in the gauss elimination method for solving a system of linear algebraic equations, triangularzation leads to

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In the Gauss elimination method for solving a system of linear algebraic equations, triangularization leads to

In the Gauss elimination method for solving a system of linear algebraic equations, triangularization leads to

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In the Gauss elimination method for solving a system of linear algebraic equations, triangularization leads to

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In the Gauss elimination method for solving a system of linear algebraic equations, triangularization leads to

A Diagonal matrix B

Lower triangular matrix

C

Upper triangular matrix

D Singular matrix Open in App Solution

The correct option is C Upper triangular matrix

Ingauss eliminatipn method for solving system of linear algebraic equation, the equation are reduce to an equivalent upper-triagular matrix.

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Solve the following pair of linear equations by the elimination method and the substitution method:

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[Solved] In the solution of simultaneous equations by the Gauss elimi

Explanation: Gauss elimination method: It is a method to solve the linear system of the form Ax = b, by bringing an augmented matrix into the upper triangle

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In the solution of simultaneous equations by the Gauss elimination method for solving equations, triangularization leads to

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singular matrix

upper triangular matrix

diagonal matrix

lower triangular matrix

Answer (Detailed Solution Below)

Option 2 : upper triangular matrix

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Explanation:Gauss elimination method:

It is a method to solve the linear system of the form Ax = b, by bringing an augmented matrix into the upper triangle matrix.

For example: Solve, y + Z = 2, 2X + 3Z = 5, X + Y + Z = 3

The augmented matrix of the given equations is

= [011220351113] = [203501121113]

Interchanging R1 and R2

= [10325201121113] R1 -----> R1/2 = [103252011201−1212] R3 -------> -R1 +R3 =

[103252011200−32−32]

R3 -------> -R2 + R3

= [10325201120011] R3 --------> -2R2/3

This is the upper triangular matrix after various operations.

So, By comparing we get

Z = 1, y +Z = 2, X+32Z=52

So, The values are x = 1, y = 1, z = 1

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