Matematika Berhingga Contoh

$[\begin{array}{c} a + 8 & 4 z + 1 & 8 m \\ 7 k & 4 & 0 \end{array}] + [\begin{array}{c} 10 a & 10 z & 7 m \\ 8 k & 5 & 0 \end{array}] = [\begin{array}{c} 41 & - 69 & 105 \\ 45 & 9 & 0 \end{array}]$

Langkah 1

Tambahkan elemen yang seletak.

$[\begin{array}{c} a + 8 + 10 a & 4 z + 1 + 10 z & 8 m + 7 m \\ 7 k + 8 k & 4 + 5 & 0 + 0 \end{array}] = [\begin{array}{c} 41 & - 69 & 105 \\ 45 & 9 & 0 \end{array}]$

Langkah 2

Simplify each element.

Ketuk untuk lebih banyak langkah...

Langkah 2.1

Tambahkan $a$ dan $10 a$ .

$[\begin{array}{c} 11 a + 8 & 4 z + 1 + 10 z & 8 m + 7 m \\ 7 k + 8 k & 4 + 5 & 0 + 0 \end{array}] = [\begin{array}{c} 41 & - 69 & 105 \\ 45 & 9 & 0 \end{array}]$

Langkah 2.2

Tambahkan $4 z$ dan $10 z$ .

$[\begin{array}{c} 11 a + 8 & 14 z + 1 & 8 m + 7 m \\ 7 k + 8 k & 4 + 5 & 0 + 0 \end{array}] = [\begin{array}{c} 41 & - 69 & 105 \\ 45 & 9 & 0 \end{array}]$

Langkah 2.3

Tambahkan $8 m$ dan $7 m$ .

$[\begin{array}{c} 11 a + 8 & 14 z + 1 & 15 m \\ 7 k + 8 k & 4 + 5 & 0 + 0 \end{array}] = [\begin{array}{c} 41 & - 69 & 105 \\ 45 & 9 & 0 \end{array}]$

Langkah 2.4

Tambahkan $7 k$ dan $8 k$ .

$[\begin{array}{c} 11 a + 8 & 14 z + 1 & 15 m \\ 15 k & 4 + 5 & 0 + 0 \end{array}] = [\begin{array}{c} 41 & - 69 & 105 \\ 45 & 9 & 0 \end{array}]$

Langkah 2.5

Tambahkan $4$ dan $5$ .

$[\begin{array}{c} 11 a + 8 & 14 z + 1 & 15 m \\ 15 k & 9 & 0 + 0 \end{array}] = [\begin{array}{c} 41 & - 69 & 105 \\ 45 & 9 & 0 \end{array}]$

Langkah 2.6

Tambahkan $0$ dan $0$ .

$[\begin{array}{c} 11 a + 8 & 14 z + 1 & 15 m \\ 15 k & 9 & 0 \end{array}] = [\begin{array}{c} 41 & - 69 & 105 \\ 45 & 9 & 0 \end{array}]$

Langkah 3

Write as a linear system of equations.

$11 a + 8 = 41$

$14 z + 1 = - 69$

$15 m = 105$

$15 k = 45$

$9 = 9$

$0 = 0$

Langkah 4

Since $0 = 0$ is always true, the matrix equation has infinite solutions.

Infinite solutions satisfying the system.