Example 1:
Using matrix method solve the following systems of linear equations 2x - y + z = -3 3x - z = - 8 2x + 6y ..
Question 3
Question: Answer: = 5 + 10 + 10 + 5 + 1 = 31 = RH..
Question: Answer: = 5 + 10 + 10 + 5 + 1 = 31 = RH..Types of Matrices
Row Matrix - A matrix having only one row is called a row-matrix. For example: A[1 3 2 -2] is a row matrix of order 1 x..
Question 5
= 1 + 3 + 3 + 1 = 8 ways The total number of combinations (31)(15)(8) = 372..
Suggested answer:
a = 2, d = -1, e = 3..
Circular Permutations
Circular Permutations - When things are arranged in places along a line with first and last place, they form a linear permutation. So far we have dealt only with linear permutations. When things are arranged in places along a closed curve or a circle, in which any place may be regarded as the first..
Circular Permutations - When things are arranged in places along a line with first and last place, they form a linear permutation. So far we have dealt only with linear permutations. When things are arranged in places along a closed curve or a circle, in which any place may be regarded as the first..Proof:
The number of permutations of n different things taken r at a time is the same as the number of ways of filling n letters and r blank spaces, supposed to be arranged in a straight line as shown above. Each blank is accommodating only one letter. We may fill the first blank with any ..
The number of permutations of n different things taken r at a time is the same as the number of ways of filling n letters and r blank spaces, supposed to be arranged in a straight line as shown above. Each blank is accommodating only one letter. We may fill the first blank with any ..Harmonic Mean (H.M.)
If three quantities are in harmonic progression, then the middle quantity is called the harmonic mean between the other two. Example: 1/3, 1/7, 1/11 are in H.P., then 1/7 is the middle term. Hence 1/7 is the harmonic mean between 1/3 and h..
Examples:
Each one of the following series form an A.P. i) 1, 3, 5, 7, ii) 3, 7, 11, 15, iii) 15, 12, 9, iv) x, x - d, x - 2d, ..... The common difference is found by subtracting any term of the series from the immediate succeeding term. In the above example, com..
Each one of the following series form an A.P. i) 1, 3, 5, 7, ii) 3, 7, 11, 15, iii) 15, 12, 9, iv) x, x - d, x - 2d, ..... The common difference is found by subtracting any term of the series from the immediate succeeding term. In the above example, com..Note:
In the above method note that To obtain D 1 , replace a 1 , a 2 , a 3 by d 1 , d 2 , d 3 in D To obtain D 2 , replace b 1 , b 2 , b 3 by d 1 , d 2 , d 3 in D To obtain D 3 , replace c 1..
In the above method note that To obtain D 1 , replace a 1 , a 2 , a 3 by d 1 , d 2 , d 3 in D To obtain D 2 , replace b 1 , b 2 , b 3 by d 1 , d 2 , d 3 in D To obtain D 3 , replace c 1..See what our Users say :
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