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Question

# Using matrices, solve the following system of equations:2x+3y+3z=5,x−2y+z=−4,3x−y−2z=3.

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Solution

## The given system of equation can be expressed can be represented in matrix form as AX = B, whereA=∣∣ ∣∣2331−213−1−2∣∣ ∣∣,X=∣∣ ∣∣xyz∣∣ ∣∣,B=∣∣ ∣∣5−43∣∣ ∣∣Now |A|=∣∣ ∣∣2331−213−1−2∣∣ ∣∣=2(4+1)−3(−2−3)+3(−1+6)⇒10+15+15=40≠0C11=(−1)1+1∣∣∣−21−1−2∣∣∣=4+1=5C12=(−1)1+2∣∣∣113−2∣∣∣=−(−2−3)=5C13=(−1)1+3∣∣∣1−23−1∣∣∣=−1+6=5C21=(−1)2+1∣∣∣33−1−2∣∣∣=−(−6+3)=3C22=(−1)2+2∣∣∣233−2∣∣∣=(−4−9)=−13C23=(−1)2+3∣∣∣233−1∣∣∣=−(−2−9)=11C31=(−1)3+1∣∣∣33−21∣∣∣=3+6=9C32=(−1)3+2∣∣∣2311∣∣∣=−(2−3)=1C33=(−1)3+3∣∣∣231−2∣∣∣=−4−3=−7AdjA=∣∣ ∣∣5553−131191−7∣∣ ∣∣T=∣∣ ∣∣5395−131511−7∣∣ ∣∣A−1=1|A|adjA=140∣∣ ∣∣5395−131511−7∣∣ ∣∣AX=B⇒X=A−1BTherefore, ⎡⎢⎣xyz⎤⎥⎦=140∣∣ ∣∣5395−131511−7∣∣ ∣∣∣∣ ∣∣5−43∣∣ ∣∣$=\frac{1}{40}\begin{bmatrix}25-12+27\\25+52+3 \\25-44-21 \end{bmatrix}$$=\frac{1}{40}\begin{bmatrix}40\\80 \\-40\end{bmatrix}$∣∣ ∣∣xyz∣∣ ∣∣=∣∣ ∣∣12−1∣∣ ∣∣Equating the corresponding elements we getx=1,y=2,z=−1.

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