Q.Using the properties of determinants, evaluate: $\begin{vmatrix} 0 & xy^2 & xz^2 \\ x^2 y & 0 & yz^2 \\ x^2 z & zy^2 & 0 \end{vmatrix}$

Key idea: every entry carries factors of $x,y,z$ in a pattern; pull them out of the columns and then the rows, leaving a plain numerical determinant.

Step 1 — factor $x$ from $C_1$, $y$ from $C_2$, $z$ from $C_3$:

Step 2 — factor $x$ from $R_1$, $y$ from $R_2$, $z$ from $R_3$:

Step 3 — expand along the first row:

So $\Delta = x^2y^2z^2\cdot 2xyz = 2x^3y^3z^3$.

Intuition

Each entry is a single monomial in $x,y,z$, so instead of a brute expansion we strip the common factors out of every column and every row. Each strip multiplies out front, and the leftover determinant is tiny.

Setting up

Working the steps

1. Factor the columns. Column 1 has common factor $x$, column 2 has $y$, column 3 has $z$:

2. Factor the rows. Now row 1 has common factor $x$, row 2 has $y$, row 3 has $z$:

3. Expand the small determinant along the first row:

4. Multiply back:

Check with $x=1,\ y=2,\ z=3$: the original determinant is $\begin{vmatrix} 0 & 4 & 9 \\ 2 & 0 & 18 \\ 3 & 12 & 0 \end{vmatrix}=432$, and $2\cdot 1^3\cdot 2^3\cdot 3^3 = 432$. ✓

Method: Factoring Common Variables Out of Rows and Columns Before Expanding

This method applies to determinants whose entries are monomials sharing common variable factors across rows and/or columns — instead of expanding a messy $3\times3$ directly, you strip out every shared factor first, leaving a tiny, easy determinant.

Steps

Step 1: Factor a common term out of each column

Scan each column for a variable common to every entry in it (treating a $0$ entry as compatible with any factor) and pull it out in front of the determinant, dividing every entry in that column by the factor as you do:

Step 2: Repeat for rows if a further common factor remains

After factoring columns, check whether each row of what's left also shares a common variable. If so, factor that out too — it's legitimate to factor rows and columns in sequence, as long as each factor is multiplied back in outside the determinant.

Step 3: Expand the small remaining determinant

What's left after two rounds of factoring is usually a determinant with simple $0$s and single variables — expand this by cofactor expansion along whichever row/column has the most zeros.

Step 4: Multiply every factored term back together

Combine all the factors pulled out in Steps 1–2 with the value of the small determinant from Step 3 to get the final answer, and sanity-check by plugging in small numeric values for the variables into both the original and final expressions.

Common Mistakes

Mistake 1: Attempting a direct cofactor expansion instead of factoring first

Why it's wrong: expanding this $3\times3$ determinant directly (without first pulling $x,y,z$ out of the columns and rows) means juggling six degree-5 monomial terms at once, which is slow and highly error-prone. Correct approach: always scan for a common monomial factor in each column (and then each row) before expanding — here $x,y,z$ factor cleanly from the three columns, then again from the three rows.

Mistake 2: Mixing up which factor belongs to which row or column

Why it's wrong: factoring happens in two separate passes (columns, then rows), and assigning the wrong variable to the wrong row/column in the second pass gives a wrong overall power of $x$, $y$, or $z$ in the final answer. Correct approach: track each factoring step explicitly — column factors give $xyz$, and the row factors on the new matrix independently give another $xyz$, for a combined $x^2y^2z^2$.

Mistake 3: Sign error in the small $3\times3$ expansion

Why it's wrong: expanding $\begin{vmatrix}0&y&z\\x&0&z\\x&y&0\end{vmatrix}$ along the first row involves a double-negative in the middle cofactor ($-y(0\cdot0-z\cdot x)=-y(-zx)=+xyz$), and dropping one of the two negative signs gives $-2xyz$ or $0$ instead of $2xyz$. Correct approach: write out $0\cdot0-z\cdot x$ explicitly before applying the cofactor's minus sign, rather than combining the signs mentally.