📖 generic · CBSE Class 12th English Medium · MATHEMATCS PART-1 · Page 19question

A Note In this chapter, we restrict ourselves to the system of linear equations · Part 2

Chapter 4: DETERMINANTS · MATHEMATCS PART-1

 = −  = − Hence x = , y = – Example Solve the following system of equations by matrix method. x – y + z = x + y – z = x – y + z = Solution The system of equations can be written in the form AX = B, where , X and B We see that A = ( – ) + ( + ) + (– – ) = – ≠ Hence, A is nonsingular and so its inverse exists. Now A = – , A = – , A = – A = – , A = – , A = A = – , A = , A = Therefore A – = So X = A B =      i.e. Hence x = , y = and z = .

Example The sum of three numbers is . If we multiply third number by and add second number to it, we get . By adding first and third numbers, we get double of the second number. Represent it algebraically and find the numbers using matrix method.

Solution Let first, second and third numbers be denoted by x , y and z , respectively. Then, according to given conditions, we have x + y + z = y + z = x + z = y or x – y + z = This system can be written as A X = B, where A = , X = and B = Here

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