Matrices Questions and Answers – Solving Equations by Crout’s Method

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This set of Matrices Multiple Choice Questions & Answers (MCQs) focuses 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) $$\frac{32}{7}$$

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

Sanfoundry Global Education & Learning Series – Matrices.