Answer:
The value of X will be the following :
[tex]\begin{bmatrix}-20\\ -5\\ -18\end{bmatrix}[/tex]
Step-by-step explanation:
So as you can tell, through substitution the equation for this problem will be as follows,
[tex]\begin{bmatrix}1&-4&-2\\ \:-2&2&5\\ \:\:\:\:\:2&-4&-2\end{bmatrix}^{^{^{^{-1}}}}\cdot \:X\:=\:\begin{bmatrix}2\\ \:\:7\\ \:-3\end{bmatrix}[/tex]
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 [tex]*[/tex] B,
[tex]X = A * B = \begin{bmatrix}1&-4&-2\\ \:\:-2&2&5\\ \:\:\:2&-4&-2\end{bmatrix}^{\:}\begin{bmatrix}2\\ 7\\ \:-3\end{bmatrix}[/tex]
To solve for X we can multiply the rows of the first matrix by the respective columns of the second matrix,
[tex]\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}[/tex]
[tex]X = \begin{bmatrix}-20\\ -5\\ -18\end{bmatrix}[/tex] - if this matrix is matrix 1, matrix 1 will be our solution
