Animated Solution for Mathematics - Matrices and Determinants: If $\Delta =\left|\begin{array}{ccc}x-2& 2x-3& 3x-4\\ 2x-3& 3x-4& 4x-5\\ 3x-5& 5x-8& 10x-17\end{array}\right|=A{x}^3+B{x}^2+Cx+D$, then $B+C$ is equal to :

We are presented with the determinant:

At first glance, this looks like a nightmare of coefficients. If you were to expand this using the standard Sarrus rule or cofactor expansion immediately, you would be writing pages of algebra, and the probability of a sign error would be nearly 100%.

But we are not here to brute force; we are here to think.

The Power of Simplification

The secret to mastering determinants lies in the property that adding or subtracting rows or columns does not change the value of the determinant. We want to reduce the 'weight' of these expressions.

If we perform the column operations $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$, the terms become much smaller:

Suddenly, the complexity has dropped. We have created common terms like $(x-1)$ and $(2x-2)$. This is the beauty of linear algebra—finding the hidden structure beneath the noise.

The Strategy of Zeros

Now that we have simplified the columns, let us target the rows. Our goal is to create zeros, because zeros are the best friends of a mathematician—they make terms vanish during expansion.

By applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$, we arrive at a beautiful state:

Look at that second row! It is now $(x-1,0,0)$. This is a gift. Expanding along this row will be effortless.

The Elegant Expansion

We expand along the second row. Remember the cofactor sign convention: the position $(2,1)$ carries a negative sign.

Thus, we have:

Now, we evaluate the $2\times 2$ minor. Notice that $2x-2=2(x-1)$ and $5x-10=5(x-2)$. Factoring these out, the determinant becomes:

This simplifies to $\Delta =-(x-1)[3(x-1)(x-2)]$, which gives us the elegant factored form:

The Final Comparison

We are almost there. We need to match this with $A{x}^3+B{x}^2+Cx+D$. Let us expand our factored form:

Multiplying this out, we get:

This simplifies to $\Delta =-3(x^3-4x^2+5x-2)$. Distributing the $-3$, we find:

By comparing coefficients, we identify $B=12$ and $C=-15$. The final step is simply:

You see? By choosing the path of simplification over the path of brute force, we turned a daunting problem into a series of logical, satisfying steps. Keep this mindset, and no determinant will ever intimidate you again.