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Exercise 4.4 · Q5

Q.Find the value of the following: [2−243]\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}

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The determinant of a 2×22\times 2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} is ad−bcad - bc. Applying this to [2−243]\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix} gives 2⋅3−(−2)⋅4=6+8=142 \cdot 3 - (-2) \cdot 4 = 6 + 8 = 14.

The determinant is a single number that captures key information about a matrix — for a 2×22\times 2 matrix, it tells you the area scaling factor of the linear transformation it represents. The formula is straightforward: multiply the top-left and bottom-right entries, then subtract the product of the top-right and bottom-left entries.

Let’s work through it:

  1. Identify the entries.

    For the matrix [2−243]\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}, we have:

    • a=2a = 2 (top-left)
    • b=−2b = -2 (top-right)
    • c=4c = 4 (bottom-left)
    • d=3d = 3 (bottom-right)
  2. Apply the determinant formula.

    The determinant is ad−bcad - bc. Substituting:

det⁡=(2)(3)−(−2)(4)\det = (2)(3) - (-2)(4)

  1. Compute carefully. First term: 2×3=62 \times 3 = 6. Second term: (−2)×4=−8(-2) \times 4 = -8. But note the minus sign in front: ad−bcad - bc means we subtract bcbc, so: det⁡=6−(−8)=6+8=14\det = 6 - (-8) = 6 + 8 = 14 …

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