Option A is correct.
Step-by-step explanation:
We given with
We have to find Element in row 1 and column 1 of Matrix C.
Matrix C is Matrix X.
So, we have
So first we find inverse of first matrix,
Inverse of matrix A is given by
let,
then,
|A| = 4 × 1 - 0 × y = 4
So,
Element at row 1 and column 1 of matrix is
Therefore, Option A is correct.
Option A is correct.
Step-by-step explanation:
We given with
We have to find Element in row 1 and column 1 of Matrix C.
Matrix C is Matrix X.
So, we have
So first we find inverse of first matrix,
Inverse of matrix A is given by
let,
then,
|A| = 4 × 1 - 0 × y = 4
So,
Element at row 1 and column 1 of matrix is
Therefore, Option A is correct.
The Matlab commands for the given index operations and corresponding outputs are given below.
Explanation:
clc % is used to clear the command window of the Matlab
clear all % is used to clear the variables stored in Matlab workspace
% We are given a 4x4 matrix
Matlab command:
M = [16 2 3 13; 5 11 10 8; 9 7 6 12; 4 14 15 1]
output:
M =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
(a) The value in the first row and second column (ie. 2)
Matlab command:
a = M(1,2)
% where a = Matrix(row_1,column_2)
output:
a =
2
(b) The value in the third row and third column (ie. 6)
Matlab command:
b = M(3,3)
% where b = Matrix(row_3,column_3)
output:
b =
6
(c) All the elements in the first row
Matlab command:
c = M(1,:)
% where c = Matrix(row_1,:)
output:
c =
16 2 3 13
(d) All the elements in the second column
Matlab command:
d = M(:,2)
% where d = Matrix(:,column_2)
output:
d =
2
11
7
14
(e) All the elements in the first 2 rows (row 1 & 2)
Matlab command:
e = M([1,2],:)
% where e = Matrix([row_1,row_2],:)
output:
e =
16 2 3 13
5 11 10 8
(f) All the elements in the last 2 columns (columns 3 & 4)
Matlab command:
f = M(:,[3,4])
% where f= Matrix(:,[column_3,column_4])
output:
f =
3 13
10 8
6 12
15 1
(g) The elements 3 13 10 8
Matlab command:
g = [M(1,3) M(1,4); M(2,3) M(2,4)]
% where g = Matrix(row_1,column_3) M(row_1,column_4); M(row_2,column_3), M(row_2,column_4)
output:
g =
3 13
10 8
The Matlab commands for the given index operations and corresponding outputs are given below.
Explanation:
clc % is used to clear the command window of the Matlab
clear all % is used to clear the variables stored in Matlab workspace
% We are given a 4x4 matrix
Matlab command:
M = [16 2 3 13; 5 11 10 8; 9 7 6 12; 4 14 15 1]
output:
M =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
(a) The value in the first row and second column (ie. 2)
Matlab command:
a = M(1,2)
% where a = Matrix(row_1,column_2)
output:
a =
2
(b) The value in the third row and third column (ie. 6)
Matlab command:
b = M(3,3)
% where b = Matrix(row_3,column_3)
output:
b =
6
(c) All the elements in the first row
Matlab command:
c = M(1,:)
% where c = Matrix(row_1,:)
output:
c =
16 2 3 13
(d) All the elements in the second column
Matlab command:
d = M(:,2)
% where d = Matrix(:,column_2)
output:
d =
2
11
7
14
(e) All the elements in the first 2 rows (row 1 & 2)
Matlab command:
e = M([1,2],:)
% where e = Matrix([row_1,row_2],:)
output:
e =
16 2 3 13
5 11 10 8
(f) All the elements in the last 2 columns (columns 3 & 4)
Matlab command:
f = M(:,[3,4])
% where f= Matrix(:,[column_3,column_4])
output:
f =
3 13
10 8
6 12
15 1
(g) The elements 3 13 10 8
Matlab command:
g = [M(1,3) M(1,4); M(2,3) M(2,4)]
% where g = Matrix(row_1,column_3) M(row_1,column_4); M(row_2,column_3), M(row_2,column_4)
output:
g =
3 13
10 8
See Explanation
Explanation:
The line by line explanation of the modified program is as follows:
(1)
min is declared and initialized to the highest possible integer and max is declared and initialized to the least possible integer
int min,max;
min = Integer.MAX_VALUE; max = Integer.MIN_VALUE;
This iterates through each row. Because each row has a different number of elements, the row length of each is calculated using data[row].length
for ( int < ; col++){
The following if statements get the minimum and the maximum elements of the array
if(data[row][col] < min){ }
if(data[row][col] > max){ }
(2):
The loop remains the same, however, the print statement is adjusted upward to make comparison between each row and the highest of each row is printed after the comparison
for ( int < row++){
for ( int < col++){
if(data[row][col] > max){ }
}
System.out.println( "max = " + max);
max = Integer.MIN_VALUE; }
See attachment for complete program that answers (1) and (2)
See Explanation
Explanation:
The line by line explanation of the modified program is as follows:
(1)
min is declared and initialized to the highest possible integer and max is declared and initialized to the least possible integer
int min,max;
min = Integer.MAX_VALUE; max = Integer.MIN_VALUE;
This iterates through each row. Because each row has a different number of elements, the row length of each is calculated using data[row].length
for ( int < ; col++){
The following if statements get the minimum and the maximum elements of the array
if(data[row][col] < min){ }
if(data[row][col] > max){ }
(2):
The loop remains the same, however, the print statement is adjusted upward to make comparison between each row and the highest of each row is printed after the comparison
for ( int < row++){
for ( int < col++){
if(data[row][col] > max){ }
}
System.out.println( "max = " + max);
max = Integer.MIN_VALUE; }
See attachment for complete program that answers (1) and (2)
Following are the code to this question:
public class ArrayMath//defining a class ArrayMath
{
public static void main(String[] args) //defining main method
{
int x[][]=new int[10][10]; //defining a 2D array
arrays(x);//calling a method arrays by passing an array
}
private static void arrays(int[][] x)//defining a method arrays
{
int odd_column_sum=0,sum_of_Elements=0,i,j,even=1;//defining integer variable
for(i=0;i<x.length;i++)//defining loop for columns
{
for(j=0;j<x[0].length;j++)//defining loop for rows
{
x[i][j]=i*j;//multiply the i and j value atore in array
}
}
System.out.println("::The 2D array :: ");//print message
for(i=0;i<x.length;i++)//defining loop for columns
{
System.out.print("\t");//use print method for line space and line break
for(j=0;j<x[0].length;j++)//defining loop for rows
{
System.out.printf("%d\t",x[i][j]); //print array values
}
System.out.println( );//print for line break
}
for(i=0;i<x.length;i++) //defining loop for Columns
{
for(j=0;j<x[0].length;j++) //defining loop for rows
{
if(even%2!=0)//defining if block for check odd number condition
{
odd_column_sum+=x[j][i];//add odd number of array
}
sum_of_Elements+=x[i][j];//add even number of array
}
even++;//increment even variable value by 1
}
System.out.println("The sum of Odd Columns:"+odd_column_sum);//print odd_column_sum value
System.out.println("The array elements :"+sum_of_Elements);//print sum_of_Elements value
}
}
Output:
please find attached file.
Explanation:
In the above-given code, a class "ArrayMath" is defined, inside the class the main method is declared, which define a 2D array "x", this stores 10 columns and 10 rows and at the last, we call the arrays method by passing an array as a variable.
In the arrays method, integer variable, "odd_column_sum, sum_of_Elements, i, j, and even" is defined, in which variable "i and j" is used in the loop for calculating value, even is used to check odd column and then store its value in the "odd_column_sum" variable, and add whole element value is added into the "sum_of_Elements" variable.
1) a. False, adding a multiple of one column to another does not change the value of the determinant.
2) d. True, column-equivalent matrices are matrices that can be obtained from each other by performing elementary column operations on the other.
Step-by-step explanation:
1) If the multiple of one column of a matrix A is added to another to form matrix B then we get: |A| = |B|. Here, the value of the determinant does not change. The correct option is A
a. False, adding a multiple of one column to another does not change the value of the determinant.
2) Two matrices can be column-equivalent when one matrix is changed to the other using a sequence of elementary column operations. Correc option is d.
d. True, column-equivalent matrices are matrices that can be obtained from each other by performing elementary column operations on the other.
1) a. False, adding a multiple of one column to another does not change the value of the determinant.
2) d. True, column-equivalent matrices are matrices that can be obtained from each other by performing elementary column operations on the other.
Step-by-step explanation:
1) If the multiple of one column of a matrix A is added to another to form matrix B then we get: |A| = |B|. Here, the value of the determinant does not change. The correct option is A
a. False, adding a multiple of one column to another does not change the value of the determinant.
2) Two matrices can be column-equivalent when one matrix is changed to the other using a sequence of elementary column operations. Correc option is d.
d. True, column-equivalent matrices are matrices that can be obtained from each other by performing elementary column operations on the other.
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