선형 대수 예제

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 1 & 0 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦ y = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 3 & 3 2 & 1 \end{matrix} ⎤ ⎥ ⎦

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

단계 1

Find the inverse of

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 1 & 0 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{array}]$ .

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단계 1.1

다시 씁니다.

∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 1 & 1 & 0 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣

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

단계 1.2

Find the determinant.

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단계 1.2.1

Choose the row or column with the most

0

$0$ elements. If there are no

0

$0$ elements choose any row or column. Multiply every element in row

1

$1$ by its cofactor and add.

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단계 1.2.1.1

Consider the corresponding sign chart.

∣ ∣ ∣ ∣ \begin{matrix} + & - & + - & + & - + & - & + \end{matrix} ∣ ∣ ∣ ∣

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

단계 1.2.1.2

The cofactor is the minor with the sign changed if the indices match a

-

$-$ position on the sign chart.

단계 1.2.1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣

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

단계 1.2.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

1 ∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣

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

단계 1.2.1.5

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣

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

단계 1.2.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

0 ∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣

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

단계 1.2.1.7

The minor for

a_{13}

$a_{13}$ is the determinant with row

1

$1$ and column

3

$3$ deleted.

∣ ∣ ∣ \begin{matrix} 1 & 1 1 & 1 \end{matrix} ∣ ∣ ∣

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

단계 1.2.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

0 ∣ ∣ ∣ \begin{matrix} 1 & 1 1 & 1 \end{matrix} ∣ ∣ ∣

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

단계 1.2.1.9

Add the terms together.

1 ∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & 1 1 & 1 \end{matrix} ∣ ∣ ∣

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

1 ∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & 1 1 & 1 \end{matrix} ∣ ∣ ∣

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

단계 1.2.2

0

$0$ 에

∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 1 & 0 \\ 1 & 1 \end{array} |$ 을 곱합니다.

1 ∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣ + 0 + 0 ∣ ∣ ∣ \begin{matrix} 1 & 1 1 & 1 \end{matrix} ∣ ∣ ∣

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

단계 1.2.3

0

$0$ 에

∣ ∣ ∣ \begin{matrix} 1 & 1 1 & 1 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 1 & 1 \\ 1 & 1 \end{array} |$ 을 곱합니다.

1 ∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣ + 0 + 0

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

단계 1.2.4

∣ ∣ ∣ \begin{matrix} 1 & 0 1 & 1 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 1 & 0 \\ 1 & 1 \end{array} |$ 의 값을 구합니다.

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단계 1.2.4.1

2 \times 2

$2 \times 2$ 행렬의 행렬식은

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 공식을 이용해 계산합니다.

1 (1 \cdot 1 - 1 \cdot 0) + 0 + 0

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

단계 1.2.4.2

행렬식을 간단히 합니다.

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단계 1.2.4.2.1

1

$1$ 에

1

$1$ 을 곱합니다.

1 (1 - 1 \cdot 0) + 0 + 0

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

단계 1.2.4.2.2

1

$1$ 에서

0

$0$ 을 뺍니다.

1 \cdot 1 + 0 + 0

$1 \cdot 1 + 0 + 0$

1 \cdot 1 + 0 + 0

$1 \cdot 1 + 0 + 0$

1 \cdot 1 + 0 + 0

$1 \cdot 1 + 0 + 0$

단계 1.2.5

행렬식을 간단히 합니다.

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단계 1.2.5.1

1

$1$ 에

1

$1$ 을 곱합니다.

1 + 0 + 0

$1 + 0 + 0$

단계 1.2.5.2

1

$1$ 를

0

$0$ 에 더합니다.

1 + 0

$1 + 0$

단계 1.2.5.3

1

$1$ 를

0

$0$ 에 더합니다.

1

$1$

1

$1$

1

$1$

단계 1.3

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

단계 1.4

Set up a

3 \times 6

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 1 & 1 & 0 & 0 & 1 & 0 1 & 1 & 1 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

단계 1.5

기약 행 사다리꼴을 구합니다.

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단계 1.5.1

Perform the row operation

R_{2} = R_{2} - R_{1}

$R_{2} = R_{2} - R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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단계 1.5.1.1

Perform the row operation

R_{2} = R_{2} - R_{1}

$R_{2} = R_{2} - R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 1 - 1 & 1 - 0 & 0 - 0 & 0 - 1 & 1 - 0 & 0 - 0 1 & 1 & 1 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

단계 1.5.1.2

R_{2}

$R_{2}$ 을 간단히 합니다.

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 0 & 1 & 0 & - 1 & 1 & 0 1 & 1 & 1 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 0 & 1 & 0 & - 1 & 1 & 0 1 & 1 & 1 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

단계 1.5.2

Perform the row operation

R_{3} = R_{3} - R_{1}

$R_{3} = R_{3} - R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

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단계 1.5.2.1

Perform the row operation

R_{3} = R_{3} - R_{1}

$R_{3} = R_{3} - R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 0 & 1 & 0 & - 1 & 1 & 0 1 - 1 & 1 - 0 & 1 - 0 & 0 - 1 & 0 - 0 & 1 - 0 \end{matrix} ⎤ ⎥ ⎦

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

단계 1.5.2.2

R_{3}

$R_{3}$ 을 간단히 합니다.

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 0 & 1 & 0 & - 1 & 1 & 0 0 & 1 & 1 & - 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 0 & 1 & 0 & - 1 & 1 & 0 0 & 1 & 1 & - 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

단계 1.5.3

Perform the row operation

R_{3} = R_{3} - R_{2}

$R_{3} = R_{3} - R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

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단계 1.5.3.1

Perform the row operation

R_{3} = R_{3} - R_{2}

$R_{3} = R_{3} - R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 0 & 1 & 0 & - 1 & 1 & 0 0 - 0 & 1 - 1 & 1 - 0 & - 1 + 1 & 0 - 1 & 1 - 0 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & - 1 & 1 & 0 \\ 0 - 0 & 1 - 1 & 1 - 0 & - 1 + 1 & 0 - 1 & 1 - 0 \end{array}]$

단계 1.5.3.2

R_{3}

$R_{3}$ 을 간단히 합니다.

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 0 & 1 & 0 & - 1 & 1 & 0 0 & 0 & 1 & 0 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 0 & 1 & 0 & - 1 & 1 & 0 0 & 0 & 1 & 0 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 0 & 1 & 0 & - 1 & 1 & 0 0 & 0 & 1 & 0 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦

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

단계 1.6

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 - 1 & 1 & 0 0 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 - 1 & 1 & 0 0 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦

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

단계 2

Multiply both sides by the inverse of

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 1 & 0 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{array}]$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 - 1 & 1 & 0 0 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 1 & 0 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦ y = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 - 1 & 1 & 0 0 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 3 & 3 2 & 1 \end{matrix} ⎤ ⎥ ⎦

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

단계 3

방정식을 간단히 합니다.

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단계 3.1

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 - 1 & 1 & 0 0 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 1 & 0 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ 0 & - 1 & 1 \end{array}] [\begin{array}{c} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{array}]$ 을 곱합니다.

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단계 3.1.1

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

$3 \times 3$ and the second matrix is

3 \times 3

$3 \times 3$ .

단계 3.1.2

첫 번째 행렬의 각 행에 두 번째 행렬의 각 열을 곱합니다.

⎡ ⎢ ⎣ \begin{matrix} 1 \cdot 1 + 0 \cdot 1 + 0 \cdot 1 & 1 \cdot 0 + 0 \cdot 1 + 0 \cdot 1 & 1 \cdot 0 + 0 \cdot 0 + 0 \cdot 1 - 1 \cdot 1 + 1 \cdot 1 + 0 \cdot 1 & - 0 + 1 \cdot 1 + 0 \cdot 1 & - 0 + 1 \cdot 0 + 0 \cdot 1 0 \cdot 1 - 1 \cdot 1 + 1 \cdot 1 & 0 \cdot 0 - 1 \cdot 1 + 1 \cdot 1 & 0 \cdot 0 - 0 + 1 \cdot 1 \end{matrix} ⎤ ⎥ ⎦ y = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 - 1 & 1 & 0 0 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 3 & 3 2 & 1 \end{matrix} ⎤ ⎥ ⎦

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

단계 3.1.3

모든 식을 전개하여 행렬의 각 원소를 간단히 합니다.

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ y = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 - 1 & 1 & 0 0 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 3 & 3 2 & 1 \end{matrix} ⎤ ⎥ ⎦

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ y = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 - 1 & 1 & 0 0 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 3 & 3 2 & 1 \end{matrix} ⎤ ⎥ ⎦

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

단계 3.2

Multiplying the identity matrix by any matrix

A

$A$ is the matrix

A

$A$ itself.

y = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 - 1 & 1 & 0 0 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 3 & 3 2 & 1 \end{matrix} ⎤ ⎥ ⎦

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

단계 3.3

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 - 1 & 1 & 0 0 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 3 & 3 2 & 1 \end{matrix} ⎤ ⎥ ⎦

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

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단계 3.3.1

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

$3 \times 3$ and the second matrix is

3 \times 2

$3 \times 2$ .

단계 3.3.2

첫 번째 행렬의 각 행에 두 번째 행렬의 각 열을 곱합니다.

y = ⎡ ⎢ ⎣ \begin{matrix} 1 \cdot 1 + 0 \cdot 3 + 0 \cdot 2 & 1 \cdot 2 + 0 \cdot 3 + 0 \cdot 1 - 1 \cdot 1 + 1 \cdot 3 + 0 \cdot 2 & - 1 \cdot 2 + 1 \cdot 3 + 0 \cdot 1 0 \cdot 1 - 1 \cdot 3 + 1 \cdot 2 & 0 \cdot 2 - 1 \cdot 3 + 1 \cdot 1 \end{matrix} ⎤ ⎥ ⎦

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

단계 3.3.3

모든 식을 전개하여 행렬의 각 원소를 간단히 합니다.

y = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 2 & 1 - 1 & - 2 \end{matrix} ⎤ ⎥ ⎦

$y = [\begin{array}{c} 1 & 2 \\ 2 & 1 \\ - 1 & - 2 \end{array}]$

y = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 2 & 1 - 1 & - 2 \end{matrix} ⎤ ⎥ ⎦

$y = [\begin{array}{c} 1 & 2 \\ 2 & 1 \\ - 1 & - 2 \end{array}]$

y = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 2 & 1 - 1 & - 2 \end{matrix} ⎤ ⎥ ⎦

$y = [\begin{array}{c} 1 & 2 \\ 2 & 1 \\ - 1 & - 2 \end{array}]$