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Use the fact that matrices A and B are row-equivalent. A = 1 2 1 0 0 2 5 1 1 0 3 7 2 2 -2 10 23 7 -2 10 1 0 3 0-4 0 1 -1 0 2

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2 1 0 0 A- 25 - - do 10 3 0-4 0 1 -1 2 O BE 37 2 - 2 lo 23 2 ملم -2 10 o 2 I OO 3 25 37 -2 2-2 10 23 7-2 10 find bases for th2 1 0 0 - -1 0 I-2 10 R: Rut5R3 I 2 - 1 4 o 1 2 نا 0 0 Ranc 3 Row .: Basis for (། ༢ » -- 0, 0) Row of = co -1, 1 (0,0,0 1,-2By eqo - - 1-1 2 R,: R, 2r, R : R2-R3 3 O- 20 - ~ - 1 2 2 с lo 3 - 2 0 1 -1 2 2 - 0 0 R,: RitzR3 1 0 3 - 4 0 2 - B 0 0 of=and %-d3 + 2x,- =0 x2+2x= 3 and 2, + 3( 22 +225)– 4%- x,+ 382 + 6 xs - 4x = 0 =0 2, + 372 +2x, a -1322 + 2x;-) , - (3^2+285)

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