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

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

Answer: c
Explanation: For the given sets of equations,
The Matrix form is given by
\(\begin{bmatrix}2&3&1\\5&1&1\\3&2&4\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}-1\\9\\11\end{bmatrix}\)
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 \(\frac{21}{8}\).