An algebra problem by Rocco Dalto

Algebra Level pending

Let f : R 4 → P 3 f: \mathbb R^4 \rightarrow \mathbb P_{3} be linear transform defined by:

f ( p 1 p 2 p 3 p 4 ) = p 1 + p 2 ∗ x + p 3 ∗ x 2 + p 4 ∗ x 3 f \left( \begin{array}{cccc} p_{1} \\ p_{2} \\ p_{3} \\ p_{4} \\ \end{array} \right) = p1 + p2 * x + p3 * x^2 + p4 * x^3

Let A = { ( 1 2 3 4 ) , ( 1 − 1 2 4 ) , ( 1 1 − 1 2 ) , ( 1 4 2 3 ) } A = \{ \left( \begin{array}{cccc} 1 \\ 2 \\ 3 \\ 4 \\ \end{array} \right), \left( \begin{array}{cccc} 1 \\ -1 \\ 2 \\ 4 \\ \end{array} \right), \left( \begin{array}{cccc} 1 \\ 1 \\ -1 \\ 2 \\ \end{array} \right), \left( \begin{array}{cccc} 1 \\ 4 \\ 2 \\ 3 \\ \end{array} \right) \} be a basis for R 4 \mathbb R^4 .

Let B = { 1 + x − x 2 + x 3 , 1 − x 2 , x − 3 ∗ x 2 − x 3 , 1 + 2 ∗ x − 3 ∗ x 2 − 4 ∗ x 3 } B = \{ 1 + x - x^2 + x^3, 1 - x^2, x - 3 * x^2 - x^3, 1 + 2 * x - 3 * x^2 - 4 * x^3 \} be a basis for P 3 \mathbb P_{3} .

If M = [ a i j ] 4 x 4 M = [a_{ij}]_{4 \: x \: 4} represents the linear transform above find S = ∑ i = 1 4 ∑ j = 1 4 a i j . \displaystyle S = \sum_{i = 1}^{4} \sum_{j = 1}^{4} a_{i j}.

Express the result to four decimal places.

If your interested you can write a program(in any language) for the General Case below and use it to find the above result.

General Case(for program)

Let f : R n → P n − 1 f: \mathbb R^n \rightarrow \mathbb P_{n - 1} be linear transform defined by:

f ( p 1 p 2 . . . p j . . . p n ) = ∑ j = 1 n p j ∗ x j − 1 f \left( \begin{array}{cccccccccc} p_{1} \\ p_{2} \\ . \\ . \\ . \\ p_{j} \\ . \\ . \\ . \\ p_{n} \\ \end{array} \right) = \sum_{j = 1}^{n} p_{j} * x^{j - 1}

Let V q = ( p 1 q p 2 q . . . p j q . . . p n q ) V_{q} = \left( \begin{array}{cccccccccc} p_{1q} \\ p_{2q} \\ . \\ . \\ . \\ p_{jq} \\ . \\ . \\ . \\ p_{nq} \\ \end{array} \right)

and A = { V q ∣ ( 1 < = q < = n ) } A = \{V_{q}| (1 <= q <= n) \} be a basis for R n \mathbb R^n

Let W q = ∑ j = 1 n p j q ∗ x j − 1 W_{q} = \sum_{j = 1}^{n} p_{jq} * x^{j - 1}

and B = { W q ∣ ( 1 < = q < = n ) } B = \{W_{q}| (1 <= q <= n) \} be a basis for P n − 1 \mathbb P_{n - 1}

Write a program in any language to find the matrix M = [ a i j ] n x n M = [a_{ij}]_{n \: x \:n} representation of the general linear transform above and the sum S = ∑ i = 1 n ∑ j = 1 n a i j \displaystyle S = \sum_{i = 1}^{n} \sum_{j = 1}^{n} a_{i j} .

Make certain A A and B B are bases for R n \mathbb R^{n} and P n − 1 \mathbb P_{n - 1} .

You can use the program written to find the matrix M = [ a i j ] 4 x 4 M = [a_{ij}]_{4 \: x \: 4} that represents the linear transform above and output S = ∑ i = 1 4 ∑ j = 1 4 a i j \displaystyle S = \sum_{i = 1}^{4} \sum_{j = 1}^{4} a_{i j} .

Let X = ( 1 3 7 − 5 ) X = \left( \begin{array}{cccc} 1 \\ 3\\ 7 \\ -5 \\ \end{array} \right)

To check the matrix M = [ a i j ] 4 x 4 M = [a_{ij}]_{4 \: x \: 4} found, first find [ f ( X ) ] B [f(X)]_B without using the matrix M = [ a i j ] 4 x 4 M = [a_{ij}]_{4 \: x \: 4} , then find [ f ( X ) ] B [f(X)]_B using the matrix M = [ a i j ] 4 x 4 M = [a_{ij}]_{4 \: x \: 4} .


The answer is 0.7143.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Apr 27, 2017

After doing the problem I supply the program I wrote for the general case which generated the output for this specific problem.

Since f : R 4 → P 3 f: \mathbb R^4 \rightarrow \mathbb P_{3} is a linear transform, for each integer j ∋ ( 1 < = j < = 4 ) j \ni (1 <= j <= 4)

f ( p 1 j p 2 j p 3 j p 4 j ) = ∑ k = 1 4 p k j ∗ x k − 1 = α 1 j ∗ ( 1 + x − x 2 + x 3 ) + α 2 j ∗ ( 1 − x 2 ) + α 3 j ∗ ( x − 3 ∗ x 2 − x 3 ) + α 4 j ∗ ( 1 + 2 ∗ x − 3 ∗ x 2 − 4 ∗ x 3 ) f \left( \begin{array}{cccc} p_{1j} \\ p_{2j} \\ p_{3j} \\ p_{4j} \\ \end{array} \right) = \sum_{k = 1}^{4} p_{kj} * x^{k - 1} = \alpha_{1j} * (1 + x - x^2 + x^3) + \alpha_{2j} * (1 - x^2) + \alpha_{3j} * (x - 3 * x^2 - x^3) + \alpha_{4j} * (1 + 2 * x - 3 * x^2 - 4 * x^3)

= [ 1 1 0 1 1 0 1 2 − 1 − 1 − 3 − 3 1 0 − 1 − 4 ] ∗ ∣ α 1 j α 2 j α 3 j α 4 j ∣ = \begin{bmatrix}{1} && {1} && {0} && {1} \\ {1} && {0} && {1} && {2} \\ {-1} && {-1} && {-3} && {-3} \\ {1} && {0} && {-1} && {-4}\end{bmatrix} * \begin{vmatrix}{\alpha_{1j}} \\{\alpha_{2j}} \\ {\alpha_{3j}} \\ {\alpha_{4j}}\end{vmatrix}

where,

f ( 1 2 3 4 ) = 1 + 2 ∗ x + 3 ∗ x 2 + 4 ∗ x 3 f \left( \begin{array}{cccc} 1 \\ 2 \\ 3 \\ 4 \\ \end{array} \right) = 1 + 2* x + 3 * x^2 + 4 * x^3

f ( 1 − 1 2 4 ) = 1 − x + 2 ∗ x 2 + 4 ∗ x 3 f \left( \begin{array}{cccc} 1 \\ -1 \\ 2 \\ 4 \\ \end{array} \right) = 1 - x + 2 * x^2 + 4 * x^3

f ( 1 1 − 1 2 ) = 1 + x − x 2 + 2 ∗ x 3 f \left( \begin{array}{cccc} 1 \\ 1 \\ -1 \\ 2 \\ \end{array} \right) = 1 + x - x^2 + 2 * x^3

f ( 1 4 2 3 ) = 1 + 4 ∗ x + 2 ∗ x 2 + 3 ∗ x 3 f \left( \begin{array}{cccc} 1 \\ 4 \\ 2 \\ 3 \\ \end{array} \right) = 1 + 4* x + 2 * x^2 + 3 * x^3

Using the above we can set up the augmented matrix below to solve for the

four 4 X 4 systems of equations.

[ 1.0000 1.0000 0.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.0000 1.0000 2.0000 2.0000 − 1.0000 1.0000 4.0000 − 1.0000 − 1.0000 − 3.0000 − 3.0000 3.0000 2.0000 − 1.0000 2.0000 1.0000 0.0000 − 1.0000 − 4.0000 4.0000 4.0000 2.0000 3.0000 ] \left[ \begin{array}{cccc|cccc} 1.0000 & 1.0000 & 0.0000 & 1.0000 & 1.0000 & 1.0000 & 1.0000 & 1.0000\\ 1.0000 & 0.0000 & 1.0000 & 2.0000 & 2.0000 & -1.0000 & 1.0000 & 4.0000\\ -1.0000 & -1.0000 & -3.0000 & -3.0000 & 3.0000 & 2.0000 & -1.0000 & 2.0000\\ 1.0000 & 0.0000 & -1.0000 & -4.0000 & 4.0000 & 4.0000 & 2.0000 & 3.0000\\ \ \end{array} \right]

For an example of row operations refer to previous problem solution. . .

Using row operations we obtain:

[ 1.0000 0.0000 0.0000 0.0000 3.1429 0.8571 1.2857 4.1429 0.0000 1.0000 0.0000 0.0000 − 2.2857 0.7857 − 0.0714 − 3.7857 0.0000 0.0000 1.0000 0.0000 − 1.4286 − 0.5714 0.1429 − 1.4286 0.0000 0.0000 0.0000 1.0000 0.1429 − 0.6429 − 0.2143 0.6429 ] \left[ \begin{array}{cccc|cccc} 1.0000 & 0.0000 & 0.0000 & 0.0000 & 3.1429 & 0.8571 & 1.2857 & 4.1429 \\ 0.0000 & 1.0000 & 0.0000 & 0.0000 & -2.2857 & 0.7857 & -0.0714 & -3.7857 \\ 0.0000 & 0.0000 & 1.0000 & 0.0000 & -1.4286 & -0.5714 & 0.1429 & -1.4286\\ 0.0000 & 0.0000 & 0.0000 & 1.0000 & 0.1429 & -0.6429 & -0.2143 & 0.6429\\ \ \end{array} \right]

⟹ \implies

M = [ 3.1429 0.8571 1.2857 4.1429 − 2.2857 0.7857 − 0.0714 − 3.7857 − 1.4286 − 0.5714 0.1429 − 1.4286 0.1429 − 0.6429 − 0.2143 0.6429 ] M = \begin{bmatrix}{3.1429} && {0.8571} && {1.2857} && {4.1429} \\ {-2.2857} && {0.7857} && {-0.0714} && {-3.7857} \\ {-1.4286} && {-0.5714} && {0.1429} && {-1.4286} \\ {0.1429} && {-0.6429} && {-0.2143} && {0.6429} \end{bmatrix}

and S = ∑ i = 1 4 ∑ j = 1 4 a i j = 0.7143 S = \sum_{i = 1}^{4} \sum_{j = 1}^{4} a_{i j} = \boxed{0.7143} .

Check [ f ( X ) ] B : [f(X)]_{B}:

Without using matrix M : M:

Let X = ( 1 3 7 − 5 ) X = \left( \begin{array}{cccc} 1 \\ 3\\ 7 \\ -5 \\ \end{array} \right)

⟹ \implies

f ( 1 3 7 − 5 ) = 1 + 3 ∗ x + 7 ∗ x 2 − 5 ∗ x 3 = α 1 ∗ ( 1 + x − x 2 + x 3 ) + α 2 ∗ ( 1 − x 2 ) + α 3 ∗ ( x − 3 ∗ x 2 − x 3 ) + α 4 ∗ ( 1 + 2 ∗ x − 3 ∗ x 2 − 4 ∗ x 3 ) f \left( \begin{array}{cccc} 1 \\ 3 \\ 7 \\ -5 \\ \end{array} \right) = 1 + 3* x + 7 * x^2 - 5 * x^3 = \alpha_{1} * (1 + x - x^2 + x^3) + \alpha_{2} * (1 - x^2) + \alpha_{3} * (x - 3 * x^2 - x^3) + \alpha_{4} * (1 + 2 * x - 3 * x^2 - 4 * x^3)

= [ 1 1 0 1 1 0 1 2 − 1 − 1 − 3 − 3 1 0 − 1 − 4 ] ∗ ∣ α 1 α 2 α 3 α 4 ∣ = \begin{bmatrix}{1} && {1} && {0} && {1} \\ {1} && {0} && {1} && {2} \\ {-1} && {-1} && {-3} && {-3} \\ {1} && {0} && {-1} && {-4}\end{bmatrix} * \begin{vmatrix}{\alpha_{1}} \\{\alpha_{2}} \\ {\alpha_{3}} \\ {\alpha_{4}}\end{vmatrix}

Using the above we can set up the augmented matrix:

[ 1.0000 1.0000 0.0000 1.0000 1.0000 1.0000 0.0000 1.0000 2.0000 3.0000 − 1.0000 − 1.0000 − 3.0000 − 3.0000 7.0000 1.0000 0.0000 − 1.0000 − 4.0000 − 5.0000 ] \left[ \begin{array}{cccc|c} 1.0000 & 1.0000 & 0.0000 & 1.0000 & 1.0000 \\ 1.0000 & 0.0000 & 1.0000 & 2.0000 & 3.0000 \\ -1.0000 & -1.0000 & -3.0000 & -3.0000 & 7.0000 \\ 1.0000 & 0.0000 & -1.0000 & -4.0000 & -5.0000 \\ \ \end{array} \right]

Using row operations on the above augmented matrix we obtain:

[ 1.0000 0.0000 0.0000 0.0000 1.8571 0.0000 1.0000 0.0000 0.0000 − 3.7143 0.0000 0.0000 1.0000 0.0000 − 4.5714 0.0000 0.0000 0.0000 1.0000 2.8571 ] \left[ \begin{array}{cccc|c} 1.0000 & 0.0000 & 0.0000 & 0.0000 & 1.8571\\ 0.0000 & 1.0000 & 0.0000 & 0.0000 & -3.7143 \\ 0.0000 & 0.0000 & 1.0000 & 0.0000 & -4.5714 \\ 0.0000 & 0.0000 & 0.0000 & 1.0000 & 2.8571\\ \ \end{array} \right]

⟹ \implies

[ f ( X ) ] B = ( 1.8571 − 3.7143 − 4.5714 2.8571 ) [f(X)]_{B} = \left( \begin{array}{ccccc} 1.8571\\ -3.7143\\ -4.5714\\ 2.8571\\ \end{array} \right)

Using matrix M M :

Using X = ( 1 3 7 − 5 ) X = \left( \begin{array}{cccc} 1 \\ 3\\ 7 \\ -5 \\ \end{array} \right) and basis A = { ( 1 2 3 4 ) , ( 1 − 1 2 4 ) , ( 1 1 − 1 2 ) , ( 1 4 2 3 ) } A = \{ \left( \begin{array}{cccc} 1 \\ 2 \\ 3 \\ 4 \\ \end{array} \right), \left( \begin{array}{cccc} 1 \\ -1 \\ 2 \\ 4 \\ \end{array} \right), \left( \begin{array}{cccc} 1 \\ 1 \\ -1 \\ 2 \\ \end{array} \right), \left( \begin{array}{cccc} 1 \\ 4 \\ 2 \\ 3 \\ \end{array} \right) \} for R 4 ⟹ \mathbb R^4 \implies

X = ( 1 3 7 − 5 ) = β 1 ∗ ( 1 2 3 4 ) + β 2 ∗ ( 1 − 1 2 4 ) + β 3 ∗ ( 1 1 − 1 2 ) + β 4 ∗ ( 1 4 2 3 ) X = \left( \begin{array}{cccc} 1 \\ 3\\ 7 \\ -5 \\ \end{array} \right) = \beta_{1} * \left( \begin{array}{cccc} 1 \\ 2\\ 3 \\ 4 \\ \end{array} \right) + \beta_{2} * \left( \begin{array}{cccc} 1 \\ -1\\ 2 \\ 4 \\ \end{array} \right) + \beta_{3} * \left( \begin{array}{cccc} 1 \\ 1\\ -1 \\ 2 \\ \end{array} \right) + \beta_{4} * \left( \begin{array}{cccc} 1 \\ 4\\ 2 \\ 3 \\ \end{array} \right)

= [ 1 1 1 1 2 − 1 1 4 3 2 − 1 2 4 4 2 3 ] ∗ ∣ α 1 α 2 α 3 α 4 ∣ = \begin{bmatrix}{1} && {1} && {1} && {1} \\ {2} && {-1} && {1} && {4} \\ {3} && {2} && {-1} && {2} \\ {4} && {4} && {2} && {3}\end{bmatrix} * \begin{vmatrix}{\alpha_{1}} \\{\alpha_{2}} \\ {\alpha_{3}} \\ {\alpha_{4}}\end{vmatrix}

Using the above we can set up the augmented matrix:

[ 1.0000 1.0000 1.0000 1.0000 1.0000 2.0000 − 1.0000 1.0000 4.0000 3.0000 3.0000 2.0000 − 1.0000 2.0000 7.0000 4.0000 4.0000 2.0000 3.0000 − 5.0000 ] \left[ \begin{array}{cccc|c} 1.0000 & 1.0000 & 1.0000 & 1.0000 & 1.0000 \\ 2.0000 & -1.0000 & 1.0000 & 4.0000 & 3.0000 \\ 3.0000 & 2.0000 & -1.0000 & 2.0000 & 7.0000 \\ 4.0000 & 4.0000 & 2.0000 & 3.0000 & -5.0000 \\ \ \end{array} \right]

Using row operations we obtain:

[ 1.0000 0.0000 0.0000 0.0000 − 163.0000 0.0000 1.0000 0.0000 0.0000 99.0000 0.0000 0.0000 1.0000 0.0000 − 56.0000 0.0000 0.0000 0.0000 1.0000 121.0000 ] \left[ \begin{array}{cccc|c} 1.0000 & 0.0000 & 0.0000 & 0.0000 & -163.0000 \\ 0.0000 & 1.0000 & 0.0000 & 0.0000 & 99.0000 \\ 0.0000 & 0.0000 & 1.0000 & 0.0000 & -56.0000 \\ 0.0000 & 0.0000 & 0.0000 & 1.0000 & 121.0000 \\ \ \end{array} \right]

⟹ [ X ] A = ( − 163.0000 99.0000 − 56.0000 121.0000 ) \implies [X]_{A} = \left( \begin{array}{ccccc} -163.0000\\ 99.0000\\ -56.0000\\ 121.0000\\ \end{array} \right)

⟹ \implies

[ f ( X ) ] B = M ∗ [ X ] A = [ 3.1429 0.8571 1.2857 4.1429 − 2.2857 0.7857 − 0.0714 − 3.7857 − 1.4286 − 0.5714 0.1429 − 1.4286 0.1429 − 0.6429 − 0.2143 0.6429 ] ∗ ( − 163.0000 99.0000 − 56.0000 121.0000 ) = ( 1.8571 − 3.7143 − 4.5714 2.8571 ) [f(X)]_B = M * [X]_A = \begin{bmatrix}{3.1429} && {0.8571} && {1.2857} && {4.1429} \\ {-2.2857} && {0.7857} && {-0.0714} && {-3.7857} \\ {-1.4286} && {-0.5714} && {0.1429} && {-1.4286} \\ {0.1429} && {-0.6429} && {-0.2143} && {0.6429} \end{bmatrix} * \left( \begin{array}{ccccc} -163.0000\\ 99.0000\\ -56.0000\\ 121.0000\\ \end{array} \right) = \left( \begin{array}{ccccc} 1.8571\\ -3.7143\\ -4.5714\\ 2.8571\\ \end{array} \right)

I used Free Pascal to write the program below:

program matrixrepresentaionoflineartransform_5;

{$R-}

{restricted to f:Rn --> P(n - 1)}

uses crt; 

                                {self contained for brillant}

const maxnum = 100;

type matrix = array[1 .. maxnum,1 .. maxnum] of extended;

var 

coeff,a,b,c,sol,prod,e:matrix;

n,l,p,v:integer;

deta,detb,product:real;

f:matrix;{for me only-check}

mywork4:text;

function power(base:real; exponent:integer):real;

var n:integer;

product:real;

begin

product:= 1;

for n:= 1 to exponent do

product:= product * base;

power:= product;

end;

procedure Getsize;

begin

writeln('Enter n for Rn');

readln(n);

writeln('Number of polynomials of degree ',n - 1,' is ', n);

end;

procedure Getaugmatrix;

var k,q:integer; 

sum:real;

begin

writeln('To Enter ', n, ' basis vectors for R',n,' : ');

for q:= 1 to n do

begin

write('Enter ',n,' elements of vector ', q, ' on one line: ');

for k:= 1 to n do

begin

read(b[k,q]);

end;

end;

writeln('To Enter ',n, ' polynomials of degree ', n - 1,' : ');

for q:= 1 to n do

begin

write('Enter ',n,' coefficients of polynomial ', q, ' on one line: ');

for k:= 1 to n do

begin

read(a[k,q]);

end;

end;

for q:= 1 to n do

begin

for k:= 1 to n do

begin

a[k,n + q]:= b[k,q];

end;

end;

writeln('Augmented matrix is:');

writeln;

for q:= 1 to n do

begin

for k:= 1 to 2 * n do

begin

write(a[q,k]:0:4,' ');

end;

writeln;

end;

end;

{For check bases}

{begin}

procedure hold(var c:matrix; e:matrix; m:integer);

var k,j:integer;

begin

for k:= 1 to m do

begin

for j:= 1 to m do

begin

c[k,j]:= e[k,j];

end;

end;

end;

{Determinants-To check Bases}

procedure switch2(var c:matrix; l,p,n:integer);

var q:integer; 

z:real;

begin{switch}

for q:= 1 to n do

begin{m} 

   z:= c[l,q];

   c[l,q]:= c[p,q];

   c[p,q]:= z;

end;{m}

end;{switch}

procedure safeguard2(var c:matrix; l:integer; var p:integer; n:integer; var v:integer; var product:real);

var q:integer;

begin{safeguard}

v:=0;

if (c[l,l] = 0) then

begin{then}

v:= 1;

q:= l;

while ((q + 1) <= n) and (v = 1) do

begin{loop}

if c[q + 1,l] <> 0 then

begin{then} 

        p:= q + 1;

        switch2(c,l,p,n);

        v:= 0;

        PRODUCT:= -1 * PRODUCT;

end;{then}

q:= q + 1;

end;{loop}

end;{then}

end;{safeguard}

procedure operation11(var a:matrix; l,n:integer; var product:real);

var q:integer; 

s:matrix;

begin{operation1}

for q:= 1 to n do

begin{q}

s[l,q]:= a[l,q]/a[l,l];

end;{q}

PRODUCT:= PRODUCT * (1/A[L,L]);

for q:= 1 to n do

begin{q}

a[l,q]:= s[l,q];

end;{q}

    end;{operation1}

procedure operation21(var a:matrix; l,n:integer);

var r,q:integer; 

x:real;

s:matrix;

begin{op2}

for q:= 1 to n do

begin{q}

if q <> l then

begin{then}

x:= -1 * a[q,l];

for r:= 1 to n do

begin{r}

s[q,r]:= x * a[l,r] + a[q,r];

end;{r}

for r:= 1 to n do

begin{r}

a[q,r]:= s[q,r];

end;{r}

end;{then}

end;{q}

end;{op2}

FUNCTION DETERMINANT(a:matrix; n:integer; PRODUCT:real):real;

var j:integer; 

DET,DIAGONAL:real;

begin{DETERMINANT}

DIAGONAL:= 1;

FOR j:= 1 TO N DO

BEGIN{LOOP}

DIAGONAL:= DIAGONAL * (a[j,j]);

END;{LOOP}

IF DIAGONAL = 1 THEN

DET:= 1/PRODUCT

ELSE

DET:= 0;

DETERMINANT:= DET;

END;{DETERMINANT}

{end-check bases}

procedure switch(var c:matrix; l,p:integer);

var q:integer; 

z:real;

begin{switch}

for q:= 1 to 2 * n do

begin{m} 

   z:= c[l,q];

   c[l,q]:= c[p,q];

   c[p,q]:= z;

end;{m}

end;{switch}

procedure safeguard(var c:matrix; l:integer; var p:integer; var v:integer);

var q:integer;

begin{safeguard}

v:=0;

if (c[l,l] = 0) then

begin{then}

v:= 1;

q:= l;

while ((q + 1) <= 2 * n) and (v = 1) do

begin{loop}

if c[q + 1,l] <> 0 then

begin{then} 

        p:= q + 1;

        switch(c,l,p);

        v:= 0;

end;{then}

q:= q + 1;

end;{loop}

end;{then}

end;{safeguard}

procedure operation1(var a:matrix; l:integer);

var q,k:integer; 

s:matrix;

x:real;
                   {s is each transformed matrix}

begin{op1}

for q:= 1 to 2 * n do

begin{m}

s[l,q]:= a[l,q]/a[l,l];

end;{m}

x:= 1/(a[l,l]);

for q:= 1 to 2 * n do

begin{m}

a[l,q]:= s[l,q];

end;{m}

writeln(mywork4, x:0:4,' * Row',l);

writeln(mywork4);

for q:= 1 to n do

begin

for k:= 1 to 2 * n do

begin

write(mywork4, a[q,k]:0:4,' ');

end;

writeln(mywork4);

end;

writeln(mywork4);

end;{op1}

procedure operation2(var a:matrix; l:integer);

var q,r,k:integer; 

x:real;

s:matrix;

begin{op2}

for q:= 1 to n do

begin{q}

if q <> l then

begin{then}

x:= a[q,l];

for r:= 1 to 2 * n do

begin{m}

s[q,r]:= -1 * x * a[l,r] + a[q,r];

end;{m}

for r:= 1 to 2 * n do

begin{m}

a[q,r]:= s[q,r];

end;{m}

writeln(mywork4, -1 * x:0:4,' * ROW ', l,' + ROW',q);

end;{then}

end;{q}

writeln(mywork4);

for q:= 1 to n do

begin

for k:= 1 to 2 * n do

begin

write(mywork4, a[q,k]:0:4,' ');

end;

writeln(mywork4);

end;

writeln(mywork4);

end;{op2}

PROCEDURE SHOWDATA;

VAR J,k:INTEGER; 

sum:extended;

myfile2:text;

BEGIN

assign(myfile2,'holdsolution4.txt');

rewrite(myfile2);

writeln;

writeln('Matrix Representation of linear transformation f: R',n,' ---> P',n - 1,' : ');

writeln('---------------------------------------------------------------------------');

writeln;

for j:= 1 to n do

begin

for k:= n + 1 to 2 * n do

begin

write(myfile2,a[j,k],' ');

end;

writeln(myfile2);

end;

reset(myfile2);

for j:= 1 to n do

begin

for k:= 1 to n do

begin

read(myfile2,sol[j,k]);

end;

end;

close(myfile2);

for j:= 1 to n do

begin

for k:= 1 to n do

begin

write(sol[j,k]:0:4,' ');

end;

writeln;

end;

{For brillant only}

sum:= 0;

for j:= 1 to n do

begin

for k:= 1 to n do

begin

sum:= sum + sol[j,k];

end;

end;

writeln;

writeln('Sum of elements of matrix = ',sum:0:4);

END;

{Last check [f(x)]B }

procedure switch3(var c:matrix; l,p,n:integer);

var r:integer;

z:real;

begin{switch3}

for r:= 1 to n + 1 do

begin{m} 

   z:= c[l,r];

   c[l,r]:= c[p,r];

   c[p,r]:= z;

end;{m}

end;{switch3}

procedure safeguard3(var c:matrix; l:integer; var p:integer; n:integer; var v:integer);

var r:integer;

begin{safeguard3}

v:=0;

if (c[l,l] = 0) then

begin{then}

v:= 1;

r:= l;

while ((r + 1) <= n + 1) and (v = 1) do

begin{loop}

if c[r + 1,l] <> 0 then

begin{then} 

        p:= r + 1;

        switch3(c,l,p,n);

        v:= 0;

end;{then}

r:= r + 1;

end;{loop}

end;{then}

end;{safeguard3}

procedure operation13(var a:matrix; l,n:integer);

var r:integer; 

s:matrix;

x:real;

                  {s is each transformed matrix}

begin{op13}

for r:= 1 to n + 1 do

begin{r}

s[l,r]:= a[l,r]/a[l,l];

end;{r}

x:= 1/(a[l,l]);

for r:= 1 to n + 1 do

begin{m}

a[l,r]:= s[l,r];

end;{r}

end;{op13}

procedure operation23(var a:matrix; l,n:integer);

var q,r:integer; 

x:real;

s:matrix;

begin{op23}

for q:= 1 to n do

begin{q}

if q <> l then

begin{then}

x:= a[q,l];

for r:= 1 to n + 1 do

begin{r}

s[q,r]:= -1 * x * a[l,r] + a[q,r];

end;{r}

for r:= 1 to n + 1 do

begin{r}

a[q,r]:= s[q,r];

end;{r}

end;{then}

end;{q}

end;{op23}

procedure solution(y:matrix; n:integer);

var q:integer;

begin{solution} 

for q:= 1 to n do

begin{q}

writeln(mywork4,y[q,n + 1]:0:4,'   ');

end;{q}

writeln(mywork4);

end;{solution}

procedure matrixmult(a,b:matrix; var summat:matrix);

var j,k,s:integer; 

   sum:extended;

begin{matrixmult}

for j:= 1 to n do

begin{j}

for k:= 1 to 1 do

begin{k} 

     sum:= 0;

for s:= 1 to n do

begin{s} 

     sum:= sum + a[j,s] * b[s,k];

end;{s} 

      summat[j,k]:= sum;

              end;{k}

writeln(mywork4,sum:0:4);

end;{j}

writeln(mywork4);

end;{matrixmult}

procedure check;

var p,j,k:integer; 

sum:real;

u:matrix;

begin

{without using matrix}

writeln(mywork4, ' Check [f(p)]B');

writeln(' -------------');

writeln(mywork4);

writeln('Enter a vector in R',n,' on one line');

for j:= 1 to n do

begin

read(f[j,n + 1]);

end;

for j:= 1 to n do

begin

e[j,n + 1]:= f[j,n + 1];

end;

for j:= 1 to n do

begin

for k:= 1 to n + 1 do

begin

write(mywork4,f[j,k]:0:4,' ');

end;

writeln(mywork4);

end;

{Next Use operations on matrix f}

writeln(mywork4);

writeln(mywork4, 'Check [f(p)]B');

writeln(mywork4);

for l:= 1 to n do

begin{l}

safeguard3(f,l,p,n,v);

if v = 0 then

begin{then}

operation13(f,l,n);

operation23(f,l,n);

end;{then}

end;{l}

writeln(mywork4);

solution(f,n);

{Now do using matrix M}

writeln(mywork4);

writeln(mywork4);

for j:= 1 to n do

begin

for k:= 1 to n + 1 do

begin

write(mywork4,e[j,k]:0:4,' ');

end;

writeln(mywork4);

end;

for l:= 1 to n do

begin{l}

safeguard3(e,l,p,n,v);

if v = 0 then

begin{then}

operation13(e,l,n);

operation23(e,l,n);

end;{then}

end;{l}

writeln(mywork4);

solution(e,n);

writeln(mywork4);

writeln(mywork4);

for j:= 1 to n do

begin

u[j,1]:= e[j,n + 1];

end;

writeln(mywork4);

writeln(mywork4);

matrixmult(sol,u,prod);

end;

begin

assign(mywork4,'mywork4.txt');

rewrite(mywork4);

getsize;

getaugmatrix;

hold(f,a,n); {for me only-check [f(X)]}

hold(e,b,n);

{do determinants here}

{To do determinants to check bases here, then done.}

{checking basis for Rn}

PRODUCT:= 1;

for l:= 1 to n do

begin{l}

safeguard2(b,l,p,n,v,PRODUCT);

if v = 0 then

begin{then}

operation11(b,l,n,PRODUCT);

operation21(b,l,n);

end;{then}

end;{l}

DETB:= DETERMINANT(b,n,PRODUCT);

writeln(detb:0:20);

while (detb >= 0) and (detb <= abs(power(1/10,16))) do {I.E; detb ~ 0}

begin

writeln('Set of matrices is not a basis for R',n);

writeln;

getaugmatrix;

PRODUCT:= 1;

for l:= 1 to n do

begin{l}

safeguard2(b,l,p,n,v,PRODUCT);

if v = 0 then

begin{then}

operation11(b,l,n,PRODUCT);

operation21(b,l,n);

end;{then}

end;{l}

DETB:= DETERMINANT(b,n,PRODUCT);

end;

{checking basis for Rn^2}

hold(c,a,n);

PRODUCT:= 1;

for l:= 1 to n do

begin{l}

safeguard2(c,l,p,n,v,PRODUCT);

if v = 0 then

begin{then}

operation11(c,l,n,PRODUCT);

operation21(c,l,n);

end;{then}

end;{l}

DETA:= DETERMINANT(c,n,PRODUCT);

writeln;

while (deta >= 0) and (deta <= abs(power(1/10,16))) do {I.E; deta ~ 0}

begin

writeln('Set of vectors is not a basis for P',n - 1);

writeln;

getaugmatrix;

hold(c,a,n);

PRODUCT:= 1;

for l:= 1 to n do

begin{l}

safeguard2(c,l,p,n,v,PRODUCT);

if v = 0 then

begin{then}

operation11(c,l,n,PRODUCT);

operation21(c,l,n);

end;{then}

end;{l}

DETA:= DETERMINANT(c,n,PRODUCT);

end;

{Only getting through if deta <> 0 and detb <> 0}

for l:= 1 to n do

begin{l}

safeguard(a,l,p,v);

if v = 0 then

begin{then}

operation1(a,l);

operation2(a,l);

end;{then}

end;{l}

showdata;

check;

close(mywork4);

readln;

readln;

end.

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