Find​ a, b,​ c, and d so that 

left bracket Start 2 By 2 Matrix 1st Row 1st Column 1 2nd Column negative 2 2nd Row 1st Column 3 2nd Column negative 1 EndMatrix right bracket

1

−2

3

−1

left bracket Start 2 By 2 Matrix 1st Row 1st Column a 2nd Column b 2nd Row 1st Column c 2nd Column d EndMatrix right bracket

a

b

c

d

equals=left bracket Start 2 By 2 Matrix 1st Row 1st Column negative 6 2nd Column 1 2nd Row 1st Column 2 2nd Column 3 EndMatrix right bracket

−6

1

2

3

.

Question content area bottom

Part 1

Select the correct choice​ and, if​ necessary, fill in the answer box within your choice.

A.

left bracket Start 2 By 2 Matrix 1st Row 1st Column a 2nd Column b 2nd Row 1st Column c 2nd Column d EndMatrix right bracket

a

b

c

d

equals=enter your response here 

​(Simplify your​ answer.)

B.

There is no solution.

To solve this, we need to determine whether we can find values for \( a \), \( b \), \( c \), and \( d \) so that:

\[
\begin{bmatrix} 1 & -2 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} -6 & 1 \\ 2 & 3 \end{bmatrix}
\]

We can expand the matrix multiplication on the left-hand side to set up a system of equations. Let’s proceed by calculating each element in the resulting matrix.

### Step 1: Set Up the Matrix Multiplication
The multiplication of the two matrices results in:

\[
\begin{bmatrix} 1 \cdot a + (-2) \cdot c & 1 \cdot b + (-2) \cdot d \\ 3 \cdot a + (-1) \cdot c & 3 \cdot b + (-1) \cdot d \end{bmatrix} = \begin{bmatrix} -6 & 1 \\ 2 & 3 \end{bmatrix}
\]

This gives us four equations by equating corresponding elements:

1. \( 1 \cdot a - 2 \cdot c = -6 \)
2. \( 1 \cdot b - 2 \cdot d = 1 \)
3. \( 3 \cdot a - 1 \cdot c = 2 \)
4. \( 3 \cdot b - 1 \cdot d = 3 \)

### Step 2: Solve the System of Equations
Let’s solve these equations step-by-step.

#### Solving for \( a \) and \( c \):
Using the first and third equations:

1. \( a - 2c = -6 \)
2. \( 3a - c = 2 \)

Multiply the first equation by 3 to align terms with the second equation:

\[
3a - 6c = -18
\]

Now subtract the second equation from this result:

\[
(3a - 6c) - (3a - c) = -18 - 2
\]
\[
-5c = -20
\]
\[
c = 4
\]

Now substitute \( c = 4 \) into the first equation:

\[
a - 2(4) = -6
\]
\[
a - 8 = -6
\]
\[
a = 2
\]

#### Solving for \( b \) and \( d \):
Using the second and fourth equations:

1. \( b - 2d = 1 \)
2. \( 3b - d = 3 \)

Multiply the first equation by 3 to align terms with the second equation:

\[
3b - 6d = 3
\]

Now subtract the second equation from this result:

\[
(3b - 6d) - (3b - d) = 3 - 3
\]
\[
-5d = 0
\]
\[
d = 0
\]

Now substitute \( d = 0 \) into the second equation:

\[
b - 2(0) = 1
\]
\[
b = 1
\]

### Solution
Thus, the values of \( a \), \( b \), \( c \), and \( d \) that satisfy the matrix equation are:

\[
\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 4 & 0 \end{bmatrix}
\]

So, the answer is:

**A.** 
\[
\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 4 & 0 \end{bmatrix}
\]

Would you like a further breakdown of any steps?

---

1. How can you verify a solution to a matrix equation?
2. What would happen if the system was inconsistent (no solution)?
3. How does matrix multiplication differ from scalar multiplication?
4. Can this method be used for any size of matrices?
5. What are real-life applications of matrix equations?

**Tip:** Always check your matrix multiplication results to ensure all terms match the target matrix, as any small error can lead to incorrect solutions.

Find a, b, c, and d so that [ [1, -2], [3, -1] ] [ [a, b], [c, d] ] = [ [-6, 1], [2, 3] ].

Discover how to solve for variables a, b, c, and d in matrix equations with step-by-step guidance. This problem focuses on matrix multiplication, setting up systems of equations, and determining values to satisfy matrix equality. Ideal for students in Grades 11-12 studying linear algebra concepts.

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