2.2 Q-12

Exercise 2.2 — Question 12

Topic: Evaluating a 3×3 Determinant Using Cofactor Expansion

This question requires evaluating a $3\times 3$ determinant by expanding along a row or column using cofactors.

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Key Concepts

Cofactor Definition: The cofactor $C_{ij}$ of element $a_{ij}$ in matrix $A$ is defined as: $C_{ij}=(-1{)}^{i+j}{M}_{ij}$ where $M_{ij}$ is the minor of $a_{ij}$ — the determinant of the $2\times 2$ submatrix obtained by deleting row $i$ and column $j$.

Cofactor Sign Pattern for a $3\times 3$ Matrix: $\left(\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right)$

Cofactor Expansion Along Row 1: $|A|=a_{11}{C}_{11}+a_{12}{C}_{12}+a_{13}{C}_{13}$ $=a_{11}\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|-a_{12}\left|\begin{array}{cc}{a}_{21}& a_{23}\\ a_{31}& a_{33}\end{array}\right|+a_{13}\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|$

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Worked Example

Evaluate the determinant: $\left|\begin{array}{ccc}1& 2& 3\\ 0& 4& 5\\ 1& 0& 6\end{array}\right|$

Step 1: Expand along Row 1.

$=1\cdot C_{11}+2\cdot C_{12}+3\cdot C_{13}$

Step 2: Compute each cofactor.

$C_{11}=(+1)\left|\begin{array}{cc}4& 5\\ 0& 6\end{array}\right|=(4)(6)-(5)(0)=24$

$C_{12}=(-1)\left|\begin{array}{cc}0& 5\\ 1& 6\end{array}\right|=-(0\cdot 6-5\cdot 1)=-(-5)=5$

$C_{13}=(+1)\left|\begin{array}{cc}0& 4\\ 1& 0\end{array}\right|=(0)(0)-(4)(1)=-4$

Step 3: Substitute back.

$|A|=1(24)+2(5)+3(-4)=24+10-12=\boxed{22}$

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Useful Properties of Determinants

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Strategy Tip

Choose the row or column with the most zeros for expansion — this minimises the number of non-zero cofactor terms and reduces computation.