Matrix Inversion Questions and Answers – Crout’s Method

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

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.

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

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.