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.
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Solution

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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:

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

Final Answer

\(\boxed{\left[\begin{array}{ll}3 & 3 \\ 1 & 4\end{array}\right]}\)

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