Math Problem Statement

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.

Solution

To solve this, we need to determine whether we can find values for aa, bb, cc, and dd so that:

[1231][abcd]=[6123]\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:

[1a+(2)c1b+(2)d3a+(1)c3b+(1)d]=[6123]\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. 1a2c=61 \cdot a - 2 \cdot c = -6
  2. 1b2d=11 \cdot b - 2 \cdot d = 1
  3. 3a1c=23 \cdot a - 1 \cdot c = 2
  4. 3b1d=33 \cdot b - 1 \cdot d = 3

Step 2: Solve the System of Equations

Let’s solve these equations step-by-step.

Solving for aa and cc:

Using the first and third equations:

  1. a2c=6a - 2c = -6
  2. 3ac=23a - c = 2

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

3a6c=183a - 6c = -18

Now subtract the second equation from this result:

(3a6c)(3ac)=182(3a - 6c) - (3a - c) = -18 - 2 5c=20-5c = -20 c=4c = 4

Now substitute c=4c = 4 into the first equation:

a2(4)=6a - 2(4) = -6 a8=6a - 8 = -6 a=2a = 2

Solving for bb and dd:

Using the second and fourth equations:

  1. b2d=1b - 2d = 1
  2. 3bd=33b - d = 3

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

3b6d=33b - 6d = 3

Now subtract the second equation from this result:

(3b6d)(3bd)=33(3b - 6d) - (3b - d) = 3 - 3 5d=0-5d = 0 d=0d = 0

Now substitute d=0d = 0 into the second equation:

b2(0)=1b - 2(0) = 1 b=1b = 1

Solution

Thus, the values of aa, bb, cc, and dd that satisfy the matrix equation are:

[abcd]=[2140]\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 4 & 0 \end{bmatrix}

So, the answer is:

A. [abcd]=[2140]\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.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Multiplication
Systems of Linear Equations

Formulas

Matrix multiplication: [A][B] = [C], where each element of C is a sum of products from A and B.
System of equations derived from matrix equality

Theorems

Matrix Equality (two matrices are equal if all corresponding elements are equal)

Suitable Grade Level

Grades 11-12