This has four parts to it. the following system oflinear equations x - y =2z=13 2x +2y -z =-6 -x + 3y +z = -7 a) Provide a coefficient matrix corresponding to the system oflinear equations b) what is the inverse of this matrix? c) What is the transpose of the matrix? d) find the determinant for this matrix?

This has four parts to it. the following system of linear equations
x - y =2z=13
2x +2y -z =-6
-x + 3y +z = -7
a) Provide a coefficient matrix corresponding to the system oflinear equations
b) what is the inverse of this matrix?
c) What is the transpose of the matrix?
d) find the determinant for this matrix?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

coolng90qo
a) $\left[\begin{array}{cccc}1& -1& -2& 13\\ 2& 2& -1& -6\\ -1& 3& 1& -7\end{array}\right]$
b) $\left[\begin{array}{cccccc}1& -1& -2& 1& 0& 0\\ 2& 2& -1& 0& 1& 0\\ -1& 3& 1& 0& 0& 1\end{array}\right]=\left[\begin{array}{cccccc}1& -1& -2& 1& 0& 0\\ 0& 4& 3& -2& 1& 0\\ -1& 3& 1& 0& 0& 1\end{array}\right]-R1+R2\to R2$
$\left[\begin{array}{cccccc}1& -1& -2& 1& 0& 0\\ 0& 4& 3& -2& 1& 0\\ 0& 2& -1& 1& 0& 1\end{array}\right]R3+R1\to R3$
$\left[\begin{array}{cccccc}1& -1& -2& 1& 0& 0\\ 0& 4& 3& -2& 1& 0\\ 0& 0& 5& -4& 1& -2\end{array}\right]-R3+R2\to R3$
$\left[\begin{array}{cccccc}1& 0& -5/4& 1/2& 1/4& 0\\ 0& 1& 3/4& -1/2& 1/4& 0\\ 0& 0& 1& -4/5& 1/5& -2/5\end{array}\right]-R1+R2\to R1$
$\left[\begin{array}{cccccc}1& 0& -5/4& 1/2& 1/4& 0\\ 0& 1& 0& 1/10& 1/10& 3/10\\ 0& 0& 1& -4/5& 1/5& -2/5\end{array}\right]-R1-\left(3/4\right)R3\to R2$
Finally
$R1+\left(5/4\right)R3\to R1$
$\left[\begin{array}{cccccc}1& 0& 0& -1/2& 1/2& -1/2\\ 0& 1& 0& 1/10& 1/10& 3/10\\ 0& 0& 1& -4/5& 1/5& -2/5\end{array}\right]$
$A=\left[\begin{array}{ccc}1& -1& -2\\ 2& 2& -1\\ -1& 3& 1\end{array}\right],{A}^{-1}=\left[\begin{array}{ccc}-1/2& 1/2& -1/2\\ 1/10& 1/10& 3/10\\ -4/5& 1/5& -2/5\end{array}\right]$
c) transpose: $A={A}^{T}$
d) Det(A)=-10
Det(A)=1(2+3)+1(2-1)-2(6+2)=-10