Questions: Find (a, b, c), and (d) so that (left[beginarrayll1 -2 3 -1endarrayright]left[beginarraylla b c dendarrayright]=left[beginarrayrr1 -5 8 5endarrayright]) Select the correct choice and, if necessary, fill in the answer box within your choice. A. (left[beginarraylla b c dendarrayright]=) (Simplify your answer.) B. There is no solution.

Find (a, b, c), and (d) so that (left[beginarrayll1 -2 3 -1endarrayright]left[beginarraylla b c dendarrayright]=left[beginarrayrr1 -5 8 5endarrayright])

Select the correct choice and, if necessary, fill in the answer box within your choice.
A. (left[beginarraylla b c dendarrayright]=) (Simplify your answer.)
B. There is no solution.

Transcript text: Find $a, b, c$, and d so that $\left[\begin{array}{ll}1 & -2 \\ 3 & -1\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{rr}1 & -5 \\ 8 & 5\end{array}\right]$ Select the correct choice and, if necessary, fill in the answer box within your choice. A. $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=$ $\square$ (Simplify your answer.) B. There is no solution.

Solution

Solution Steps

To find the values of $a, b, c,$ and $d$, we need to solve the matrix equation:

\[ \left[\begin{array}{ll}1 & -2 \\ 3 & -1\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{rr}1 & -5 \\ 8 & 5\end{array}\right] \]

We can do this by performing matrix multiplication on the left-hand side and equating the resulting matrix to the right-hand side matrix. This will give us a system of linear equations that we can solve for $a, b, c,$ and $d$.

Step 1: Define the Matrix Equation

We start with the matrix equation:

\[ \left[\begin{array}{ll}1 & -2 \\ 3 & -1\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{rr}1 & -5 \\ 8 & 5\end{array}\right] \]

Step 2: Perform Matrix Multiplication

Perform the matrix multiplication on the left-hand side:

\[ \left[\begin{array}{ll}1 & -2 \\ 3 & -1\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] = \left[\begin{array}{ll}a - 2c & b - 2d \\ 3a - c & 3b - d\end{array}\right] \]

Step 3: Set Up the System of Equations

Equate the resulting matrix to the right-hand side matrix:

\[ \left[\begin{array}{ll}a - 2c & b - 2d \\ 3a - c & 3b - d\end{array}\right] = \left[\begin{array}{rr}1 & -5 \\ 8 & 5\end{array}\right] \]

This gives us the following system of equations:

\[ \begin{cases} a - 2c = 1 \\ b - 2d = -5 \\ 3a - c = 8 \\ 3b - d = 5 \end{cases} \]

Step 4: Solve the System of Equations

Solving the system of equations, we get:

\[ \begin{cases} a = 3 \\ b = 3 \\ c = 1 \\ d = 4 \end{cases} \]