1.4

Question and Solution Template Learning Attribute(s) Included in Question : 1.4.10 Find the inverse of a matrix if it exists and use it to solve systems of linear equations (using technology for matrices of dimension 3x3 or greater). Calculator Active? Yes Question : Solve the below system using matrix inverse methods. x + y + z = 5 − x + 3 y + 2 z = 2 2 x + 2 y + z = 1 A) ( x , y , z ) = ( 1,2,3 ) B) ( x , y , z ) = ( − 1 , − 5,9 ) C) ( x , y , z ) = ( 1,5,9 ) D) ( x , y , z )=( 1 , − 5,9 ) Correct Answer: D

Equation Upload (Please write the text of the question along with the LaTeX Code): Solve the below system using matrix inverse methods. $x+y+z=5$ $-x+3y+2z=2$ $2x+2y+z =1$ A) $(x,y,z)=(1,2,3)$ B) $(x,y,z)=(-1,-5,9$) C) $(x,y,z)=(1,5,9)$ D) $(x,y,z)=(1,-5,9$) On a scale of 1-10, how difficult would you estimate your question to be (1=easy, 10=extremely difficult): 10 i Solution : Step 1 : The coefficient matrix of the system is: A = [ 1 1 1 − 1 3 2 2 2 1 ] Find the inverse of A using your graphing calculator. A − 1 = 1 4 [ 1 − 1 1 − 5 1 3 8 0 − 4 ] Equation Upload (Please write the text of the question along with the LaTeX Code): \textbf{Step 1}:

This shows x = 1 , y =− 5 , and z = 9 . Equation Upload (Please write the text of the question along with the LaTeX Code): \textbf{Step 2}: Solve: $X=A^{-1}B$ $$ \begin{bmatrix} x \\ y \\ y \end{bmatrix}= \frac{1}{4} \begin{bmatrix} 1 & -1 & 1\\ -5 & 1 & 3\\ 8 & 0 & -4 \end{bmatrix} \begin{bmatrix} 5 \\ 2 \\ 1 \end{bmatrix}$$

$$= \frac{1}{4}\begin{bmatrix} 5-2+1 \\ -25+2+3 \\ 40+0-4 \end{bmatrix}$$ $$=\begin{bmatrix} 1 \\ -5 \\ 9 \end{bmatrix}$$ This shows $x=1, $y=-5$, and $z=9$.