If A = ■(1&2@1&4) & ■(2&1@2&1) Find the product matrix AB & hence find its inverse matrix, if it exists?

What is a singular matrix?

A singular matrix is a matrix which has no inverse because its determinant is zero. If you recall, the inverse of a matrix is1/ ad-bc multiplied by:[ d -b ][-c a ]If ad-bc = 0, then the inverse matrix would not exist because 1/0 is undefined, and hence it would be a singular matrix.E.g.[ 1 3][ 2 6]Is a singular matrix because 6x1-3x2 = 0.

What is the inverse sine of 0.75?

arcsin(.75)≈0.848062079

What is the inverse log of 2?

inverse log of 2= 1/(log{10}2)= 1/(log2)=1/0.3010299=3.3219. hence answer is 3.3219

What is the additive inverse of 9.1?

We are talking group theory here. A group with addition has an additive inverse. A group with multiplication has a multiplicative inverse. The additive inverse of a number x is a y with x + y = 0. The additive inverse of x is written -x. Hence, the additive inverse of 9.1 equals -9.1. The reason that this question can arise is that beyond groups, there are rings and fields. Rings and fields have, besides addition, also multiplication. An element can have an additive inverse and a multiplicative inverse at the same time.

Does titanium exist as an atom or molecule?

Since it is a pure metal, hence an element, it exists as atoms.

Why is it necessary to dry your product after esterication?

Water will be the byproduct of esterification and hence the product should be dried (to remove water).

A multiplicative inverse of 5 module 7 is?

A multiplicative inverse of 5 mod7 would be a number n ( not a unique one) such that 5n=1Let's look at the possible numbers5x1=5mode 75x2=10=3 mod 75x3=15=1 mod 7 THAT WILL DO IT3 is the multiplicative inverse of 5 mod 7.What about the others? 5x4=20, that is -1 mod 7 or 65x5=25 which is 4 mod 75x6=30 which is -5 or 2 mod 7How did we know it existed? Because 7 is a prime. For every prime number p and positive integer n, there exists a finite field with pn elements. This is an important theorem in abstract algebra. Since it is a field, it must have a multiplicative inverse. So the numbers mod 7 make up a field and hence have a multiplicative inverse.

How do you show that a square matrix A is similar to its transpose?

First we will handle the diagonalizable case.Assume A is diagonalizable, A=VDV-1.Thus AT=(V-1)TDVT,and D= VT AT(V-1)T.Finally we have that A= VVT AT(V-1)TV-1, hence A is similar to ATwith matrix VVT.If A is not diagonalizable, then we must consider its Jordan canonical form,A=VJV-1, where J is block diagonal with Jordan blocks along the diagonal.Recall that a Jordan block of size m with eigenvalue at L is a mxm matrix having L along the diagonal and ones along the superdiagonal.A Jordan block is similar to its transpose via the permutation that has ones along the antidiagonal, and zeros elsewhere.With this in mind we proceed as in the diagonalizable case,AT=(V-1)TJTVT.There exists a block diagonal permutation matrix P such thatJT=PJPT, thus J=PTVT AT(V-1)TP.Finally we have that A= VPTVT AT(V-1)TPV-1, hence A is similar to ATwith matrix VPTVT.Q.E.D.

What is the multiplication inverse of -4?

inverse means one over the number. i.e. the inverse of 4 is 1/4. These multiplied together gives 1 (4/4)

Write a c program to find inverse of a matrix?

#include<stdio.h> #include<math.h> float detrm(float[][],float); void cofact(float[][],float); void trans(float[][],float[][],float); main() { float a[25][25],k,d; int i,j; printf("ENTER THE ORDER OF THE MATRIX:\n"); scanf("%f",&k); printf("ENTER THE ELEMENTS OF THE MATRIX:\n"); for(i=0;i<k;i++) { for(j=0;j<k;j++) { scanf("%f",&a[i][j]); } } d=detrm(a,k); printf("THE DETERMINANT IS=%f",d); if(d==0) printf("\nMATRIX IS NOT INVERSIBLE\n"); else cofact(a,k); } /******************FUNCTION TO FIND THE DETERMINANT OF THE MATRIX************************/ float detrm(float a[25][25],float k) { float s=1,det=0,b[25][25]; int i,j,m,n,c; if(k==1) { return(a[0][0]); } else { det=0; for(c=0;c<k;c++) { m=0; n=0; for(i=0;i<k;i++) { for(j=0;j<k;j++) { b[i][j]=0; if(i!=0&&j!=c) { b[m][n]=a[i][j]; if(n<(k-2)) n++; else { n=0; m++; } } } } det=det+s*(a[0][c]*detrm(b,k-1)); s=-1*s; } } return(det); } /*******************FUNCTION TO FIND COFACTOR*********************************/ void cofact(float num[25][25],float f) { float b[25][25],fac[25][25]; int p,q,m,n,i,j; for(q=0;q<f;q++) { for(p=0;p<f;p++) { m=0; n=0; for(i=0;i<f;i++) { for(j=0;j<f;j++) { b[i][j]=0; if(i!=q&&j!=p) { b[m][n]=num[i][j]; if(n<(f-2)) n++; else { n=0; m++; } } } } fac[q][p]=pow(-1,q+p)*detrm(b,f-1); } } trans(num,fac,f); } /*************FUNCTION TO FIND TRANSPOSE AND INVERSE OF A MATRIX**************************/ void trans(float num[25][25],float fac[25][25],float r) { int i,j; float b[25][25],inv[25][25],d; for(i=0;i<r;i++) { for(j=0;j<r;j++) { b[i][j]=fac[j][i]; } } d=detrm(num,r); inv[i][j]=0; for(i=0;i<r;i++) { for(j=0;j<r;j++) { inv[i][j]=b[i][j]/d; } } printf("\nTHE INVERSE OF THE MATRIX:\n"); for(i=0;i<r;i++) { for(j=0;j<r;j++) { printf("\t%f",inv[i][j]); } printf("\n"); } } ALTERNATIVE CODE: #include<iostream.h> #include<stdio.h> #include<conio.h> #include<process.h> #include<math.h> // Written by Ran // There have been enough comments to help the reader easily understand this program // Helpfulness of COMMENTS in this program and Pre-requisites:- // a. However it's assumed that the reader is familiar with the basics of C++ // b. It is also assumed that the reader knows the basic mathematics involving matrices. // c. Since this program focusses on how to find inverse of a matrix, the comments // in the program are sufficient for understanding this. // It is assumed that the reader is familiar with basics of matrices in C++ (like input, display, // addition,transpose,etc. of matrices) // The comments in this program aim to explain the reader how to find inverse // d. Hence through comments, the reader will also be taught how to find determinant, // adjoint, cofactor,etc. However as said in the previous lines, there won't be comments // for explaining basics like input,display,etc of a matrix using C++. // NOTES: // 1. float datatype has been used for matrix, determinant. // 2. To have consistency between mathematics and C++, this program considers a[1][1] as the first element // i.e row and column indices begin with 1 same as mathematics. // Define a structure matrix with a matrix (2D array of type float) and size n // Declare the objects of this structure used in this program struct matrix { float a[25][25]; int n; }obj,c_obj,trans_obj,obj_cof,obj_adj,obj_inv; // Prototypes of the functions used in this program void input(matrix&); void display(matrix&); matrix reduced(matrix &, int ,int ); float determinant(matrix); float cofactor(matrix,int,int); matrix transpose(matrix); matrix adjoint(matrix); matrix inverse(matrix obj); // Begining of Main function int main() { // Getting dimensions input by the user int r,c; again: cout<<"Enter the order of the matrix: "<<endl; cout<<"Enter Row dimension: "; cin>>r; cout<<"Enter Column dimension: "; cin>>c; // Check dimensions for square matrix so that inverse can be found // If user enters different dimensions for row and column, ask to re-enter or quit program if(r!=c) { char ans; cout<<"Inverse can be found out only for a square matrix. Enter same dimension for row and column. Do you want to enter the dimensions again? Press Y for yes"<<endl; cin>>ans; if(ans=='y') goto again; else cout<<"Program exit"; getch(); exit(0); } // If it's a square matrix, proceed else if(r==c) {obj.n=r;} cout<<endl; input(obj); // call input function to input the matrix elements from the user display(obj); // display the matrix got as input now // Following lines were used to test parts/sections/segments of the code and hence commented /* char ans2; cout<<"do u want to check reduce matrix? Press y to check reduce matrix and press any char to skip this"<<endl; cin>>ans2; if(ans2=='y') { int i,j; cout<<"Enter row i and col j to get reduced matrix"<<endl; cin>>i>>j; // i=i-1; // j=j-1; c_obj=reduced(obj,i,j); char ans1; cout<<"Do you want to display the reduced matrix? If yes, Press y "<<endl; cin>>ans1; if(ans1=='y') { cout<<"Displaying reduced matrix..."<<endl; display(c_obj); } }*/ //Find Determinant cout<<"Finding determinant......"<<endl; cout<<"The determinant is"<<determinant(obj)<<endl; //Find Cofactor if user wishes to char ans3; cout<<"Do you want to find cofactor? Press y if yes"<<endl; cin>>ans3; while(ans3=='y') { int i,j; cout<<"Finding cofactor. Enter row and column"<<endl; cin>>i>>j; cout<<"Cofactor of a["<<i<<"]["<<j<<"] is "<<cofactor(obj,i,j)<<endl; cout<<"want of find cofactor of another element? Press y for yes"<<endl; cin>>ans3; } // Following lines were to meant to test ONLY the transpose function and hence commented /* cout<<"Printing Transpose of the matrix "<<endl; matrix trans1; trans1=transpose(obj); display(trans1); */ // Find Inverse cout<<"\n\n\n Finding Inverse. . .\n\n"; matrix obj_inv2; obj_inv2=inverse(obj); display(obj_inv2); // Display the matrix inverse getch(); return 0; } void input(matrix &obj) { // This function gets elements of a matrix input by the user // Parameter is the structure object obj (used throughout the program) // Parameter is "passed by reference" so as to reflect the changes made by this function, to other functions that call it cout<<"Enter the matrix "<<endl; for (int i=1;i<=obj.n;i++) { for (int j=1;j<=obj.n;j++) { cout<<"Enter the element a["<<i<<"]["<<j<<"] : "; cout<<endl; cin>>obj.a[i][j]; } } } void display(matrix &obj) { // This function displays elements of a matrix passed to it as a parameter // Parameter is the structure object obj (used throughout the program) // Parameter is "passed by reference" but may be "passed by value" also. if(obj.n==0) return; else{ cout<<"The matrix is: "<<endl; for (int i=1;i<=obj.n;i++) { for (int j=1;j<=obj.n;j++) { cout<<obj.a[i][j]<<" "; } cout<<endl; } }} matrix reduced(matrix &obj, int i,int j) { // This function reduces the matrix passed as input to it // The 'reduction' requirement is like this: // Eliminate the row i and column j from the given matrix to get the reduced matrix // This is done by the following logic: // a is given matrix. c_obj is desired reduced matrix // i. Using two for loops (iterating with p and q here) as usual, we scan the given matrix. // row and col represent the current location pointer of row and column of the required reduced matrix. // ii. All elements from given matrix are copied to reduced matrix except for those corresponding to // row i and column j // iii. The reduced matrix has its dimensions one less than that of given matrix int row=1,col=1; for(int p=1;p<=obj.n;p++) // outer loop traverses through rows as usual { for(int q=1;q<=obj.n;q++) // inner loop traverses through columns as usual { if((p!=i)&&(q!=j)) // Skip the elements corresponding to row i OR column j of the given matrix { c_obj.a[row][col]=obj.a[p][q]; col=col+1; } if(col>=obj.n) // When column 'col' of reduced matrix reaches (or exceeds n), reset it to 1 { // and increment 'row'. This means current row of reduced matrix got filled and // we need to begin filling a new row. col=1; row=row+1; if (row>=obj.n) //This represents the case when both 'col' and 'row of reduce matrix reach (or // exceed) n. This means the reduced matrix has been filled up.Break out of the loops. break; } } } c_obj.n=obj.n-1; // Fix the dimension of the reduced matrix one less than the given input matrix return c_obj; // Return the reduced matrix to the calling function. } float determinant(matrix obj) { // This function is called recursively until we get dimension = 1 where the only element in the matrix gets returned. float det=0; if(obj.n==1) {return obj.a[1][1]; } else { for(int scan=1;scan<=obj.n;scan++)//Fix the first row and vary the column in this row using for loop iteration variable 'scan' { det=det+obj.a[1][scan]*int(pow(-1,(1+scan)))*determinant(reduced(obj,1,scan)); // det is calculated to be the sum of the following // i. prev value stored in det. // ii. current element in the first row (i.e. a[1][scan]) MULTIPLIED by -1^(i+j) [i is 1 for 1st row and j is nothing but scan here MULTIPLIED by the reduced matrix corresponding to this i (1) and j (scan) // PLEASE UNDERSTAND BY COMPARING THIS WITH THE MATHEMATICAL WAY OF CALCULATING DETERMINANT // - It's computed in a similar way here. } return det; } } float cofactor(matrix obj,int i,int j) { // The computation done here is like this: // If the matrix (passed as paramenter) has dimension = 1, return the only element as cofactor // Else, return determinant of the reduced matrix corresponding to i and j passed with a // multiplication factor -1^(i+j) float cofact; if(obj.n==1) { return obj.a[1][1]; } else { cofact=int(pow(-1,(i+j)))*determinant(reduced(obj,i,j)); } return cofact; } matrix transpose(matrix obj) { // Transpose matrix is the given matrix with its rows and columns interchanged. // Just invert the elements during storing when scanning through the for loops // trans_obj is the transposed matrix, returned by the function. // obj is the input matrix passed to this function. trans_obj.n=obj.n; for(int i=1;i<=obj.n;i++) { for(int j=1;j<=obj.n;j++) { trans_obj.a[i][j]=obj.a[j][i]; } } return trans_obj; } matrix adjoint(matrix obj) { // obj_adj is adjoint matrix and obj_cof is cofactor matrix // both have dimensions n // Cofactor matrix of a given matrix is a matrix whose elements are the cofactors of the respective // elements of the given matrix // Adjoint matrix is transpose of cofactor matrix. Return this obj_adj.n=obj.n; obj_cof.n=obj.n; for(int i=1;i<=obj.n;i++) { for(int j=1;j<=obj.n;j++) { obj_cof.a[i][j]=cofactor(obj,i,j); } } obj_adj=transpose(obj_cof); return obj_adj; } matrix inverse(matrix obj) { // Formula : Inverse of a matrix is = adj(matrix)/its determinant float d=determinant(obj); // First find determinant of the given matrix matrix obj_null; obj_null.n=0; // Display error message if determinant is 0 if(d==0) { cout<<"Inverse can be found only if the determninant of the matrix is non-zero"<<endl; return obj_null; } // Determinant is non-zero - Proceed finding inverse using the above formula else { matrix obj_adj1=adjoint(obj); obj_adj1.n=obj.n; obj_inv.n=obj.n; for(int i=1;i<=obj.n;i++) { for(int j=1;j<=obj.n;j++) { obj_inv.a[i][j]=(obj_adj1.a[i][j])/d; } } return obj_inv; } }

Is a negative number an integer?

Integers are all of the real numbers that can be written without using a decimal or a fraction; i.e, they're the set of all positive and negative whole numbers plus zero.Integers have the following properties:i) 0 exists, and is the additive identity element; 0 + a = a for all a.ii) 1 exists, and is the multiplicative identity element; 1a = a for all a.iii) Addition and multiplication are both associative and commutative.iv) An additive inverse exists for all elements a, denoted as -a.v) Distributivity.vi) Adding or multiplying two integers yields another integer.From iv), we know that for each a > 0, there exists an additive inverse, -a, which must be less than 0, hence a negative number.So, there are integers that are negative numbers.But, not every negative number is an integer. There are negative numbers that must be expressed as a decimal or a fraction. For example, -0.5 isn't an integer, because, were you to not write it as a decimal, you would have to write it as a fraction: -1/2

What is the relationship between marginal productivity and marginal cost?

The marginal product curve is 'n' shaped because of the law of diminishing returns. As you add more units of a variable factor, at first, the marginal product rises, (this is because the fixed factor is under-utilised, so adding more units of the variable factor will increase the output from each additional unit). But after a certain point, the marginal product begins to fall, as the fixed factor input becomes diluted amongst workers and so you get less from each additional unit of the variable factor. For an example, re-read the above paragraph and replace the word variable factor with labour and fixed factor with capital. The marginal cost curve is the inverse of the marginal product curve - hence it is shaped like a 'u' or a 'Nike tick'. This is because if your marginal product is high - then your marginal costs are low. For example, if a firm must pay electricity for the time it takes to produce a unit, if the firm can produce the unit quicker (i.e. has a high marginal product) then the cost of electricity will be lower. Hence the inverse relationship between marginal cost and marginal product.