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If $a,b,c$ and $d$ satisfy the equations

  • $a+7b+3c+5d =16$
  • $8a+4b+6c+2d = -16$
  • $2a+6b+4c+8d = 16$
  • $5a+3b+7c+d= -16$

Then $(a+d)(b+c)$ equals

  1. $-4$
  2. $0$
  3. $16$
  4. $-16$
in Linear Algebra by Active (1.5k points)
edited by | 434 views
0
Is the answer is $0$?

4 Answers

+9 votes
$\color{maroon}{a+7b+3c+5d=16}$   

$\color{maroon}{2a+6b+4c+8d=16}$  

Or, $(a+a)+(7b-b)+(3c+c)+ (5d+3d) = 16$

Or, $(a+7b+3c+5d)+(a-b+c+3d) = 16$

Or, $a-b+c+3d = 0 \qquad\qquad ∵\big[\color{blue}{ a+7b+3c+5d=16}\big]$ ------------- 1)

$\color{maroon}{8a+4b+6c+2d = -16}$

$\color{maroon}{5a+3b+7c+d = -16}$

Or, $(5a+3a)+(3b+b)+(7c-c)+(d+d) = -16$

Or, $(5a+3b+7c+d)+(3a+b-c+d) = -16$

Or, $3a+b-c+d = 0 \qquad \qquad \big[∵\color{blue}{5a+3b+7c+d = -16}\big]$ ------------- 2)

Adding 1) & 2)

$3a+b-c+d=0\\a-b+c+3d=0$

Or, $4a +4d = 0$

Or, $a+d = 0$

Now, we've to find $\color{maroon}{(a+d)(b+c) = ?}$

$(a+d)(b+c) = (0)\times(b+c) \qquad \qquad ∵ \big[\color{blue}{a+d=0}\big]$

∴$\color{red}{(a+d)(b+c) = 0}$

$\color{green}{\text{Hence, option }} \color{orange}{ B)} \color{green}{ \text{ is the right answer}}$.
by Boss (15.4k points)
edited by
0
In 3rd line it is $7b -b$ rt?
0
Oh. yes

Done mistake
+1
Answer Updated
0
Nice. Just a question - suppose if someone ask you to write a program to solve such a question how will you do? Any different method exist which can work for an idiot like computer?
0
This is my code for idiot computer to check One ,Two ,Three variables linear equations.Please Check Once and comment on me.

#include<stdio.h>
#include<stdlib.h>
#include<math.h>

static float W,X,Y,Z;
int f1(int *a,int *b,int *c);
int f2(int *a,int *b,int *c,int *d);
int f3(int *a,int *b,int *c,int *d,int *e);

int main()
{
    int n,a[4],b[4],c[4],d[4],e[4],i;
    printf(" 1: aX+bY=c \n 2: aX+bY+cZ=d\n 3: aX+bY+cZ+dW=e \n");
    scanf("%d",&n);
    if(n==1)
    {
        for(i=0;i<2;i++)
        {
          printf("Values of %d th equation :",n);
          scanf("%d",&a[i]);
          scanf("%d",&b[i]);
          scanf("%d",&c[i]);
       }
       f1(a,b,c);
    }
    else if(n==2)
    {
        for(i=0;i<3;i++)
        {
          printf("Values of %d th equation :",n);
          scanf("%d",&a[i]);
          scanf("%d",&b[i]);
          scanf("%d",&c[i]);
          scanf("%d",&d[i]);
       }
       f2(a,b,c,d);
    }
    else
    {
        for(i=0;i<4;i++)
        {
         printf(" Values of %d th equation :",n);
         scanf("%d",&a[i]);
         scanf("%d",&b[i]);
         scanf("%d",&c[i]);
         scanf("%d",&d[i]);
         scanf("%d",&e[i]);
        }
        f3(a,b,c,d,e);
    }
    printf("\nFinish");
    return 0;
}

int f1(int *a,int *b,int *c)
{
    float r1,r2;    
    r1=((b[1]*c[0])-(b[0]*c[1]));
    r2=((a[0]*b[1])-(b[0]*a[1]));
    r1=r1/r2;
    r2=(c[0]-(a[0]*r1))/b[0];
    X=r1;
    Y=r2;
    printf("X=%f,Y=%f",X,Y);
    return 0;
}

int f2(int *a,int *b,int *c,int *d)
{
    int A[3],B[3],C[3];
    A[0]=((a[0]*c[1])-(a[1]*c[0]));
    B[0]=((b[0]*c[1])-(b[1]*c[0]));
    C[0]=((c[1]*d[0])-(c[0]*d[1]));
    A[1]=((a[1]*c[2])-(a[2]*c[1]));
    B[1]=((b[1]*c[2])-(b[2]*c[1]));
    C[1]=((c[2]*d[1])-(c[1]*d[2]));
    f1(A,B,C);
    Z=((d[0]-(a[0]*X))-(b[0]*Y));
    Z=Z/c[0];
    printf("Z=%f",Z);
    return 0;
}

int f3(int *a,int *b,int *c,int *d,int *e)
{
    int A[4],B[4],C[4],D[4];
    A[0]=((a[0]*d[1])-(a[1]*d[0]));
    B[0]=((b[0]*d[1])-(b[1]*d[0]));
    C[0]=((c[0]*d[1])-(c[1]*d[0]));
    D[0]=((d[1]*e[0])-(d[0]*e[1]));
    
    A[1]=((a[1]*d[2])-(a[2]*d[1]));
    B[1]=((b[1]*d[2])-(b[2]*d[1]));
    C[1]=((c[1]*d[2])-(c[2]*d[1]));
    D[1]=((d[2]*e[1])-(d[1]*e[2]));
    
    A[2]=((a[2]*d[3])-(a[3]*d[2]));
    B[2]=((b[2]*d[3])-(b[3]*d[2]));
    C[2]=((c[2]*d[3])-(c[3]*d[2]));
    D[2]=((d[3]*e[2])-(d[2]*e[3]));
    f2(A,B,C,D);
    W=((e[0]-(a[0]*X))-(b[0]*Y)-(Z*c[0]));
    W=W/d[0];
    printf("W=%f",W);
    return 0;
}
+5 votes
$a+7b+3c+5d=16 ..............eq.1$

$8a+4b+6c+2d=−16.................eq.2$

$2a+6b+4c+8d=16...................eq.3$

$5a+3b+7c+d=−16....................eq.4$

add $eq.2$ and $eq.3$, we get,

$10a + 10b + 10c + 10d = 0$

$a + b + c + d = 0.....................eq.5$

add $eq.1$ and $eq.4$, we get,

$6a + 10b + 10c + 6d = 0$

$6(a+d) + 10(b+c) = 0$

on putting the value of $(a+d)$ from eq.5, we get

$-6(b+c) + 10(b+c) = 0$

$b+c = 0$

$a+d = 0$ is also 0

So, $(a+d)*(b+c) = 0 $
by Boss (10.8k points)
0
Any alternate method which can work systematically?
0

Arjun, Sir,  maybe, but I will post if got something.

0 votes
a=-2,b=2,c=-2,d=2
answer is 0
(B)
by Active (1.5k points)
0 votes
a+7b+3c+5d=16....(1)

8a+4b+6c+2d=-16......(2)

2a+6b+4c+8d=16......(3)

5a+3b+7c+d=-16......(4)

add (1) and (4) and add(2) and (3)

u will get

a+d=-(b+c)

and a+d=-10/6(b+c)

subtract both u will get (b+c)=0

so (a+d)(b+c)=0
by Active (1.4k points)

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