is (n + 6), since (n + 5) we know is even, (n + 6) must be odd. Since (n + 6) is the smallest integer that is greater than (n + 4) and is odd, so (n + 6) is the next odd integer.
What is the proof for if n cubed is odd then n squared plus one is even?
No even values of n will give give an odd value of n3, so n must be odd.When n is odd, n2 is also odd, so n2+1 must be even. ■
Definition of inverse matrix?
The inverse of a non-singular, n*n matrix, A Is another n*n matrix, A' such that A*A' = A'*A =I(n), the n*n identity matrix.Singular square matrices do not have inverses, nor do non-square matrices.
If n is an odd integer is n plus 6 odd as well?
Yes. If n is odd, then n + c where c is an even constant will be odd. n + d where d is an odd constant will be even.
What is the meaning of inverse matrices?
If, for an n*n matrix, A, there exists a matrix B such that AB = I, where I is the n*n identity matrix, then the matrix B is said to be the inverse of A. In that case, BA = I (in general, with matrices, AB ≠BA) I is an n*n matrix consisting of 1 on the principal diagonal and 0s elsewhere.
What are the next two consecutive odd integers if n is an odd integer?
If n is an odd integer then the next two consecutive odd integers are n+2 and n+4.
What is the formula for matrices?
A matrix is an array of elements in m rows and n columns.
Choose the true biconditional statement that can be formed from the conditional statement If a natural number n is odd then n2 is odd and its converse.?
An integer n is odd if and only if n^2 is odd.
If n plus 4 represents an odd integer the next larger number odd integer is represented by?
Every integer is either even (divisible by 2) or odd (not divisible by 2). Since an even number plus 1 is odd and an odd number plus one is even, because 1 does not divide 2. We know (n + 4) is odd. The next integer is (n + 4 + 1) = (n + 5), because an odd number plus 1 is even, (n + 5) is even. The integer after (n + 5) is (n + 6), since (n + 5) we know is even, (n + 6) must be odd. Since (n + 6) is the smallest integer that is greater than (n + 4) and is odd, so (n + 6) is the next odd integer.
What will be dimension of vector space of all symmetric matrices of order n x n with real entries and trace zero?
(n^2-n)/2-1
What is the proof for if n cubed is odd then n squared plus one is even?
No even values of n will give give an odd value of n3, so n must be odd.When n is odd, n2 is also odd, so n2+1 must be even. ■
C program to accept a sequence of number terminated by zero and sum all odd nos and even nos?
#include <stdio.h> main() { int n, odd=0, even=0; while (scanf("%d",&n)&&(n!=0)) (n%2)?++odd:++even; printf("odd: %d even: %d\n",odd,even); }
What has the author Elmer A Hoyer written?
Elmer A. Hoyer has written: 'Digital computer synthesis of admittance matrices of N+1 nodes' -- subject(s): Electric networks, Matrices, Computer programs
Can you prove n2 - n is even for any natural number n?
There are two cases here: one where n is even and one where n is odd. Let's consider the case where n is even: If n is even then n2 has to be even (since multiple of an even number must always be even.) In this case, we are subtracting an even number from and even number, the result must be even. This proves than n2 - n is even when n is even. Now let's consider the case where n is odd: If n is odd, then n2 must be odd. This is because an odd number times an odd number is always odd. (You could think if this as an odd number times and even number and then adding an odd number. For example, say that n is odd. n-1 is then even, and n2 = n(n-1) + n. n(n-1) must be even, since it is a multiple of an even number. And even number plus and odd number then has to be odd.) Now we know we have and odd number minus and odd number, which has to be even. So this proves that n2 - n is even when n is odd. Since we have proved that n2 - n is even for both when n is even and when n is odd, and there are no other cases, n2 - n must be even for any natural number n. or Let n be a natural number. Then n can be even or odd. We want to show that n2 - n = 2m where m is any positive nteger (by the def. of even). Case 1: Let n be even. Then n = 2k (def. of even), where k is any positive integer. Then, n2 - n = (2k)2 - (2k) = 2(2k2 - k); let 2k2 - k = m = 2m Therefore, n2 - n is even. Case 2: Let n be odd. Then n = 2k +1 (def. of odd), where k is any positive integer. Then, n2 - n = (2k - 1)2 - (2k - 1) = 4k2 - 4k + 1 - 2k + 1 = 4k2 - 6k + 2 = 2(2k2 - 3k + 1); let 2k2 -3k + 1 = m = 2m Therefore, n2 - n is even. Therefore, for any natural number n, n2 - n is even.
