Sistem Persamaan Linear Homogen

Sistem Persamaan Linear Homogen

Yaitu sistem persamaan linear (SPL) yang semua suku konstan atau nilai ruas kanannya adalah nol.

Bentuk umum:

a11x1  a12x2  ... a1nxn = 0
a21x1  a22x2  ... a2nxn = 0
am1x1  am2x2  ... amnxn = 0

Sistem Persamaan Linear Homogen 3 Persamaan dan 3 VariabelSunting

a11x1  a12x2  a13x3 = 0
a21x1  a22x2  a23x3 = 0
a31x1  a32x2  a33x3 = 0

SPL Homogen dapat diselesaikan dengan metode Operasi Baris Elementer. Maka, SPL Homogen tersebut diubah menjadi matriks:

{\displaystyle \left[{\begin{array}{rrr|r}a_{11}&a_{12}&a_{13}&0\\a_{21}&a_{22}&a_{32}&0\\a_{31}&a_{32}&a_{33}&0\\\end{array}}\right]}{\displaystyle \left[{\begin{array}{rrr|r}a_{11}&a_{12}&a_{13}&0\\a_{21}&a_{22}&a_{32}&0\\a_{31}&a_{32}&a_{33}&0\\\end{array}}\right]}

SPL Homogen ini mempunyai dua kemungkinan solusi, yaitu solusi trivial dan non trivial.

Solusi Trivial

Contoh: {\displaystyle \left[{\begin{array}{rrr|r}1&2&1&0\\1&3&2&0\\2&1&2&0\\\end{array}}\right]}{\displaystyle \left[{\begin{array}{rrr|r}1&2&1&0\\1&3&2&0\\2&1&2&0\\\end{array}}\right]}

Penyelesaian:

{\displaystyle \left[{\begin{array}{rrr|r}1&2&1&0\\0&1&1&0\\0&-3&0&0\\\end{array}}\right]}{\displaystyle \left[{\begin{array}{rrr|r}1&2&1&0\\0&1&1&0\\0&-3&0&0\\\end{array}}\right]} B2 - B1, B3 - 2.B1

{\displaystyle \left[{\begin{array}{rrr|r}1&2&1&0\\0&1&1&0\\0&0&3&0\\\end{array}}\right]}{\displaystyle \left[{\begin{array}{rrr|r}1&2&1&0\\0&1&1&0\\0&0&3&0\\\end{array}}\right]} B3 3.B2

Det = 1 x 1 x 3 = 3

Karena det ≠ 0, solusi SPL Homogen tersebut trivial yaitu x1 = x2 = x3 = 0.

Solusi Non Trivial

Contoh: {\displaystyle \left[{\begin{array}{rrr|r}1&2&1&0\\1&3&2&0\\2&1&-1&0\\\end{array}}\right]}{\displaystyle \left[{\begin{array}{rrr|r}1&2&1&0\\1&3&2&0\\2&1&-1&0\\\end{array}}\right]}

Penyelesaian:

{\displaystyle \left[{\begin{array}{rrr|r}1&2&1&0\\0&1&1&0\\0&-3&-3&0\\\end{array}}\right]}{\displaystyle \left[{\begin{array}{rrr|r}1&2&1&0\\0&1&1&0\\0&-3&-3&0\\\end{array}}\right]} B2 - B1, B3 - 2.B1

{\displaystyle \left[{\begin{array}{rrr|r}1&2&1&0\\0&1&1&0\\0&0&0&0\\\end{array}}\right]}{\displaystyle \left[{\begin{array}{rrr|r}1&2&1&0\\0&1&1&0\\0&0&0&0\\\end{array}}\right]} B3 3.B2

{\displaystyle \left[{\begin{array}{rrr|r}1&0&-1&0\\0&1&1&0\\0&0&0&0\\\end{array}}\right]}{\displaystyle \left[{\begin{array}{rrr|r}1&0&-1&0\\0&1&1&0\\0&0&0&0\\\end{array}}\right]} B1 - 2.B2

Det = 1 x 1 x 0 = 0

Maka, {\displaystyle {\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\\end{bmatrix}}}{\displaystyle {\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\\end{bmatrix}}} = {\displaystyle {\begin{bmatrix}1\\-1\\1\\\end{bmatrix}}}{\displaystyle {\begin{bmatrix}1\\-1\\1\\\end{bmatrix}}}t


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