studEE16A
  • Introduction
  • Linear Algebra
    • Linear Equations
      • Description
      • Example Problems
    • Vector Spaces
      • Description
      • Example Problems
    • Inner Products
      • Description
      • Example Problems
    • Determinants
      • Description
      • Example Problems
    • Eigen-everything
      • Description
      • Example Problems
    • Matrices
      • Description
      • Example Problems
    • Least Squares
      • Description
      • Example Problems
    • Gram-Schmidt
      • Description
      • Example Problems
    • Basis
      • Description
      • Example Problems
    • Page Rank
  • Circuits
    • Circuit Basics
    • Capacitance
    • Nodal Analysis
    • Superposition
    • Thevenin and Norton
    • What, When, Where, and Why?
    • Op Amps
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  1. Linear Algebra
  2. Determinants

Example Problems

Calculations

We can either use the properties listed above (mainly derived property 1 and 3) or these two equation to solve for determinants in specific cases.

For 2 x 2 matrices: det([abcd])=ad−bc\text{det}\left(\begin{bmatrix}a & b\\ c & d\end{bmatrix}\right) = ad - bcdet([ac​bd​])=ad−bc

For 3 x 3 matrices: det([abcdefghi])=a⋅det([efhi])−b⋅det([dfgi])+c⋅det([degh])\text{det}\left(\begin{bmatrix}a & b & c\\ d & e & f\\ g & h & i\end{bmatrix}\right) = a \cdot \text{det}\left(\begin{bmatrix}e & f \\ h & i\end{bmatrix}\right) - b \cdot \text{det}\left(\begin{bmatrix}d & f \\ g & i\end{bmatrix}\right) + c \cdot \text{det}\left(\begin{bmatrix}d & e\\ g & h\end{bmatrix}\right)det​​adg​beh​cfi​​​=a⋅det([eh​fi​])−b⋅det([dg​fi​])+c⋅det([dg​eh​])

Example

Let A=([5625])A = \left(\begin{bmatrix}5 & 6 \\ 2 & 5\end{bmatrix}\right)A=([52​65​]). Using the equation for 2 x 2 matrices: det(A)=5⋅5−2⋅6=13\text{det}(\textbf{A}) = 5 \cdot 5 - 2 \cdot 6 = 13det(A)=5⋅5−2⋅6=13

Using the properties listed above:

By derived property 2: det(A)=det([562−25−125])=det([560135])\text{det}(\textbf{A}) = \text{det}\left(\begin{bmatrix}5 & 6 \\ 2 - 2 & 5 - \frac{12}{5}\end{bmatrix}\right) =\text{det}\left(\begin{bmatrix}5 & 6 \\ 0 & \frac{13}{5}\end{bmatrix}\right)det(A)=det([52−2​65−512​​])=det([50​6513​​])

By derived property 3: det([560135])=5⋅135=13\text{det}\left(\begin{bmatrix}5 & 6 \\ 0 & \frac{13}{5}\end{bmatrix}\right) = 5 \cdot \frac{13}{5} = 13det([50​6513​​])=5⋅513​=13

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Last updated 5 years ago

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