Animated Solution for Mathematics - Matrices and Determinants: Let $A$ be a $3\times 3$ matrix such that $adjA=\left(\begin{array}{ccc}2& -1& 1\\ -1& 0& 2\\ 1& -2& -1\end{array}\right)$ and $B=adj(adjA)$ If $|A|=\lambda$ and $|(B^{-1}{)}^T|=\mu$, then the ordered pair, $(|\lambda |,\mu )$ is equal to

Welcome, fellow traveler on the JEE journey! Today, we are going to demystify a problem that often intimidates students: the double adjoint of a matrix.

When you first look at a $3\times 3$ matrix like

your instinct might be to start calculating cofactors immediately. But wait! In the world of JEE Advanced, the secret is rarely in the brute force; it is in the elegant properties hidden beneath the surface.

Our first mission is to find $\lambda =|A|$. We are given $adjA$.

We know the powerful property: $|adjA|=|A{|}^{n-1}$. Since our matrix is $3\times 3$, $n=3$, so $|adjA|=|A{|}^{3-1}=|A{|}^2$.

Now, let us calculate the determinant of the given matrix $adjA$ by expanding along the first row:

Calculating these minors:

Thus, $|A{|}^2=9$. This confirms that $|A|=\pm 3$, and consequently, $|\lambda |=3$. We have successfully cracked the first part of the code!

Next, we encounter $B=adj(adjA)$. A student might panic here, thinking they need to find the adjoint of the adjoint.

But remember the property: $|adj(adjA)|=|A{|}^{(n-1{)}^2}$. For $n=3$, this becomes:

Since we already know $|A{|}^2=9$, we can simply write:

See how the property turns a mountain of work into a simple square?

Finally, we need $\mu =|(B^{-1}{)}^T|$. Let us break this down.

First, the transpose property: $|(B^{-1}{)}^T|=|B^{-1}|$. Second, the inverse property: $|B^{-1}|=\frac{1}{|B|}$.

Therefore,

We have arrived at our destination: the ordered pair $(|\lambda |,\mu )$ is $(3,\frac{1}{81})$.

Always remember, the beauty of linear algebra lies in these connections. When you see adjoints, think of determinants. When you see inverses, think of reciprocals. Keep practicing, stay curious, and you will master these concepts in no time!