Write an equation of the perpendicular bisector of the segment with endpoints M(1,5) and N(7,−1).

Mathematics

1 answer:

solong [7]3 years ago

6 0

Answer:

Step-by-step explanation:

find the perpendicular bisector of a line segment with endpoints

(ii) Find a point on the perpendicular bisector (the midpoint of the given line segment) using the midpoint formula:

( x 3 , y 3 ) = ( x 1 + x 2 2 , y 1 + y 2 2 )

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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

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