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How is the determinant of the 2×22\times 2 matrix A=[[3,1],[1,2]]A = [[3, 1], [1, 2]] computed in this clip?

The determinant is computed using the standard 2×22\times 2 rule ad−bcad - bc. For the matrix A=[3112]A = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}, this means multiplying the main diagonal entries (3⋅23 \cdot 2) and subtracting the product of the off-diagonal entries (1⋅11 \cdot 1). The calculation yields 6−1=56 - 1 = 5.

Conditions

  • The matrix is 2×22\times 2.
  • Entries are real numbers.

Reasoning, step by step

  1. Identify the matrix entries: a=3,b=1,c=1,d=2a=3, b=1, c=1, d=2.
  2. Apply the formula det⁡(A)=ad−bc\det(A) = ad - bc.
  3. Compute the main diagonal product: 3⋅2=63 \cdot 2 = 6.
  4. Compute the off-diagonal product: 1⋅1=11 \cdot 1 = 1.
  5. Subtract the off-diagonal product from the main diagonal product: 6−1=56 - 1 = 5.

Example

The video explicitly shows the calculation on screen: '|A| = 6−1=56 - 1 = 5'.

Common misconceptions

  • Adding the products instead of subtracting them.
  • Confusing rows with columns in the cross-multiplication process.

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.