Fractions and Decimals

Discrete Mathematics

Differential Equations

Solved on Feb 10, 2024

Solve the linear system $Ax=b$ where $A$ is a $3\times 3$ matrix, $x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$ , and $b=\begin{bmatrix}5\\0\\-5\end{bmatrix}$ . If $A^{-1}=\begin{bmatrix}1&4&-1\\2&-3&2\\0&5&-4\end{bmatrix}$ , then $x_3=$

STEP 1

Assumptions
1. $A$ is a $3 \times 3$ matrix.
2. $x$ is a column vector with elements $x_1, x_2, x_3$ .
3. $b$ is a column vector with elements $5, 0, -5$ .
4. $A^{-1}$ is given as $\left[\begin{array}{ccc}1 & 4 & -1 \\ 2 & -3 & 2 \\ 0 & 5 & -4\end{array}\right]$ .
5. We need to find the value of $x_3$ .

STEP 2

To find the vector $x$ , we can use the inverse of matrix $A$ and multiply it by vector $b$ .
$x = A^{-1} b$

STEP 3

Plug in the given values for $A^{-1}$ and $b$ into the equation.
$x = \left[\begin{array}{ccc}1 & 4 & -1 \\ 2 & -3 & 2 \\ 0 & 5 & -4\end{array}\right] \left[\begin{array}{c}5 \\ 0 \\ -5\end{array}\right]$

STEP 4

Perform the matrix multiplication to find the vector $x$ .
$x = \left[\begin{array}{ccc}1 & 4 & -1 \\ 2 & -3 & 2 \\ 0 & 5 & -4\end{array}\right] \left[\begin{array}{c}5 \\ 0 \\ -5\end{array}\right] = \left[\begin{array}{c}1 \cdot 5 + 4 \cdot 0 - 1 \cdot (-5) \\ 2 \cdot 5 - 3 \cdot 0 + 2 \cdot (-5) \\ 0 \cdot 5 + 5 \cdot 0 - 4 \cdot (-5)\end{array}\right]$

STEP 5

Calculate the elements of the vector $x$ .
$x = \left[\begin{array}{c}5 + 0 + 5 \\ 10 + 0 - 10 \\ 0 + 0 + 20\end{array}\right]$

STEP 6

Simplify the elements of the vector $x$ .
$x = \left[\begin{array}{c}10 \\ 0 \\ 20\end{array}\right]$

STEP 7

The third element of vector $x$ , which is $x_3$ , is 20.
Therefore, the unknown $x_3 = 20$ .
The answer is c. 20.

Was this helpful?

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Contact Influencer program Policy Terms