Animated Solution for Mathematics - Matrices and Determinants: Let $a,b,c$ be such that $b(a+c)\ne 0$. If $\left|\begin{array}{ccc}a& a+1& a-1\\ -b& b+1& b-1\\ c& c-1& c+1\end{array}\right|+\left|\begin{array}{ccc}a+1& b+1& c-1\\ a-1& b-1& c+1\\ (-1{)}^{n+2}a& (-1{)}^{n+1}b& (-1{)}^{n}c\end{array}\right|=0$, then the value of $n$ is:

When you first look at the equation involving two determinants, it is natural to feel a surge of anxiety. You see variables $a,b,c$, you see powers of $(-1{)}^n$, and you see a $3\times 3$ matrix structure.

The instinct is to grab your pen and start expanding, calculating cofactors, and drowning in a sea of algebraic terms. But stop. Take a breath.

In the world of competitive mathematics, the most powerful tool is not speed, but observation. Let us dissect this problem with the precision of a surgeon.

We are given ${\Delta }_1+{\Delta }_2=0$. Our goal is to make ${\Delta }_2$ look exactly like ${\Delta }_1$.

Look at the columns of ${\Delta }_2$. They are jumbled. But wait! The transpose property tells us that $|A|=|A^T|$.

This is our first move. By transposing ${\Delta }_2$, we turn its rows into columns. Suddenly, the structure starts to emerge. We are not changing the value of the determinant; we are simply changing our perspective.

Now that we have transposed ${\Delta }_2$, we have columns that are almost identical to ${\Delta }_1$, but they are in the wrong order. This is where the 'Dance of Columns' begins.

We need to swap columns to align them with ${\Delta }_1$. Remember the golden rule: every time you swap two columns, you must multiply the determinant by $-1$.

We perform the first swap, $C_1\leftrightarrow C_3$, and a negative sign appears outside. Then, we perform the second swap, $C_2\leftrightarrow C_3$, and another negative sign appears. Two negatives make a positive!

The determinant remains unchanged in value, but its columns are now perfectly aligned with ${\Delta }_1$. This is the beauty of determinant properties—they allow us to manipulate the structure without altering the essence of the expression.

Now, we have two determinants with identical second and third columns. This is the moment of truth.

We use the linearity property of determinants. We can add them together by simply adding their first columns. The expression becomes a single determinant where the first column is the sum of the first columns of ${\Delta }_1$ and ${\Delta }_2$.

When you look at this new first column, you will see a common factor: $(1+(-1{)}^n)$. We pull this factor out, and what remains is exactly ${\Delta }_1$. We have reduced the entire problem to:

We are almost at the finish line. We have a product of two terms equal to zero. We must evaluate ${\Delta }_1$.

Expanding it carefully—and I mean carefully, watching every sign—we find that:

Now, look at the problem statement again. It explicitly tells us that $b(a+c) eq 0$. This is the 'trap catcher.'

Because this term cannot be zero, the only way for the product to be zero is if the other factor, $(1+(-1{)}^n)$, is zero. This leads us to:

For a power of $-1$ to result in $-1$, the exponent $n$ must be an odd integer. And there it is. We didn't just solve a problem; we navigated a maze of properties to find the elegant truth hidden within.