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Express the system as AX = B; then solve using matrix inverses

1. 3x – 5y = 2 –x + 2y = 0

2. x + y – z = 2 x + z = 7 2x + y + z = 13

User Chopss
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1 Answer

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Answer with explanation:

1. The given equations are

3x -5 y=2

-x+2 y= 0

The matrix in the form of , AX=B, is


A=\left[\begin{array}{cc}3&-5\\-1&2\end{array}\right] ,\\\\ X=\left[\begin{array}{c}x&y\end{array}\right],\\\\B=\left[\begin{array}{c}2&0\end{array}\right]


\rightarrow X=A^(-1)B\\\\\rightarrow X=(Adj.A)/(|A|)* B

Adj.A=Transpose of cofactor of Matrix A


Adj.A=\left[\begin{array}{cc}2&1\\5&3\end{array}\right] ,\\\\ |A|=6-5\\\\|A|=1\\\\\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{cc}2&5\\1&3\end{array}\right] * \left[\begin{array}{c}2&0\end{array}\right]\\\\x=4, y=2

2.

The given equations are

x+y-z=2

x+z=7

2 x +y+z=13

The matrix in the form of , AX=B, is


A=\left[\begin{array}{ccc}1&1&-1\\1&0&1\\2&1&1\end{array}\right]\\\\ X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\\\\B= \left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\\rightarrow X=A^(-1)B\\\\\rightarrow X=(Adj.A)/(|A|)* B\\\\a_(11)=-1,a_(12)=1,a_(13)=1,a_(21)=-2,a_(22)=3,a_(23)=1,a_(31)=1,a_(32)=-2,a_(33)=-1\\\\|A|=1*(0-1)-1*(1-2)-1*(1-0)\\\\=-1+1-1\\\\|A|=-1\\\\Adj.A=\left[\begin{array}{ccc}-1&-2&1\\1&3&-2\\1&1&-1\end{array}\right]


(Adj.A)/(|A|)=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\\\\X=A^(-1)B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]*\left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\x=3,y=3,z=4

User Tarun Dugar
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