Diketahui matriks-matriks T = [tex]\left[\begin{array}{ccc}-3a&&a-2b\\b+c&&2d+c\\e-2d&&e-3f\end{array}\right][/tex] R = [tex]\left[\begin{array}{ccc}8&4&0\\

         [-3a        b + c     e - 2d
a) T =[a - 2b    2d + c   e - 3f]

b) [8   2]     [-3a      a - 2b]
    [4  10] = [b + c   2d + c]
    [0  -1]     [e - 2d  e - 3f]

- 3a = 8
a = - 8/3

2 = - 8/3 - 2b
14/3 = - 2b
b = -7/3

c - 7/3 = 4
c = 19/3

2d + c = 10
2d + 19/3 = 10
2d = 11/3
d = 11/6

e - 2d = 0
e - 22/6 = 0
e = 22/6

e - 3f = - 1
22/6 + 1 = 3f
28/6 = 3f
f = 28/18

a)
[tex]T ^{t} = \left[\begin{array}{ccc}-3a&b+c&e-2d\\a-2b&2d+c&e-3f\end{array}\right] [/tex]
b)
[tex]R ^{t} = \left[\begin{array}{cc}8&2\\4&10\\0&-1\end{array}\right] = T = \left[\begin{array}{cc}-3a&a-2d\\b+c&2d+c\\e-2d&e-3f\end{array}\right][/tex]
diperoleh persamaan:
-3a = 8 ...(i)
a - 2b = 2 ... (ii)
b + c = 4 ... (iii)
2d + c = 10 ...(iv)
e - 2d = 0 ... (v)
e - 3f = -1 ...(vi)
maka
-3a = 8 ...(i)
a = 2 + 2b ...(ii)
(i)(ii) substitusi
-3(2+2b) = 8
-6 + 6b = 8
6b = -14
  b - 14/6 = -2[tex] \frac{2}{3} [/tex]
(iii) b + c = 4
     2[tex] \frac{2}{3} [/tex] + c = 4
(i) a = 2b + 2
      = 2(-2[tex] \frac{2}{3} [/tex]) + 2
     = - 2[tex] \frac{2}{3} [/tex]
(iii) b + c = 4
     -2[tex] \frac{2}{3} [/tex] + c = 4
     c = 4 + 2[tex] \frac{2}{3} [/tex]
(iv) 2d + c = 10
     2d + 6[tex] \frac{2}{3} [/tex] = 3[tex] \frac{1}{3} [/tex]
       d = 1[tex] \frac{2}{3} [/tex]
(v) e - 2d = 0
    e - 2(1[tex] \frac{2}{3} [/tex] = 0
    e = 3[tex] \frac{1}{3} [/tex]
(vi) e - 2f = -1
     3[tex] \frac{1}{3} [/tex] -3f = -1
     f = 4[tex] \frac{2}{3} [/tex]