Aljabar Linear Contoh

$x + y + z = 2$ , $4 x + 5 y + z = 12$ , $2 x = - 4$

Langkah 1

Tentukan $A X = B$ dari sistem persamaan tersebut.

$[\begin{array}{c} 1 & 1 & 1 \\ 4 & 5 & 1 \\ 2 & 0 & 0 \end{array}] \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 2 \\ 12 \\ - 4 \end{array}]$

Langkah 2

Tentukan balikan dari matriks koefisien.

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Langkah 2.1

Find the determinant.

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Langkah 2.1.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in row $3$ by its cofactor and add.

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Langkah 2.1.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Langkah 2.1.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Langkah 2.1.1.3

The minor for $a_{31}$ is the determinant with row $3$ and column $1$ deleted.

$| \begin{array}{c} 1 & 1 \\ 5 & 1 \end{array} |$

Langkah 2.1.1.4

Multiply element $a_{31}$ by its cofactor.

$2 | \begin{array}{c} 1 & 1 \\ 5 & 1 \end{array} |$

Langkah 2.1.1.5

The minor for $a_{32}$ is the determinant with row $3$ and column $2$ deleted.

$| \begin{array}{c} 1 & 1 \\ 4 & 1 \end{array} |$

Langkah 2.1.1.6

Multiply element $a_{32}$ by its cofactor.

$0 | \begin{array}{c} 1 & 1 \\ 4 & 1 \end{array} |$

Langkah 2.1.1.7

The minor for $a_{33}$ is the determinant with row $3$ and column $3$ deleted.

$| \begin{array}{c} 1 & 1 \\ 4 & 5 \end{array} |$

Langkah 2.1.1.8

Multiply element $a_{33}$ by its cofactor.

$0 | \begin{array}{c} 1 & 1 \\ 4 & 5 \end{array} |$

Langkah 2.1.1.9

Add the terms together.

$2 | \begin{array}{c} 1 & 1 \\ 5 & 1 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ 4 & 1 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ 4 & 5 \end{array} |$

Langkah 2.1.2

Kalikan $0$ dengan $| \begin{array}{c} 1 & 1 \\ 4 & 1 \end{array} |$ .

$2 | \begin{array}{c} 1 & 1 \\ 5 & 1 \end{array} | + 0 + 0 | \begin{array}{c} 1 & 1 \\ 4 & 5 \end{array} |$

Langkah 2.1.3

Kalikan $0$ dengan $| \begin{array}{c} 1 & 1 \\ 4 & 5 \end{array} |$ .

$2 | \begin{array}{c} 1 & 1 \\ 5 & 1 \end{array} | + 0 + 0$

Langkah 2.1.4

Evaluasi $| \begin{array}{c} 1 & 1 \\ 5 & 1 \end{array} |$ .

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Langkah 2.1.4.1

Determinan dari matriks $2 \times 2$ dapat dicari menggunakan rumus $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$2 (1 \cdot 1 - 5 \cdot 1) + 0 + 0$

Langkah 2.1.4.2

Sederhanakan determinannya.

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Langkah 2.1.4.2.1

Sederhanakan setiap suku.

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Langkah 2.1.4.2.1.1

Kalikan $1$ dengan $1$ .

$2 (1 - 5 \cdot 1) + 0 + 0$

Langkah 2.1.4.2.1.2

Kalikan $- 5$ dengan $1$ .

$2 (1 - 5) + 0 + 0$

Langkah 2.1.4.2.2

Kurangi $5$ dengan $1$ .

$2 \cdot - 4 + 0 + 0$

Langkah 2.1.5

Sederhanakan determinannya.

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Langkah 2.1.5.1

Kalikan $2$ dengan $- 4$ .

$- 8 + 0 + 0$

Langkah 2.1.5.2

Tambahkan $- 8$ dan $0$ .

$- 8 + 0$

Langkah 2.1.5.3

Tambahkan $- 8$ dan $0$ .

$- 8$

Langkah 2.2

Since the determinant is non-zero, the inverse exists.

Langkah 2.3

Set up a $3 \times 6$ matrix where the left half is the original matrix and the right half is its identity matrix.

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 4 & 5 & 1 & 0 & 1 & 0 \\ 2 & 0 & 0 & 0 & 0 & 1 \end{array}]$

Langkah 2.4

Tentukan bentuk eselon baris yang dikurangi.

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Langkah 2.4.1

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Langkah 2.4.1.1

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 4 - 4 \cdot 1 & 5 - 4 \cdot 1 & 1 - 4 \cdot 1 & 0 - 4 \cdot 1 & 1 - 4 \cdot 0 & 0 - 4 \cdot 0 \\ 2 & 0 & 0 & 0 & 0 & 1 \end{array}]$

Langkah 2.4.1.2

Sederhanakan $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - 3 & - 4 & 1 & 0 \\ 2 & 0 & 0 & 0 & 0 & 1 \end{array}]$

Langkah 2.4.2

Perform the row operation $R_{3} = R_{3} - 2 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Langkah 2.4.2.1

Perform the row operation $R_{3} = R_{3} - 2 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - 3 & - 4 & 1 & 0 \\ 2 - 2 \cdot 1 & 0 - 2 \cdot 1 & 0 - 2 \cdot 1 & 0 - 2 \cdot 1 & 0 - 2 \cdot 0 & 1 - 2 \cdot 0 \end{array}]$

Langkah 2.4.2.2

Sederhanakan $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - 3 & - 4 & 1 & 0 \\ 0 & - 2 & - 2 & - 2 & 0 & 1 \end{array}]$

Langkah 2.4.3

Perform the row operation $R_{3} = R_{3} + 2 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Langkah 2.4.3.1

Perform the row operation $R_{3} = R_{3} + 2 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - 3 & - 4 & 1 & 0 \\ 0 + 2 \cdot 0 & - 2 + 2 \cdot 1 & - 2 + 2 \cdot - 3 & - 2 + 2 \cdot - 4 & 0 + 2 \cdot 1 & 1 + 2 \cdot 0 \end{array}]$

Langkah 2.4.3.2

Sederhanakan $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - 3 & - 4 & 1 & 0 \\ 0 & 0 & - 8 & - 10 & 2 & 1 \end{array}]$

Langkah 2.4.4

Multiply each element of $R_{3}$ by $- \frac{1}{8}$ to make the entry at $3, 3$ a $1$ .

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Langkah 2.4.4.1

Multiply each element of $R_{3}$ by $- \frac{1}{8}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - 3 & - 4 & 1 & 0 \\ - \frac{1}{8} \cdot 0 & - \frac{1}{8} \cdot 0 & - \frac{1}{8} \cdot - 8 & - \frac{1}{8} \cdot - 10 & - \frac{1}{8} \cdot 2 & - \frac{1}{8} \cdot 1 \end{array}]$

Langkah 2.4.4.2

Sederhanakan $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - 3 & - 4 & 1 & 0 \\ 0 & 0 & 1 & \frac{5}{4} & - \frac{1}{4} & - \frac{1}{8} \end{array}]$

Langkah 2.4.5

Perform the row operation $R_{2} = R_{2} + 3 R_{3}$ to make the entry at $2, 3$ a $0$ .

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Langkah 2.4.5.1

Perform the row operation $R_{2} = R_{2} + 3 R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 + 3 \cdot 0 & 1 + 3 \cdot 0 & - 3 + 3 \cdot 1 & - 4 + 3 (\frac{5}{4}) & 1 + 3 (- \frac{1}{4}) & 0 + 3 (- \frac{1}{8}) \\ 0 & 0 & 1 & \frac{5}{4} & - \frac{1}{4} & - \frac{1}{8} \end{array}]$

Langkah 2.4.5.2

Sederhanakan $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & - \frac{1}{4} & \frac{1}{4} & - \frac{3}{8} \\ 0 & 0 & 1 & \frac{5}{4} & - \frac{1}{4} & - \frac{1}{8} \end{array}]$

Langkah 2.4.6

Perform the row operation $R_{1} = R_{1} - R_{3}$ to make the entry at $1, 3$ a $0$ .

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Langkah 2.4.6.1

Perform the row operation $R_{1} = R_{1} - R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 0 & 1 - 0 & 1 - 1 & 1 - \frac{5}{4} & 0 + \frac{1}{4} & 0 + \frac{1}{8} \\ 0 & 1 & 0 & - \frac{1}{4} & \frac{1}{4} & - \frac{3}{8} \\ 0 & 0 & 1 & \frac{5}{4} & - \frac{1}{4} & - \frac{1}{8} \end{array}]$

Langkah 2.4.6.2

Sederhanakan $R_{1}$ .

$[\begin{array}{c} 1 & 1 & 0 & - \frac{1}{4} & \frac{1}{4} & \frac{1}{8} \\ 0 & 1 & 0 & - \frac{1}{4} & \frac{1}{4} & - \frac{3}{8} \\ 0 & 0 & 1 & \frac{5}{4} & - \frac{1}{4} & - \frac{1}{8} \end{array}]$

Langkah 2.4.7

Perform the row operation $R_{1} = R_{1} - R_{2}$ to make the entry at $1, 2$ a $0$ .

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Langkah 2.4.7.1

Perform the row operation $R_{1} = R_{1} - R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 0 & 1 - 1 & 0 - 0 & - \frac{1}{4} + \frac{1}{4} & \frac{1}{4} - \frac{1}{4} & \frac{1}{8} + \frac{3}{8} \\ 0 & 1 & 0 & - \frac{1}{4} & \frac{1}{4} & - \frac{3}{8} \\ 0 & 0 & 1 & \frac{5}{4} & - \frac{1}{4} & - \frac{1}{8} \end{array}]$

Langkah 2.4.7.2

Sederhanakan $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} \\ 0 & 1 & 0 & - \frac{1}{4} & \frac{1}{4} & - \frac{3}{8} \\ 0 & 0 & 1 & \frac{5}{4} & - \frac{1}{4} & - \frac{1}{8} \end{array}]$

Langkah 2.5

The right half of the reduced row echelon form is the inverse.

$[\begin{array}{c} 0 & 0 & \frac{1}{2} \\ - \frac{1}{4} & \frac{1}{4} & - \frac{3}{8} \\ \frac{5}{4} & - \frac{1}{4} & - \frac{1}{8} \end{array}]$

Langkah 3

Kalikan kedua sisi persamaan matriks dengan matriks balikan.

$([\begin{array}{c} 0 & 0 & \frac{1}{2} \\ - \frac{1}{4} & \frac{1}{4} & - \frac{3}{8} \\ \frac{5}{4} & - \frac{1}{4} & - \frac{1}{8} \end{array}] \cdot [\begin{array}{c} 1 & 1 & 1 \\ 4 & 5 & 1 \\ 2 & 0 & 0 \end{array}]) \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 0 & 0 & \frac{1}{2} \\ - \frac{1}{4} & \frac{1}{4} & - \frac{3}{8} \\ \frac{5}{4} & - \frac{1}{4} & - \frac{1}{8} \end{array}] \cdot [\begin{array}{c} 2 \\ 12 \\ - 4 \end{array}]$

Langkah 4

Semua matriks akan selalu bernilai $1$ jika dikalikan dengan balikannya. $A \cdot A^{- 1} = 1$ .

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 0 & 0 & \frac{1}{2} \\ - \frac{1}{4} & \frac{1}{4} & - \frac{3}{8} \\ \frac{5}{4} & - \frac{1}{4} & - \frac{1}{8} \end{array}] \cdot [\begin{array}{c} 2 \\ 12 \\ - 4 \end{array}]$

Langkah 5

Kalikan $[\begin{array}{c} 0 & 0 & \frac{1}{2} \\ - \frac{1}{4} & \frac{1}{4} & - \frac{3}{8} \\ \frac{5}{4} & - \frac{1}{4} & - \frac{1}{8} \end{array}] [\begin{array}{c} 2 \\ 12 \\ - 4 \end{array}]$ .

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Langkah 5.1

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is $3 \times 3$ and the second matrix is $3 \times 1$ .

Langkah 5.2

Kalikan setiap baris pada matriks pertama dengan setiap kolom pada matriks kedua.

$[\begin{array}{c} 0 \cdot 2 + 0 \cdot 12 + \frac{1}{2} \cdot - 4 \\ - \frac{1}{4} \cdot 2 + \frac{1}{4} \cdot 12 - \frac{3}{8} \cdot - 4 \\ \frac{5}{4} \cdot 2 - \frac{1}{4} \cdot 12 - \frac{1}{8} \cdot - 4 \end{array}]$

Langkah 5.3

Sederhanakan setiap elemen dalam matriks dengan mengalikan semua pernyataannya.

$[\begin{array}{c} - 2 \\ 4 \\ 0 \end{array}]$

Langkah 6

Sederhanakan sisi kiri dan kanan.

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - 2 \\ 4 \\ 0 \end{array}]$

Langkah 7

Tentukan penyelesaiannya.

$x = - 2$

$y = 4$

$z = 0$