189k views
3 votes
-50 points- matrix system

-50 points- matrix system-example-1
User Somi Meer
by
8.4k points

2 Answers

5 votes

Answer:

The value of X will be the following :


\begin{bmatrix}-20\\ -5\\ -18\end{bmatrix}

Explanation:

So as you can tell, through substitution the equation for this problem will be as follows,


\begin{bmatrix}1&-4&-2\\ \:-2&2&5\\ \:\:\:\:\:2&-4&-2\end{bmatrix}^{^{^{^(-1)}}}\cdot \:X\:=\:\begin{bmatrix}2\\ \:\:7\\ \:-3\end{bmatrix}

Therefore to isolate X, we have to multiply the inverse of the inverse of the co - efficient of X on either side, such that X = A
* B,


X = A * B = \begin{bmatrix}1&-4&-2\\ \:\:-2&2&5\\ \:\:\:2&-4&-2\end{bmatrix}^(\:)\begin{bmatrix}2\\ 7\\ \:-3\end{bmatrix}

To solve for X we can multiply the rows of the first matrix by the respective columns of the second matrix,


\begin{bmatrix}1&-4&-2\\ -2&2&5\\ 2&-4&-2\end{bmatrix}\begin{bmatrix}2\\ 7\\ -3\end{bmatrix} = \begin{bmatrix}1\cdot \:2+\left(-4\right)\cdot \:7+\left(-2\right)\left(-3\right)\\ \left(-2\right)\cdot \:2+2\cdot \:7+5\left(-3\right)\\ 2\cdot \:2+\left(-4\right)\cdot \:7+\left(-2\right)\left(-3\right)\end{bmatrix} = \begin{bmatrix}-20\\ -5\\ -18\end{bmatrix}


X = \begin{bmatrix}-20\\ -5\\ -18\end{bmatrix} - if this matrix is matrix 1, matrix 1 will be our solution

User Barif
by
7.8k points
3 votes

Answer:

-20

-5

-18

Explanation:

AX = B to find x

A^-1 AX = A^-1 B

X = 1 -4 -2 2

-2 2 5 * 7

2 -4 -2 -3

We multiply across and down

-1 *2 + -4 *7 -2 *-3 = -20

-2 * 2 + 2 * 7 + 5 * -3 = -5

2 * 2 -4 * 7 -2 * -3 = -18

The matrix is

-20

-5

-18

User Grosser
by
8.0k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories