Question

Difficulty: EasyMatrices and Matrix Operations

The matrices AA and BB are defined as follows:

A=[2013],B=[3102]A = \begin{bmatrix} 2 & 0 \\ 1 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix}

What is the product matrix ABAB?

  1. [6237]\begin{bmatrix} 6 & 2 \\ 3 & 7 \end{bmatrix}Answer
  2. B
    [6006]\begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}
  3. C
    [7326]\begin{bmatrix} 7 & 3 \\ 2 & 6 \end{bmatrix}
  4. D
    [5115]\begin{bmatrix} 5 & 1 \\ 1 & 5 \end{bmatrix}
  5. E
    [6066]\begin{bmatrix} 6 & 0 \\ 6 & 6 \end{bmatrix}

Answer

[6237]\begin{bmatrix} 6 & 2 \\ 3 & 7 \end{bmatrix}
The correct matrix is obtained by performing row-by-column matrix multiplication, which yields row 1 entries of 6 and 2, and row 2 entries of 3 and 7.

Step-by-Step Solution

1
Multiply the first row of AA by the first column of BB to find the element in the first row, first column of the product matrix.
2(3)+0(0)=62(3) + 0(0) = 6
The row-by-column rule dictates that the entry at row ii, column jj of the product is the dot product of row ii of the first matrix and column jj of the second matrix.
2
Multiply the first row of AA by the second column of BB to find the element in the first row, second column.
2(1)+0(2)=22(1) + 0(2) = 2
This determines the top-right entry of the resulting matrix.
3
Multiply the second row of AA by the first column of BB to find the element in the second row, first column.
1(3)+3(0)=31(3) + 3(0) = 3
This determines the bottom-left entry of the resulting matrix.
4
Multiply the second row of AA by the second column of BB to find the element in the second row, second column.
1(1)+3(2)=1+6=71(1) + 3(2) = 1 + 6 = 7
This determines the bottom-right entry of the resulting matrix.

Key Concept

Matrix multiplication involves computing the dot product of the rows of the first matrix with the columns of the second matrix.
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