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University of California, Berkeley *

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Mathematics

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Apr 3, 2024

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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}:
The coefficient matrix of the system is: $A= \begin{bmatrix} 1 & 1 & 1\\ -1 & 3 & 2\\ 2 & 2 & 1 \end{bmatrix}$ Find the inverse of $A$ using your graphing calculator. $$ A^{-1}= \frac{1}{4} \begin{bmatrix} -1 & 0 & 1\\ -5 & 1 & 3\\ 8 & 0 & -4 \end{bmatrix}$$ Step 2 : Solve: X = A 1 B [ x y z ] = 1 4 [ 1 1 1 5 1 3 8 0 4 ][ 5 2 1 ] ¿ 1 4 [ 5 2 + 1 25 + 2 + 3 40 + 0 4 ] ¿ [ 1 5 9 ]
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