Exercise 2.3 — Question 3

Topic: Evaluating 3×3 Determinants Using Cofactors

Key Concepts

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

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

Cofactor Expansion along Row 1: $|A|=a_{11}{C}_{11}+a_{12}{C}_{12}+a_{13}{C}_{13}$

Singular vs Non-Singular Matrix:

• If $|A|=0$: matrix is singular (no inverse exists)
• If $|A|\ne 0$: matrix is non-singular (inverse exists)

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Worked Example

Evaluate the determinant of: $A=\left(\begin{array}{ccc}1& 2& 3\\ 0& 4& 5\\ 1& 0& 6\end{array}\right)$

Step 1: Expand along Row 1.

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

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.

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

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Properties of Determinants (Key for Q-3)

1. If two rows (or columns) are identical, then $|A|=0$.
2. If a row (or column) consists entirely of zeros, then $|A|=0$.
3. Interchanging two rows changes the sign of the determinant.
4. Multiplying a row by scalar $k$ multiplies the determinant by $k$.
5. Adding a multiple of one row to another does not change the determinant.

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