Answer :

The given system can be written in matrix form as:


or A X = B


A = , X = and B =


Now, |A| = 5


= 5(4 – 6) – 3(8 – 3) + 1(4 – 2)


= – 10 – 15 + 3


= – 22


So, the above system has a unique solution, given by


X = A – 1B


Cofactors of A are:


C11 = (– 1)1 + 1 (4 – 6) = – 2


C21 = (– 1)2 + 1(12 – 2) = – 10


C31 = (– 1)3 + 1(9 – 1) = 8


C12 = (– 1)1 + 2 (8 – 3) = – 5


C22 = (– 1)2 + 1 20 – 1 = 19


C32 = (– 1)3 + 1 15 – 2 = – 13


C13 = (– 1)1 + 2 (4 – 2) = 2


C23 = (– 1)2 + 1 10 – 3 = – 7


C33 = (– 1)3 + 1 5 – 6 = – 1


adj A =


=


A – 1 =


Now, X = A – 1B =


X =


X =


X =


Hence, X = 1,Y = 2 and Z = 5


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

If <span lang="ENMathematics - Board Papers

Using matrices soMathematics - Board Papers

<span lang="EN-USMathematics - Board Papers

Using properties Mathematics - Board Papers

Using matrices, sMathematics - Board Papers

If A is square maMathematics - Exemplar

If <span lang="ENMathematics - Exemplar

The management coMathematics - Board Papers

If A is an invertMathematics - Board Papers

If <iMathematics - Board Papers