Using matrices, solve the following system of equations:2x+3y+3z=5,x𕒶y+z=𕒸,3x−y𕒶z=3.Using matrices, solve the following system of equations:2x+3y+3z=5,x𕒶y+z=𕒸,3x−y𕒶z=3.2x+3y+3z=5,x𕒶y+z=𕒸,3x−y𕒶z=3.2x+3y+3z=5,x𕒶y+z=𕒸,3x−y𕒶z=3.2x+3y+3z=5,x𕒶y+z=𕒸,3x−y𕒶z=3.22xx++33yy++33zz==55,,xx−𕒶2yy++zz==−󔼴,,33xx−−yy−𕒶2zz==3.3.?

The given system of equation can be expressed can be represented in matrix form as AX = B, whereA=∣∣∣∣2331𕒶13𕒵𕒶∣∣∣∣,X=∣∣∣∣xyz∣∣∣∣,B=∣∣∣∣5󔼳∣∣∣∣Now|A|=∣∣∣∣2331𕒶13𕒵𕒶∣∣∣∣=2(4+1)𕒷(𕒶𕒷)+3(𕒵+6)⇒10+15+15=40≠0C11=(𕒵)1+1∣∣∣𕒶1𕒵𕒶∣∣∣=4+1=5C12=(𕒵)1+2∣∣∣113𕒶∣∣∣=−(𕒶𕒷)=5C13=(𕒵)1+3∣∣∣1𕒶3𕒵∣∣∣=𕒵+6=5C21=(𕒵)2+1∣∣∣33𕒵𕒶∣∣∣=−(𕒺+3)=3C22=(𕒵)2+2∣∣∣233𕒶∣∣∣=(𕒸𕒽)=󔼕C23=(𕒵)2+3∣∣∣233𕒵∣∣∣=−(𕒶𕒽)=11C31=(𕒵)3+1∣∣∣33𕒶1∣∣∣=3+6=9C32=(𕒵)3+2∣∣∣2311∣∣∣=−(2𕒷)=1C33=(𕒵)3+3∣∣∣231𕒶∣∣∣=𕒸𕒷=𕒻AdjA=∣∣∣∣5553�𕒻∣∣∣∣T=∣∣∣∣5395�𕒻∣∣∣∣A𕒵=1|A|adjA=140∣∣∣∣5395�𕒻∣∣∣∣AX=B⇒X=A𕒵BTherefore,⎡⎢⎣xyz⎤⎥⎦=140∣∣∣∣5395�𕒻∣∣∣∣∣∣∣∣5󔼳∣∣∣∣=1/40[ 25-12+275+52+3 5-44-21 ]=1/40[ 400 ­40 ]∣∣∣∣xyz∣∣∣∣=∣∣∣∣12𕒵∣∣∣∣Equating the corresponding elements we getx=1,y=2,z=𕒵.