Amari Flowers

2021-03-06

Solve the systems of equations using matrices.
$4x+y+z=3$
$-x+y-2z=-11$
$x+2y+2z=-1$

gotovub

Step 1
Write all the coefficients of x in one column, all the coefficients of y in one column and all the coefficients of z in one column to obtain the A matrix.
Wite x ,y ,z in one column of a matrix to obtain the matrix of variables be X matrix.
write all the constants in a single column of matrix to get the matrix of all constants let it be B matrix.
Take the dot product of A matrix and X matrix and equate it to B matrix to get it in the form $AX=B$.
$\left[\begin{array}{ccc}4& 1& 1\\ -1& 1& -2\\ 1& 2& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}3\\ -11\\ -1\end{array}\right]$
Step 2
Find the inverse of A matrix and multiply it with B matrix to obtain the value of x,y,z.
Find the matrix of minors and apply alternate negative sign on elements of the matrix to obtain the matrix of cofactors.
take the transpose of the matrix of cofactors and multiply it by the determinant of A matrix to obtain the inverse matrix.
$\left[\begin{array}{ccc}2+4& -2+2& -2-2\\ 2-2& 8-1& 8-1\\ -2-1& -8+1& 4+1\end{array}\right]=\left[\begin{array}{ccc}6& 0& -3\\ 0& 7& 7\\ -3& -7& 5\end{array}\right]$
$=\left[\begin{array}{ccc}\left(+\right)6& \left(-\right)0& \left(+\right)-3\\ \left(-\right)0& \left(+\right)7& \left(-\right)7\\ \left(+\right)-3& \left(-\right)7& \left(+\right)5\end{array}\right]$
$=\left[\begin{array}{ccc}6& 0& -3\\ 0& 7& -7\\ -3& 7& 5\end{array}\right]$
$AdjA=\left[\begin{array}{ccc}6& 0& -3\\ 0& 7& 7\\ -3& -7& 5\end{array}\right]$
${A}^{-1}=\left[\begin{array}{ccc}6& 0& -3\\ 0& 7& 7\\ -3& -7& 5\end{array}\right]\left(\frac{1}{21}\right)$
$=\left[\begin{array}{ccc}\frac{2}{7}& 0& -\frac{1}{3}\\ 0& \frac{1}{3}& \frac{1}{3}\\ -\frac{1}{7}& -\frac{1}{3}& \frac{5}{21}\end{array}\right]$
Step 3 Substitute the value of ${A}^{-1}$ in $X={A}^{-1}$ B to obtain the values of x,y,z.
$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{ccc}\frac{2}{7}& 0& -\frac{1}{7}\\ 0& \frac{1}{3}& \frac{1}{3}\\ -\frac{1}{7}& -\frac{1}{3}& \frac{5}{21}\end{array}\right]\left[\begin{array}{c}3\\ -11\\ -1\end{array}\right]$
$=\left[\begin{array}{c}\frac{6}{7}+0+\frac{1}{7}\\ 0-\frac{11}{3}-\frac{1}{3}\\ -\frac{3}{7}-\frac{11}{3}-\frac{5}{21}\end{array}\right]$
$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ -4\\ 3\end{array}\right]$
thus the value of x is 1, y is -4 and z is 3.

Jeffrey Jordon

Answer is given below (on video)

Jeffrey Jordon