Using matrices, solve the following system of equations:3x+4y+7z=4,2x−y+3z=𕒷x+2y𕒷z=8Using matrices, solve the following system of equations:3x+4y+7z=4,2x−y+3z=𕒷x+2y𕒷z=83x+4y+7z=4,2x−y+3z=𕒷x+2y𕒷z=83x+4y+7z=4,2x−y+3z=𕒷x+2y𕒷z=83x+4y+7z=4,2x−y+3z=𕒷x+2y𕒷z=833xx++44yy++77zz==44,,22xx−−yy++33zz==−󔼩;xx++22yy−󔼩zz==88?

The given system of equation can be expressed can be represented in matrix form as AX = B, whereA=∣∣∣∣3472�𕒷∣∣∣∣,X=∣∣∣∣xyz∣∣∣∣,B=∣∣∣∣4󔼮∣∣∣∣Now|A|=∣∣∣∣3472�𕒷∣∣∣∣=3(3𕒺)𕒸(𕒺𕒷)+7(4+1)⇒𕒽+36+35=62≠0C11=(𕒵)1+1∣∣∣�𕒷∣∣∣=3𕒺=𕒷C12=(𕒵)1+2∣∣∣231𕒷∣∣∣=−(𕒺𕒷)=9C13=(𕒵)1+3∣∣∣2�∣∣∣=4+1=5C21=(𕒵)2+1∣∣∣472𕒷∣∣∣=−(󔼔󔼖)=26C22=(𕒵)2+2∣∣∣371𕒷∣∣∣=𕒽𕒻=󔼘C23=(𕒵)2+3∣∣∣3412∣∣∣=−(6𕒸)=𕒶C31=(𕒵)3+1∣∣∣47󔼕∣∣∣=12+7=19C32=(𕒵)3+2∣∣∣3723∣∣∣=−(9󔼖)=5C33=(𕒵)3+3∣∣∣342𕒵∣∣∣=𕒷𕒼=󔼓AdjA=∣∣∣∣�󔼘𕒶195󔼓∣∣∣∣T=∣∣∣∣��𕒶󔼓∣∣∣∣A𕒵=1|A|adjA=162∣∣∣∣��𕒶󔼓∣∣∣∣AX=B⇒X=A𕒵BTherefore,⎡⎢⎣xyz⎤⎥⎦=162∣∣∣∣��𕒶󔼓∣∣∣∣∣∣∣∣4󔼮∣∣∣∣=162⎡⎢⎣󔼔󔽖+15236+48+4020+6󔽠⎤⎥⎦=162⎡⎢⎣62124𕒺2⎤⎥⎦∣∣∣∣xyz∣∣∣∣=∣∣∣∣12𕒵∣∣∣∣Equating the corresponding elements we getx=1,y=2,z=𕒵.