Question

Linear Algebra.

Given a system of the form -M1X1 + X2 = 61 -m2X1 + X2 = b 2 where my, m2, 64, and bz are constants, (a) Show that the system will have a unique solution if m4 m2. (b) Show that if m1 = m2, then the system will be consistent only if b1 = b2 (c) Give a geometric interpretation of parts (a) and (b).

Answer 1

Answer :- -P7,7, &la=bi -M₂2, A4, = b 2 ni 1 - TO -mini + R2 - bo o ang, & X₂ - b2 = 0 solving by cross-multiplication, we have 22 ixtbo) - {xtbi) trix (ob, )-(m.) (b.) -Mix!-(-) ra -bztb, maba-mabi om,tm. X biobz mibg-mabi mem, we observe that a unique value of x, and X2 does exist if m₂-m, to m2 fm, proved 희 12

b) Now, mam,im (let) -mxit X2 = 6 -mx2 + x₂ = 62 -mx + x₂-bioo -mxl + x2. -52=0 solving by cross-multiplication, TA howe XI mm) of -batby mbaombi x X2 berbe mlb2-b.) we observe that when birba then system solution is possible that means system u consistent. otherwise it will in consistent means no solution exist. proved

AX2-axis x = mix, tb, a bil x-axis be m2 has *2= miatt, X2=m. x₂ + b 2 het my has positive X₂ = mq x₂ + b 2 value ie le positive slope has unique system Solution and negative volue ie negative stope when miem,om X₂ = mx, tb, and bi=b2 x₂ = mathi ie system is consistent and have infinite no. of Solution. X2 b Xi