If $$ A^{-1}=\left[\begin{array}{ll}1 & -1 \ 3 & -2\end{array} ight] $$ and $$ b=\left[\begin{array}{c}1 \ -2\end{array} ight] $$ find $$ x $$, where $$ A x=b $$

Question

If $$ A^{-1}=\left[\begin{array}{ll}1 & -1 \ 3 & -2\end{array}ight] $$ and $$ b=\left[\begin{array}{c}1 \ -2\end{array}ight] $$ find $$ x $$, where $$ A x=b $$

UpStudy AI · Accepted Answer

To find $$ x $$ in the equation $$ A x = b $$, we can use the inverse of matrix $$ A $$. Given that $$ A^{-1} = \begin{bmatrix} 1 & -1 \ 3 & -2 \end{bmatrix} $$ and $$ b = \begin{bmatrix} 1 \ -2 \end{bmatrix} $$, we can solve for $$ x $$ by multiplying both sides of the equation $$ A x = b $$ by $$ A^{-1} $$:

$$
x = A^{-1} b
$$

Now, we will perform the matrix multiplication $$ A^{-1} b $$:

$$
A^{-1} b = \begin{bmatrix} 1 & -1 \ 3 & -2 \end{bmatrix} \begin{bmatrix} 1 \ -2 \end{bmatrix}
$$

Calculating the first component of the resulting vector:

$$
1 \cdot 1 + (-1) \cdot (-2) = 1 + 2 = 3
$$

Calculating the second component of the resulting vector:

$$
3 \cdot 1 + (-2) \cdot (-2) = 3 + 4 = 7
$$

Thus, we have:

$$
x = \begin{bmatrix} 3 \ 7 \end{bmatrix}
$$

Therefore, the solution is:

$$
\boxed{\begin{bmatrix} 3 \ 7 \end{bmatrix}}
$$
The solution is $$ x = \begin{bmatrix} 3 \ 7 \end{bmatrix} $$.

If \( A^{-1}=\left[\begin{array}{ll}1 & -1 \\ 3 & -2\end{array}\right] \) and \( b=\left[\begin{array}{c}1 \\ -2\end{array}\right] \) find \( x \), where \( A x=b \)

Real Tutor Solution

Answer

Solution

Related Questions

Latest Algebra Questions