Math Problem StatementDiketahui persamaan matriks: 2(a 2; -3 1) + (4 -1; 0 b) = (3 2; c 4)(2 d; 1 3). Nilai dari a + b + c + d adalah .....
SolutionThe image contains a matrix equation that reads:
2 ( a 2 − 3 1 ) + ( 4 − 1 0 b ) = ( 3 2 c 4 ) ( 2 d 1 3 ) 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 \\ 1 & 3 \end{pmatrix} 2 ( a − 3 2 1 ) + ( 4 0 − 1 b ) = ( 3 c 2 4 ) ( 2 1 d 3 )
We are tasked with finding the values of a + b + c + d a + b + c + d a + b + c + d .
Step 1: Simplifying the Left Side
First, expand the matrices on the left-hand side.
Multiply the matrix ( a 2 − 3 1 ) \begin{pmatrix} a & 2 \\ -3 & 1 \end{pmatrix} ( a − 3 2 1 ) by 2:
2 ( a 2 − 3 1 ) = ( 2 a 4 − 6 2 ) 2 \begin{pmatrix} a & 2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} 2a & 4 \\ -6 & 2 \end{pmatrix} 2 ( a − 3 2 1 ) = ( 2 a − 6 4 2 )
Add the matrix ( 4 − 1 0 b ) \begin{pmatrix} 4 & -1 \\ 0 & b \end{pmatrix} ( 4 0 − 1 b ) to the result:
( 2 a 4 − 6 2 ) + ( 4 − 1 0 b ) = ( 2 a + 4 4 − 1 − 6 + 0 2 + b ) = ( 2 a + 4 3 − 6 2 + 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} = \begin{pmatrix} 2a + 4 & 3 \\ -6 & 2 + b \end{pmatrix} ( 2 a − 6 4 2 ) + ( 4 0 − 1 b ) = ( 2 a + 4 − 6 + 0 4 − 1 2 + b ) = ( 2 a + 4 − 6 3 2 + b )
Step 2: Simplifying the Right Side
Now, perform matrix multiplication on the right-hand side:
( 3 2 c 4 ) ( 2 d 1 3 ) \begin{pmatrix} 3 & 2 \\ c & 4 \end{pmatrix} \begin{pmatrix} 2 & d \\ 1 & 3 \end{pmatrix} ( 3 c 2 4 ) ( 2 1 d 3 )
Multiply the matrices:
( ( 3 × 2 + 2 × 1 ) ( 3 × d + 2 × 3 ) ( c × 2 + 4 × 1 ) ( c × d + 4 × 3 ) ) = ( 6 + 2 3 d + 6 2 c + 4 c d + 12 ) = ( 8 3 d + 6 2 c + 4 c d + 12 ) \begin{pmatrix} (3 \times 2 + 2 \times 1) & (3 \times d + 2 \times 3) \\ (c \times 2 + 4 \times 1) & (c \times d + 4 \times 3) \end{pmatrix} = \begin{pmatrix} 6 + 2 & 3d + 6 \\ 2c + 4 & cd + 12 \end{pmatrix} = \begin{pmatrix} 8 & 3d + 6 \\ 2c + 4 & cd + 12 \end{pmatrix} ( ( 3 × 2 + 2 × 1 ) ( c × 2 + 4 × 1 ) ( 3 × d + 2 × 3 ) ( c × d + 4 × 3 ) ) = ( 6 + 2 2 c + 4 3 d + 6 c d + 12 ) = ( 8 2 c + 4 3 d + 6 c d + 12 )
Step 3: Equating Both Sides
We now have the following equation:
( 2 a + 4 3 − 6 2 + b ) = ( 8 3 d + 6 2 c + 4 c d + 12 ) \begin{pmatrix} 2a + 4 & 3 \\ -6 & 2 + b \end{pmatrix} = \begin{pmatrix} 8 & 3d + 6 \\ 2c + 4 & cd + 12 \end{pmatrix} ( 2 a + 4 − 6 3 2 + b ) = ( 8 2 c + 4 3 d + 6 c d + 12 )
By comparing the elements of both matrices:
2 a + 4 = 8 2a + 4 = 8 2 a + 4 = 8 → 2 a = 4 2a = 4 2 a = 4 → a = 2 a = 2 a = 2
3 = 3 d + 6 3 = 3d + 6 3 = 3 d + 6 → 3 d = − 3 3d = -3 3 d = − 3 → d = − 1 d = -1 d = − 1
− 6 = 2 c + 4 -6 = 2c + 4 − 6 = 2 c + 4 → 2 c = − 10 2c = -10 2 c = − 10 → c = − 5 c = -5 c = − 5
2 + b = c d + 12 = ( − 5 ) ( − 1 ) + 12 = 5 + 12 = 17 2 + b = cd + 12 = (-5)(-1) + 12 = 5 + 12 = 17 2 + b = c d + 12 = ( − 5 ) ( − 1 ) + 12 = 5 + 12 = 17 → b = 15 b = 15 b = 15
Step 4: Calculating a + b + c + d a + b + c + d a + b + c + d
Now, we calculate:
a + b + c + d = 2 + 15 + ( − 5 ) + ( − 1 ) = 2 + 15 − 5 − 1 = 11 a + b + c + d = 2 + 15 + (-5) + (-1) = 2 + 15 - 5 - 1 = 11 a + b + c + d = 2 + 15 + ( − 5 ) + ( − 1 ) = 2 + 15 − 5 − 1 = 11
Thus, the value of a + b + c + d a + b + c + d a + b + c + d is 11 .
Would you like further explanation on any steps? Here are 5 additional questions related to this:
How do you multiply two matrices step by step?
Why does matrix addition require the same dimensions for both matrices?
What is the determinant of a 2x2 matrix?
How would this problem change if the matrices were 3x3?
Can matrix multiplication be commutative, and if not, why?
Tip: Always double-check matrix dimensions before attempting operations to avoid errors.