# Matrix Inversion Questions and Answers – Crout’s Method

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

Answer: d
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
View Answer

Answer: a
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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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn