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Zorluk: OrtaMatrices and Determinants

Given the matrices A=(2103)A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} and B=(1421)B = \begin{pmatrix} 1 & 4 \\ 2 & -1 \end{pmatrix}, what is the product matrix ABAB?

  1. (4763)\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}Cevap
  2. B
    (2403)\begin{pmatrix} 2 & 4 \\ 0 & -3 \end{pmatrix}
  3. C
    (21341)\begin{pmatrix} 2 & 13 \\ 4 & -1 \end{pmatrix}
  4. D
    (4963)\begin{pmatrix} 4 & 9 \\ 6 & -3 \end{pmatrix}

Cevap

The matrix product ABAB is (4763)\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}.
To find ABAB, each element cijc_{ij} is formed by taking the dot product of row ii of matrix AA and column jj of matrix BB. Performing these steps gives c11=2(1)+1(2)=4c_{11} = 2(1) + 1(2) = 4, c12=2(4)+1(1)=7c_{12} = 2(4) + 1(-1) = 7, c21=0(1)+3(2)=6c_{21} = 0(1) + 3(2) = 6, and c22=0(4)+3(1)=3c_{22} = 0(4) + 3(-1) = -3, yielding (4763)\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}.

Adım Adım Çözüm

1
Multiply the first row of AA by the first column of BB to find element (1,1)(1,1).
2(1)+1(2)=2+2=42(1) + 1(2) = 2 + 2 = 4
Matrix multiplication rule requires taking the dot product of rows from the first matrix and columns from the second matrix.
2
Multiply the first row of AA by the second column of BB to find element (1,2)(1,2).
2(4)+1(1)=81=72(4) + 1(-1) = 8 - 1 = 7
Evaluates the element in row 1, column 2 of the resulting matrix.
3
Multiply the second row of AA by the first column of BB to find element (2,1)(2,1).
0(1)+3(2)=0+6=60(1) + 3(2) = 0 + 6 = 6
Evaluates the element in row 2, column 1 of the resulting matrix.
4
Multiply the second row of AA by the second column of BB to find element (2,2)(2,2).
0(4)+3(1)=03=30(4) + 3(-1) = 0 - 3 = -3
Evaluates the element in row 2, column 2 of the resulting matrix.

Anahtar Kavram

Matrix Multiplication
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