Answer:
Both the answers are as in the solution.
Step-by-step explanation:
As the given matrix is not in the readable form, a similar question is found online and the solution of which is attached herewith.
Part a:
Given matrix is : A = [tex]\left[\begin{array}{ccc}0&3&4\\1&2&3\\-3&-7&8\end{array}\right][/tex]
Here,
[tex]det(A) =\left|\begin{array}{ccc}0&3&4\\1&2&3\\-3&-7&8\end{array}\right| = -55 \neq 0.[/tex]
Then, A is non-singular matrix.
Here, A₁₁= 0.
If we write A as LU with L lower triangular matrix and U upper triangular matrix, then A₁₁=L₁₁U₁₁.
So, As
A₁₁ = 0 gives L₁₁U₁₁= 0 ,
This indicates that either L₁₁= 0 or U₁₁ = 0.
If L₁₁= 0 or U₁₁ = 0, this would made the corresponding matrix singular, which contradicts the condition as A is non-singular.
Therefore, A has no LU decomposition.
Part b:
By the implementation of the various row operations
interchange R1 and R2
[tex]\left[\begin{array}{ccc}1&2&3\\0&3&4\\-3&-7&8\end{array}\right][/tex]
R3+3R1=R3
[tex]\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&-1&17\end{array}\right][/tex]
R3+(1/3)R2 = R3
[tex]\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right][/tex]
Therefore, U = [tex]\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right][/tex].
Here, LP = E₁₂=E₃₁=-3 &E₃₂=-1/3
[tex]LP=\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\-3&0&1\end{array}\right]\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&-1/3&1\end{array}\right][/tex]
[tex]LP=\left[\begin{array}{ccc}0&1&0\\1&0&0\\-3&-1/3&1\end{array}\right][/tex]
[tex]LP=\left[\begin{array}{ccc}1&0&0\\0&1&0\\-1/3&-3&1\end{array}\right]\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right][/tex]
So now U is given as
[tex]U=\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right]\\L=\left[\begin{array}{ccc}1&0&0\\0&1&0\\-1/3&-3&1\end{array}\right]\\P=\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right]\\[/tex]
