Math Problem Statement

Jika 2(a 2 -3 1) + (4 -1 0 b) = (3 2 c 4)(2 d 2 3), nilai a + b + c + d = ...

Solution

We have the following matrix equation based on the image you provided:

2(a231)+(410b)=(32c4)(2d23)2 \begin{pmatrix} a & 2 \\ -3 & 1 \end{pmatrix} + \begin{pmatrix} 4 & -1 \\ 0 & b \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ c & 4 \end{pmatrix} \begin{pmatrix} 2 & d \\ 2 & 3 \end{pmatrix}

We will first simplify the left-hand side and then solve the matrix multiplication on the right-hand side to compare and find the values of aa, bb, cc, and dd.

Left-hand side simplification:

First, multiply the matrix by 2:

2(a231)=(2a462)2 \begin{pmatrix} a & 2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} 2a & 4 \\ -6 & 2 \end{pmatrix}

Now, add the matrix (410b)\begin{pmatrix} 4 & -1 \\ 0 & b \end{pmatrix}:

(2a462)+(410b)=(2a+4416+02+b)\begin{pmatrix} 2a & 4 \\ -6 & 2 \end{pmatrix} + \begin{pmatrix} 4 & -1 \\ 0 & b \end{pmatrix} = \begin{pmatrix} 2a + 4 & 4 - 1 \\ -6 + 0 & 2 + b \end{pmatrix}

This results in:

(2a+4362+b)\begin{pmatrix} 2a + 4 & 3 \\ -6 & 2 + b \end{pmatrix}

Right-hand side simplification:

Now, we compute the matrix product on the right-hand side:

(32c4)(2d23)\begin{pmatrix} 3 & 2 \\ c & 4 \end{pmatrix} \begin{pmatrix} 2 & d \\ 2 & 3 \end{pmatrix}

Performing matrix multiplication:

((3×2+2×2)(3×d+2×3)(c×2+4×2)(c×d+4×3))\begin{pmatrix} (3 \times 2 + 2 \times 2) & (3 \times d + 2 \times 3) \\ (c \times 2 + 4 \times 2) & (c \times d + 4 \times 3) \end{pmatrix}

This gives:

(6+43d+62c+8cd+12)=(103d+62c+8cd+12)\begin{pmatrix} 6 + 4 & 3d + 6 \\ 2c + 8 & cd + 12 \end{pmatrix} = \begin{pmatrix} 10 & 3d + 6 \\ 2c + 8 & cd + 12 \end{pmatrix}

Equating both sides:

We now compare the left-hand and right-hand sides:

(2a+4362+b)=(103d+62c+8cd+12)\begin{pmatrix} 2a + 4 & 3 \\ -6 & 2 + b \end{pmatrix} = \begin{pmatrix} 10 & 3d + 6 \\ 2c + 8 & cd + 12 \end{pmatrix}

From this, we have the following system of equations:

  1. 2a+4=102a + 4 = 10
  2. 3=3d+63 = 3d + 6
  3. 6=2c+8-6 = 2c + 8
  4. 2+b=cd+122 + b = cd + 12

Solving the system:

  1. From 2a+4=102a + 4 = 10, we solve for aa: 2a=6    a=32a = 6 \implies a = 3

  2. From 3=3d+63 = 3d + 6, we solve for dd: 3d=3    d=13d = -3 \implies d = -1

  3. From 6=2c+8-6 = 2c + 8, we solve for cc: 2c=14    c=72c = -14 \implies c = -7

  4. Substituting d=1d = -1 into the equation 2+b=cd+122 + b = cd + 12: 2+b=(7)(1)+12    2+b=7+12=192 + b = (-7)(-1) + 12 \implies 2 + b = 7 + 12 = 19 So, b=17b = 17

Sum of a+b+c+da + b + c + d:

a+b+c+d=3+1771=12a + b + c + d = 3 + 17 - 7 - 1 = 12

It appears the sum is 12, but none of the options match this exactly. Could you clarify if there is an adjustment needed, or if there may be an issue with the provided question?

Do you need further details or clarifications?

Here are five relative questions:

  1. How can you verify matrix multiplication results step-by-step?
  2. What happens if there are multiple matrix solutions?
  3. How do determinant and inverse matrices play a role in such equations?
  4. What is the significance of non-commutative property in matrix multiplication?
  5. How do we check if a matrix equation has no solution?

Tip: Always check matrix dimensions when performing multiplication to ensure compatibility.

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Math Problem Analysis

Mathematical Concepts

Matrix Equations
Matrix Multiplication
Algebra

Formulas

Matrix addition: A + B = (Aij + Bij) for matrices A and B
Matrix multiplication: A * B = C, where Cij = Σ Aik * Bkj
System of linear equations

Theorems

Matrix Multiplication Properties
System of Equations

Suitable Grade Level

Grades 10-12