Prakalkulus Contoh

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

Langkah 1

Find the determinant.

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Langkah 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 $2$ by its cofactor and add.

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Langkah 1.1.1

Consider the corresponding sign chart.

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

Langkah 1.1.2

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

Langkah 1.1.3

The minor for $a_{21}$ is the determinant with row $2$ and column $1$ deleted.

$| \begin{array}{c} 0 & 3 \\ 2 & 4 \end{array} |$

Langkah 1.1.4

Multiply element $a_{21}$ by its cofactor.

$- 3 | \begin{array}{c} 0 & 3 \\ 2 & 4 \end{array} |$

Langkah 1.1.5

The minor for $a_{22}$ is the determinant with row $2$ and column $2$ deleted.

$| \begin{array}{c} 2 & 3 \\ 0 & 4 \end{array} |$

Langkah 1.1.6

Multiply element $a_{22}$ by its cofactor.

$0 | \begin{array}{c} 2 & 3 \\ 0 & 4 \end{array} |$

Langkah 1.1.7

The minor for $a_{23}$ is the determinant with row $2$ and column $3$ deleted.

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

Langkah 1.1.8

Multiply element $a_{23}$ by its cofactor.

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

Langkah 1.1.9

Add the terms together.

$- 3 | \begin{array}{c} 0 & 3 \\ 2 & 4 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ 0 & 4 \end{array} | + 0 | \begin{array}{c} 2 & 0 \\ 0 & 2 \end{array} |$

Langkah 1.2

Kalikan $0$ dengan $| \begin{array}{c} 2 & 3 \\ 0 & 4 \end{array} |$ .

$- 3 | \begin{array}{c} 0 & 3 \\ 2 & 4 \end{array} | + 0 + 0 | \begin{array}{c} 2 & 0 \\ 0 & 2 \end{array} |$

Langkah 1.3

Kalikan $0$ dengan $| \begin{array}{c} 2 & 0 \\ 0 & 2 \end{array} |$ .

$- 3 | \begin{array}{c} 0 & 3 \\ 2 & 4 \end{array} | + 0 + 0$

Langkah 1.4

Evaluasi $| \begin{array}{c} 0 & 3 \\ 2 & 4 \end{array} |$ .

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Langkah 1.4.1

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

$- 3 (0 \cdot 4 - 2 \cdot 3) + 0 + 0$

Langkah 1.4.2

Sederhanakan determinannya.

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Langkah 1.4.2.1

Sederhanakan setiap suku.

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Langkah 1.4.2.1.1

Kalikan $0$ dengan $4$ .

$- 3 (0 - 2 \cdot 3) + 0 + 0$

Langkah 1.4.2.1.2

Kalikan $- 2$ dengan $3$ .

$- 3 (0 - 6) + 0 + 0$

Langkah 1.4.2.2

Kurangi $6$ dengan $0$ .

$- 3 \cdot - 6 + 0 + 0$

Langkah 1.5

Sederhanakan determinannya.

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Langkah 1.5.1

Kalikan $- 3$ dengan $- 6$ .

$18 + 0 + 0$

Langkah 1.5.2

Tambahkan $18$ dan $0$ .

$18 + 0$

Langkah 1.5.3

Tambahkan $18$ dan $0$ .

$18$

Langkah 2

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

Langkah 3

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

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

Langkah 4

Tentukan bentuk eselon baris yang dikurangi.

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Langkah 4.1

Multiply each element of $R_{1}$ by $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

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Langkah 4.1.1

Multiply each element of $R_{1}$ by $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

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

Langkah 4.1.2

Sederhanakan $R_{1}$ .

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

Langkah 4.2

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

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Langkah 4.2.1

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

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

Langkah 4.2.2

Sederhanakan $R_{2}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & - \frac{9}{2} & - \frac{3}{2} & 1 & 0 \\ 0 & 2 & 4 & 0 & 0 & 1 \end{array}]$

Langkah 4.3

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 2 & 4 & 0 & 0 & 1 \\ 0 & 0 & - \frac{9}{2} & - \frac{3}{2} & 1 & 0 \end{array}]$

Langkah 4.4

Multiply each element of $R_{2}$ by $\frac{1}{2}$ to make the entry at $2, 2$ a $1$ .

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Langkah 4.4.1

Multiply each element of $R_{2}$ by $\frac{1}{2}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ \frac{0}{2} & \frac{2}{2} & \frac{4}{2} & \frac{0}{2} & \frac{0}{2} & \frac{1}{2} \\ 0 & 0 & - \frac{9}{2} & - \frac{3}{2} & 1 & 0 \end{array}]$

Langkah 4.4.2

Sederhanakan $R_{2}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 & \frac{1}{2} \\ 0 & 0 & - \frac{9}{2} & - \frac{3}{2} & 1 & 0 \end{array}]$

Langkah 4.5

Multiply each element of $R_{3}$ by $- \frac{2}{9}$ to make the entry at $3, 3$ a $1$ .

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Langkah 4.5.1

Multiply each element of $R_{3}$ by $- \frac{2}{9}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 & \frac{1}{2} \\ - \frac{2}{9} \cdot 0 & - \frac{2}{9} \cdot 0 & - \frac{2}{9} (- \frac{9}{2}) & - \frac{2}{9} (- \frac{3}{2}) & - \frac{2}{9} \cdot 1 & - \frac{2}{9} \cdot 0 \end{array}]$

Langkah 4.5.2

Sederhanakan $R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & \frac{1}{3} & - \frac{2}{9} & 0 \end{array}]$

Langkah 4.6

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

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Langkah 4.6.1

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

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 - 2 \cdot 0 & 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 0 - 2 (\frac{1}{3}) & 0 - 2 (- \frac{2}{9}) & \frac{1}{2} - 2 \cdot 0 \\ 0 & 0 & 1 & \frac{1}{3} & - \frac{2}{9} & 0 \end{array}]$

Langkah 4.6.2

Sederhanakan $R_{2}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & - \frac{2}{3} & \frac{4}{9} & \frac{1}{2} \\ 0 & 0 & 1 & \frac{1}{3} & - \frac{2}{9} & 0 \end{array}]$

Langkah 4.7

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

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Langkah 4.7.1

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

$[\begin{array}{c} 1 - \frac{3}{2} \cdot 0 & 0 - \frac{3}{2} \cdot 0 & \frac{3}{2} - \frac{3}{2} \cdot 1 & \frac{1}{2} - \frac{3}{2} \cdot \frac{1}{3} & 0 - \frac{3}{2} (- \frac{2}{9}) & 0 - \frac{3}{2} \cdot 0 \\ 0 & 1 & 0 & - \frac{2}{3} & \frac{4}{9} & \frac{1}{2} \\ 0 & 0 & 1 & \frac{1}{3} & - \frac{2}{9} & 0 \end{array}]$

Langkah 4.7.2

Sederhanakan $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & \frac{1}{3} & 0 \\ 0 & 1 & 0 & - \frac{2}{3} & \frac{4}{9} & \frac{1}{2} \\ 0 & 0 & 1 & \frac{1}{3} & - \frac{2}{9} & 0 \end{array}]$

Langkah 5

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

$[\begin{array}{c} 0 & \frac{1}{3} & 0 \\ - \frac{2}{3} & \frac{4}{9} & \frac{1}{2} \\ \frac{1}{3} & - \frac{2}{9} & 0 \end{array}]$

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