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3 years ago

Help please I got to pass

Mathematics

1 answer:

mixer [17]3 years ago

4 0

Answer:

A'(5,3)

Step-by-step explanation:

First, you must understand that A(5,3) is the pre-image and that A' is what we are looking for which is the image.

With that in mind, you translate the preimage by adding or subtracting from the x and y values.

(x+4, y-3)

To find the x value of the pre-image, you will add 4 to the preimages' x value

Pre-image: A(1,6)

To find the y value, you subtract the preimages' y value by 3.

Hope this helps!

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[10 points] Given matrix A =  2 2 3, −6 −7 8 (a) (5 points). Show that A has no LU decomposition. (b) (5 points). Find the dec

ch4aika [34]

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 = $\left[\begin{array}{ccc}0&3&4\\1&2&3\\-3&-7&8\end{array}\right]$

Here,

$det(A) =\left|\begin{array}{ccc}0&3&4\\1&2&3\\-3&-7&8\end{array}\right| = -55 \neq 0.$

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

$\left[\begin{array}{ccc}1&2&3\\0&3&4\\-3&-7&8\end{array}\right]$

R3+3R1=R3

$\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&-1&17\end{array}\right]$

R3+(1/3)R2 = R3

$\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right]$

Therefore, U = $\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right]$ .

Here, LP = E₁₂=E₃₁=-3 &E₃₂=-1/3

$LP=\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\-3&0&1\end{array}\right]\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&-1/3&1\end{array}\right]$

$LP=\left[\begin{array}{ccc}0&1&0\\1&0&0\\-3&-1/3&1\end{array}\right]$

$LP=\left[\begin{array}{ccc}1&0&0\\0&1&0\\-1/3&-3&1\end{array}\right]\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right]$

So now U is given as

$U=\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right]\\L=\left[\begin{array}{ccc}1&0&0\\0&1&0\\-1/3&-3&1\end{array}\right]\\P=\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right]\\$

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