Question

# The sum of three numbers is 6. If we multiply the third number by 3 and add it to the second number we get 11. By adding first and third numbers we get a number, which is double than the second number. Use this information and find a system of linear equations. Find these three numbers using matrices.

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Solution

## Let the first , second & third number be x,y,z respectivelyGiven, ∴x+y+z=6 y+3z=11 x+z=2y or x−2y+z=0 Step 1 Write equation as AX=B A=⎡⎢⎣1110131−21⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦,B=⎡⎢⎣6110⎤⎥⎦ Hence A=⎡⎢⎣1110131−21⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦ & B=⎡⎢⎣6110⎤⎥⎦ Step 2Calculate |A| |A|=⎡⎢⎣1110131−21⎤⎥⎦ =1(1+6)−0(1+2)+1(3+1) =7+2 =9SO , |A|≠0 ∴ The system of equation is consistent & has a unique solutionsNow AX=B X=A−1B Hence A=⎡⎢⎣1110131−21⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦ & B=⎡⎢⎣6110⎤⎥⎦ =1(1+6)−0(1+2)+1(3−1)=7+2 =9≠0 Since determinant is not equal to O, A−1 exists. Now find adj (A) Now A=B X=A−1B Step 3 Calculating X=A−1B Calculating A−1 Now A−1=1|A|ajd(A)adj A =⎡⎢⎣A11A12A13A21A22A23A31A32A33⎤⎥⎦′=⎡⎢⎣A11A21A31A12A22A32A13A23A33⎤⎥⎦ A=⎡⎢⎣1−1234−52−13⎤⎥⎦ A11=1×1−3×(2)=1+6=7 A12=−[0×1−3×1]=−(−3)=3 A13=−0×(−2)−1×1=−1 A21=[1×1−(−2)×1]=−[1+2]=−3 A22=1×1−1×1=1−1=0 A23=[1×(−2)−1×1]=−[−2−1]=−(−3)=3 A31=1×3−1×1=3−1=2 A32=−[1×3−0×1]=−[3−0]=−3 A33=1×1−1×0=1−0=1 Hence, adj(A)=⎡⎢⎣7−3230−3−131⎤⎥⎦ Now , A−1=1|A|adj(A) A−1=19⎡⎢⎣7−3230−3−131⎤⎥⎦ Solution of given system of equations is X=A−1B ⎡⎢⎣xyz⎤⎥⎦=19⎡⎢⎣7−3230−3−131⎤⎥⎦⎡⎢⎣6110⎤⎥⎦ ⎡⎢⎣xyz⎤⎥⎦=19⎡⎢⎣42−33+018+0+0−6+33+0⎤⎥⎦ ⎡⎢⎣xyz⎤⎥⎦=19⎡⎢⎣91827⎤⎥⎦ ⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣123⎤⎥⎦ ∴x=1,y=2,z=3

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