$$ A = \left[ \begin{array} { c c c c c } { 25 } & { 16 } & { 0 } & { 0 } & { 1 } \ { 0 } & { 8 } & { 12 } & { 0 } & { 0 } \ { 0 } & { 0 } & { 1 } & { 0 } & { 0 } \ { 8 } & { - 2 } & { 0 } & { 8 } & { 9 } \ { 1 } & { 48 } & { 1 } & { 0 } & { 0 } \end{array} ight] , x = \left[ \begin{array} { l } { x _ { 1 } } \ { x _ { 2 } } \ { x _ { 3 } } \ { x _ { 4 } } \ { x _ { 5 } } \end{array} ight] , A x = \left[ \begin{array} { c } { 1 } \ { 8 } \ { 0 } \ { 0 } \ { 16 } \end{array} ight] $$

Question

$$ A = \left[ \begin{array} { c c c c c } { 25 } & { 16 } & { 0 } & { 0 } & { 1 } \ { 0 } & { 8 } & { 12 } & { 0 } & { 0 } \ { 0 } & { 0 } & { 1 } & { 0 } & { 0 } \ { 8 } & { - 2 } & { 0 } & { 8 } & { 9 } \ { 1 } & { 48 } & { 1 } & { 0 } & { 0 } \end{array} ight] , x = \left[ \begin{array} { l } { x _ { 1 } } \ { x _ { 2 } } \ { x _ { 3 } } \ { x _ { 4 } } \ { x _ { 5 } } \end{array} ight] , A x = \left[ \begin{array} { c } { 1 } \ { 8 } \ { 0 } \ { 0 } \ { 16 } \end{array} ight] $$

UpStudy AI · Accepted Answer

To solve the system of linear equations represented by $$ A \mathbf{x} = \mathbf{b} $$, where:

$$
A = \begin{bmatrix}
25 & 16 & 0 & 0 & 1 \
0 & 8 & 12 & 0 & 0 \
0 & 0 & 1 & 0 & 0 \
8 & -2 & 0 & 8 & 9 \
1 & 48 & 1 & 0 & 0
\end{bmatrix},
\quad
\mathbf{x} = \begin{bmatrix}
x_1 \
x_2 \
x_3 \
x_4 \
x_5
\end{bmatrix},
\quad
\mathbf{b} = \begin{bmatrix}
1 \
8 \
0 \
0 \
16
\end{bmatrix},
$$

we can write out the system of equations as follows:

$$
\begin{cases}
25x_1 + 16x_2 + x_5 = 1 \quad &(1) \
8x_2 + 12x_3 = 8 \quad &(2) \
x_3 = 0 \quad &(3) \
8x_1 - 2x_2 + 8x_4 + 9x_5 = 0 \quad &(4) \
x_1 + 48x_2 + x_3 = 16 \quad &(5)
\end{cases}
$$

Let's solve the system step by step:

1. **From Equation (3):**
   $$
   x_3 = 0
   $$

2. **Substitute $$ x_3 = 0 $$ into Equation (2):**
   $$
   8x_2 = 8 \implies x_2 = 1
   $$

3. **Substitute $$ x_2 = 1 $$ and $$ x_3 = 0 $$ into Equation (5):**
   $$
   x_1 + 48(1) = 16 \implies x_1 = 16 - 48 \implies x_1 = -32
   $$

4. **Substitute $$ x_1 = -32 $$, $$ x_2 = 1 $$, and $$ x_3 = 0 $$ into Equation (1):**
   $$
   25(-32) + 16(1) + x_5 = 1 \implies -800 + 16 + x_5 = 1 \implies x_5 = 785
   $$

5. **Substitute $$ x_1 = -32 $$, $$ x_2 = 1 $$, and $$ x_5 = 785 $$ into Equation (4):**
   $$
   8(-32) - 2(1) + 8x_4 + 9(785) = 0 \implies -256 - 2 + 8x_4 + 7065 = 0
   $$
   $$
   6807 + 8x_4 = 0 \implies 8x_4 = -6807 \implies x_4 = -850.875
   $$
   $$
   x_4 = -850\frac{7}{8}
   $$

**Final Solution:**

$$
\begin{cases}
x_1 = -32, \
x_2 = 1, \
x_3 = 0, \
x_4 = -850.875 \quad (\text{or } -850\frac{7}{8}), \
x_5 = 785.
\end{cases}
$$

**Verification:**

To ensure the solution is correct, substitute the values back into all five original equations. Each equation should hold true, confirming the validity of the solution.

---

**Summary:**

The solution to $$ A \mathbf{x} = \mathbf{b} $$ is:
$$
\mathbf{x} = \begin{bmatrix}
-32 \
1 \
0 \
-850.875 \
785
\end{bmatrix}
$$
The solution to the system is: $$ x_1 = -32, \quad x_2 = 1, \quad x_3 = 0, \quad x_4 = -850.875, \quad x_5 = 785 $$

Solución de tutoría real

Responder

Solución

Bonus Knowledge

preguntas relacionadas

Latest Algebra Questions