Question

+ 3y - 5z = bi Consider the linear system of equations: 3 + 4y - 8z = 62 -I - 2y + 2z = b3 (a) Show that A is not onto. (b) Using R.R.E.F., find a solvability condition on (b1,b2, 63) which guarantees a solution Adjoint Theorem. (c) Reproduce your solvability condition in (b) using the adjoint theorem.

Answer 1

The given system of equations is, x + 3y +53 - bi x + 4y - 82 = be -X - dy + 23 - b3 We can write this system in [Aib] ; ſi 3.-5°: 6,7 augmented form, Here ſi 3-57 . A = li u sl Hi-2 2] Now we find the rank of A by applying elementary roue operations, Rg Ro-Ro , Rg > Pzt R, ſi 3 -57 10 3 -57 Ryt fy-fz To ! -3 11 To 0 -3 | Lo I -3] Looo] rank of matrix A = 2 < No. of columns. → Matrix K is not a full rank matrix. So I connot be onto.

ir 11.3 -5; 6; 7 A:b] : Rot-R, (- 22 : 63 l l l3 Ti 3 -5 : bi 1 To I - båbil Rz > R3-R2 ? Ti 3 -5 : 6,7 11 3-5:6, 1 Tot -3 : bz-bil ~ loi -3: bab, I lo o o b, the lato lo o 0:26, + bg ba] The augmented mateix has a solution when gank [A: b) - vonk CAJ But here rand [Ab] = 3 and san&CA)= d. So in this case there is no solution. Of 26,4b3-ba :0 = 20, - be- bz be = ba-b? 21 then It there the are infinitely many solution. solvability condition. is