# Matrix Inversion Questions and Answers – Crout’s Method

This set of Numerical Methods Multiple Choice Questions & Answers (MCQs) focuses on “Crout’s Method”.

1. Solve the following equations using Crout’s Method to find the value of x.

x+y+z=7
x+2y+3z=16
x+3y+4z=22

a) 3
b) 7
c) 0
d) 1

Explanation: From the above question, we get the matrix equation as –
$$\begin{bmatrix}1&1&1\\1&2&3\\1&3&4\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}7\\16\\22\end{bmatrix}$$
Let A=L*U
Here L is the lower triangular matrix
L=$$\begin{bmatrix}a&0&0\\b&c&0\\d&e&f\end{bmatrix}$$
U is the lower triangular matrix
U=$$\begin{bmatrix}1&g&h\\0&1&I\\0&0&1\end{bmatrix}$$
A=$$\begin{bmatrix}a&ag&ah\\b&bg+c&bh+cI\\d&dg+e&dh+eI+f\end{bmatrix}$$
A=LU
$$\begin{bmatrix}1&1&1\\1&2&3\\1&3&4\end{bmatrix}$$=$$\begin{bmatrix}a&ag&ah\\b&bg+c&bh+cI\\d&dg+e&dh+eI+f\end{bmatrix}$$
From this substitutions and comparisons, we get the following values,
a=1
b=1
c=1
d=1
e=2
f=-1
g=1
h=1
I=2
Thus from the above values,
L=$$\begin{bmatrix}1&0&0\\1&1&0\\1&2&-1\end{bmatrix}$$ and U=$$\begin{bmatrix}1&1&1\\0&1&2\\0&0&1\end{bmatrix}$$
Since AX=B i.e. LUX=B
Assume V=UX
V = $$\begin{bmatrix}v1\\v2\\v3\end{bmatrix}$$
We can say that LV=B
$$\begin{bmatrix}1&0&0\\1&1&0\\1&2&-1\end{bmatrix} \begin{bmatrix}v1\\v2\\v3\end{bmatrix} = \begin{bmatrix}7\\16\\22\end{bmatrix}$$
From this we get,
v1=7
v2=9
v3=3
UX=V
$$\begin{bmatrix}1&1&1\\0&1&2\\0&0&1\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}7\\9\\3\end{bmatrix}$$
$$\begin{bmatrix}x+y+z\\y+2z\\z\end{bmatrix} = \begin{bmatrix}7\\9\\3\end{bmatrix}$$
Hence, x=1, y=3, z=3
Thus, the value of x is 1.

2. Solve the following equations using Crout’s Method to find the value of z.

x-2y+3z=6
x-y+2z=9
3x+2y-z=16

a) 13
b) 7
c) 10
d) -12

Explanation: From the above question, we get the matrix equation as-
$$\begin{bmatrix}1&-2&3\\1&-1&2\\3&2&-1\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}6\\9\\16\end{bmatrix}$$
Let A=L*U
Here L is the lower triangular matrix
L=$$\begin{bmatrix}a&0&0\\b&c&0\\d&e&f\end{bmatrix}$$
U is the lower triangular matrix
U=$$\begin{bmatrix}1&g&h\\0&1&I\\0&0&1\end{bmatrix}$$
A=$$\begin{bmatrix}a&ag&ah\\b&bg+c&bh+cI\\d&dg+e&dh+eI+f\end{bmatrix}$$
A=LU
$$\begin{bmatrix}1&-2&3\\1&-1&2\\3&2&-1\end{bmatrix}$$=$$\begin{bmatrix}a&ag&ah\\b&bg+c&bh+cI\\d&dg+e&dh+eI+f\end{bmatrix}$$
From this substitutions and comparisons, we get the following values,
a=1
b=1
c=1
d=3
e=8
f=-2
g=-2
h=3
I=-1
Thus from the above values,
L=$$\begin{bmatrix}1&0&0\\1&1&0\\3&8&-2\end{bmatrix}$$ and U=$$\begin{bmatrix}1&-2&3\\0&1&-1\\0&0&1\end{bmatrix}$$
Since AX=B i.e. LUX=B
Assume V=UX
V = $$\begin{bmatrix}v1\\v2\\v3\end{bmatrix}$$
We can say that LV=B
$$\begin{bmatrix}1&0&0\\1&1&0\\3&8&-2\end{bmatrix} \begin{bmatrix}v1\\v2\\v3\end{bmatrix} = \begin{bmatrix}6\\9\\16\end{bmatrix}$$
From this we get,
v1=6
v2=3
v3=13
UX=V
$$\begin{bmatrix}1&-2&3\\0&1&-1\\0&0&1\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}6\\3\\13\end{bmatrix}$$

$$\begin{bmatrix}x-2y+3z\\y-z\\z\end{bmatrix} = \begin{bmatrix}6\\3\\13\end{bmatrix}$$
Hence, x=-1, y=16, z=13
Thus, the value of z is 13.

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