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Consider a lower triangular matrix. When this lower triangular matrix is stored in array format then only the elements a[i][j] with i≥j are stored in array i.e only the elements present in lower triangular matrix are stored. Hence less size is consumed to store the array. Consider a lower triangular matrix as [25----100 , 25----100] with base address as 1000 and size of each element in matrix is 10, If the array is stored in column major order then find the address of the element a [80][45] stored in array ________________.
Correct answer is '15210'. Can you explain this answer?
Verified Answer
Consider a lower triangular matrix. When this lower triangular matrix...
In general , if the array is a [lb1---ub1 , lb2----ub2]
Base address = BA
Size of element = c
Number of rows = nr = ub1- lb1 + 1
Number of columns = nc = ub2 - lb2 + 1
and lower triangular matrix stored in column major order then
=
So,
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Consider a lower triangular matrix. When this lower triangular matrix...
Given:
- Lower triangular matrix
- Matrix stored in array format
- Only elements with i≥j are stored
- Base address: 1000
- Size of each element: 10
- Matrix: [25----100 , 25----100]

To find:
Address of a[80][45] in the array stored in column major order

Solution:
Step 1: Calculate the number of elements in each row before a[80][45]
- The number of elements in each row before a[80][45] is given by the sum of the arithmetic series from 1 to 79 (excluding 80).
- Number of elements in each row = 1 + 2 + 3 + ... + 79
- Using the formula for the sum of an arithmetic series: Sn = (n/2)(a + l)
- n = number of terms = 79
- a = first term = 1
- l = last term = 79
- Number of elements in each row = (79/2)(1 + 79) = 3160

Step 2: Calculate the number of elements in each column before column 45
- The number of elements in each column before column 45 is given by the sum of the arithmetic series from 1 to 44 (excluding 45).
- Number of elements in each column = 1 + 2 + 3 + ... + 44
- Using the formula for the sum of an arithmetic series: Sn = (n/2)(a + l)
- n = number of terms = 44
- a = first term = 1
- l = last term = 44
- Number of elements in each column = (44/2)(1 + 44) = 990

Step 3: Calculate the address of a[80][45] in the array
- The address of a[80][45] can be calculated using the formula:
- Address = Base address + (Number of elements in each column * Size of each element) + (Number of elements in each row * Size of each element)
- Base address = 1000
- Number of elements in each column = 990
- Size of each element = 10
- Number of elements in each row = 3160
- Address = 1000 + (990 * 10) + (3160 * 10) = 1000 + 9900 + 31600 = 42100

Answer:
The address of a[80][45] stored in the array is 42100.
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Consider a lower triangular matrix. When this lower triangular matrix is stored in array format then only the elements a[i][j] with i≥j are stored in array i.e only the elements present in lower triangular matrix are stored. Hence less size is consumed to store the array. Consider a lower triangular matrix as [25----100 , 25----100] with base address as 1000 and size of each element in matrix is 10, If the array is stored in column major order then find the address of the element a [80][45] stored in array ________________.Correct answer is '15210'. Can you explain this answer?
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Consider a lower triangular matrix. When this lower triangular matrix is stored in array format then only the elements a[i][j] with i≥j are stored in array i.e only the elements present in lower triangular matrix are stored. Hence less size is consumed to store the array. Consider a lower triangular matrix as [25----100 , 25----100] with base address as 1000 and size of each element in matrix is 10, If the array is stored in column major order then find the address of the element a [80][45] stored in array ________________.Correct answer is '15210'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Consider a lower triangular matrix. When this lower triangular matrix is stored in array format then only the elements a[i][j] with i≥j are stored in array i.e only the elements present in lower triangular matrix are stored. Hence less size is consumed to store the array. Consider a lower triangular matrix as [25----100 , 25----100] with base address as 1000 and size of each element in matrix is 10, If the array is stored in column major order then find the address of the element a [80][45] stored in array ________________.Correct answer is '15210'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a lower triangular matrix. When this lower triangular matrix is stored in array format then only the elements a[i][j] with i≥j are stored in array i.e only the elements present in lower triangular matrix are stored. Hence less size is consumed to store the array. Consider a lower triangular matrix as [25----100 , 25----100] with base address as 1000 and size of each element in matrix is 10, If the array is stored in column major order then find the address of the element a [80][45] stored in array ________________.Correct answer is '15210'. Can you explain this answer?.
Solutions for Consider a lower triangular matrix. When this lower triangular matrix is stored in array format then only the elements a[i][j] with i≥j are stored in array i.e only the elements present in lower triangular matrix are stored. Hence less size is consumed to store the array. Consider a lower triangular matrix as [25----100 , 25----100] with base address as 1000 and size of each element in matrix is 10, If the array is stored in column major order then find the address of the element a [80][45] stored in array ________________.Correct answer is '15210'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE. Download more important topics, notes, lectures and mock test series for GATE Exam by signing up for free.
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