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Worked Examples · Example 15

Q.Find ABAB, if A=[0−102]A = \begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix} and B=[3500]B = \begin{bmatrix} 3 & 5 \\ 0 & 0 \end{bmatrix}.

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Matrix multiplication ABAB is defined only when the number of columns in AA equals the number of rows in BB. Here AA is 2×22 \times 2 and BB is 2×22 \times 2, so the product exists and is a 2×22 \times 2 matrix. The result is AB=[0000]AB = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.

Why this works: Matrix Multiplication Compatibility

Matrix multiplication isn't just about multiplying numbers — it's about combining the rows of the first matrix with the columns of the second. For ABAB to be defined, the number of columns in AA must match the number of rows in BB. Here:

  • AA is 2×22 \times 2 (2 rows, 2 columns)
  • BB is 2×22 \times 2 (2 rows, 2 columns)

The inner dimensions (2 and 2) match, so the product exists. The outer dimensions (2 and 2) tell us the result is also 2×22 \times 2.

Watch out

A common mistake is to multiply corresponding entries directly (like A11×B11A_{11} \times B_{11}, etc.). That's not matrix multiplication — that's element-wise multiplication, which is a different operation. In matrix multiplication, each entry is a dot product of a row from AA and a column from BB.

Step-by-step calculation

1. Entry (1,1)(1,1) — first row of AA, first column of BB:

  • Row 1 of AA: [0,−1][0, -1]
  • Column 1 of BB: [30]\begin{bmatrix} 3 \\ 0 \end{bmatrix}
  • Dot product: (0)(3)+(−1)(0)=0+0=0(0)(3) + (-1)(0) = 0 + 0 = 0

2. Entry (1,2)(1,2) — first row of AA, second column of BB:

  • Row 1 of AA: [0,−1][0, -1]
  • Column 2 of BB: [50]\begin{bmatrix} 5 \\ 0 \end{bmatrix}
  • Dot product: (0)(5)+(−1)(0)=0+0=0(0)(5) + (-1)(0) = 0 + 0 = 0

3. Entry (2,1)(2,1) — second row of AA, first column of BB:

  • Row 2 of AA: [0,2][0, 2]
  • Column 1 of BB: [30]\begin{bmatrix} 3 \\ 0 \end{bmatrix} …

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