Animated Solution for Mathematics - Matrices and Determinants: Let $P=[a_{ij}]$ be a $3\times 3$ matrix and let $Q=[b_{ij}]$, where $b_{ij}=2^{i+j}{a}_{ij}$ for $1\le i,j\le 3$. If the determinant of $P$ is 2, then the determinant of the matrix $Q$ is

Welcome, future engineer. Today, we are going to peel back the layers of a matrix problem that, at first glance, looks like a simple exercise in algebra. But beneath the surface, it is a beautiful lesson in the nature of determinants.

When we look at a matrix, we often see just a grid of numbers. But a determinant is not just a number; it is a measure of the 'volume' or the 'scaling factor' of the linear transformation that the matrix represents. When we manipulate the elements of a matrix, we are fundamentally changing the geometry of that transformation.

Defining Our Transformation

We are given a $3\times 3$ matrix $P=[a_{ij}]$. We then construct a new matrix $Q=[b_{ij}]$ where each element is defined by the relation $b_{ij}=2^{i+j}{a}_{ij}$.

This is not a simple scalar multiplication. If we were multiplying the entire matrix $P$ by a scalar $k$, we would write $Q=kP$, and the determinant would scale by $k^3$. But here, the scaling factor depends on the position of the element.

To understand what happens to the determinant, we must visualize the matrix $Q$ explicitly:

Now, we need to find $|Q|$. This is where students often stumble. They try to pull out a single factor of 2 from the whole matrix.

But remember, the determinant is a multilinear function of its rows. This means we can factor out a common term from a single row or a single column at a time. We cannot factor out a term from the entire matrix as if it were a scalar product.

Let us look at the first row of $|Q|$. Every term contains a factor of $2^2$. We can pull that out. The second row contains a factor of $2^3$, and the third row contains a factor of $2^4$:

Calculating the exponent outside, we get $2^{2+3+4}=2^9$. We are left with a new determinant, but we are not finished yet.

Look closely at the remaining determinant. The first column has no common factor. The second column, however, has a common factor of $2^1$, and the third column has a common factor of $2^2$.

Let us extract these as well:

Look at what remains inside the determinant. It is exactly the original matrix $P$! We have successfully transformed the determinant of $Q$ into a scalar multiple of the determinant of $P$.

The total power of 2 we extracted is $2^9\cdot 2^1\cdot 2^2=2^{9+1+2}=2^{12}$. So, we have the elegant relationship:

Since the problem gives us $|P|=2$, we simply substitute this value:

And there it is. The complexity of the element-wise scaling collapses into a simple power of 2. This is the beauty of linear algebra—when you understand the properties of the structures you are working with, even the most intimidating problems become a series of logical, satisfying steps.