250+ TOP MCQs on Solving Equations by Crout’s Method and Answers

Matrices Multiple Choice Questions on “Solving Equations by Crout’s Method”.

1. Solve the given equations using Crout’s Method to get value of z.

2x + 3y + z = -1
5x + y + z = 9
3x + 2y + 4z = 11

a) (frac{22}{7})
b) 8
c) (frac{21}{8})
d) (frac{32}{7})
Answer: c
Explanation: For the given sets of equations,
The Matrix form is given by
(begin{bmatrix}2&3&1\5&1&1\3&2&4end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}-1\9\11end{bmatrix})
As the form of AX=B
Let A be assumed to be LU
(begin{bmatrix}2&3&1\5&1&1\3&2&4end{bmatrix}=begin{bmatrix}a&0&0\b&c&0\d&e&fend{bmatrix}
begin{bmatrix}1&g&h\0&1&i\0&0&1end{bmatrix}=begin{bmatrix}a&ag&ah\b&bg+c&bh+ci\d&dg+e&dh+ei+fend{bmatrix})
By comparing both sides
a=2, b=5, d=3
g=(frac{3}{2}), c=(frac{-13}{2}), e=(frac{-5}{2}), h=(frac{1}{2}), f=(frac{40}{13}) and i=(frac{3}{13})
Thus,
L=(begin{bmatrix}2&0&0\5&frac{-13}{2}&0\3&frac{-5}{2}&frac{40}{13}end{bmatrix}) and U=(begin{bmatrix}1& frac{3}{2}&frac{1}{2}\0&1&frac{3}{13}\0&0&1end{bmatrix})
Now LY=B where Y=UX
(begin{bmatrix}2&0&0\5&frac{-13}{2}&0\3&frac{-5}{2}&frac{40}{13}end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}-1\9\11end{bmatrix})
Comparing Directly,
y₁=(frac{-1}{2}) y₂=(frac{-23}{13}) y₃=(frac{21}{8})
Assume UX=Y
(begin{bmatrix}1& frac{3}{2}&frac{1}{2}\0&1&frac{3}{13}\0&0&1end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}frac{-1}{2}\frac{-23}{12}\frac{21}{8}end{bmatrix})
Comparing both sides we get.
Z=(frac{21}{8})
Thus, the value of y is (frac{21}{8}).

Global Education & Learning Series – Matrices.

To practice all areas of Matrices,