Q.If 𝑨 and 𝑩 are non-singular matrices of same order with 𝒅𝒆𝒕(𝑨) = 𝟓, then [𝒅𝒆𝒕(𝑩−𝟏𝑨𝑩)]² is equal to
(A) 5
(B) 25
(C) 45
(D) 55

The problem gives a relation involving the determinant of a scalar multiple of the adjugate of a $3 \times 3$ matrix $A$. To solve it, you need to connect $|\text{Adj} A|$ back to $|A|$ itself, and also handle the scalar multiplication correctly.

For any square matrix of order $n$, the adjugate satisfies $|\text{Adj} A| = |A|^{n-1}$. This is a standard result that comes from the relation $A \cdot \text{Adj} A = |A| I_n$, and taking determinants on both sides gives $|A| \cdot |\text{Adj} A| = |A|^n$, so for $|A| \neq 0$ we get $|\text{Adj} A| = |A|^{n-1}$. (If $|A| = 0$, the formula still holds, but here we'll see $|A|$ is non-zero.)

Also, when you multiply a matrix by a scalar $k$, each of its $n$ rows gets multiplied by $k$, so the determinant gets multiplied by $k^n$. That is, $|k M| = k^n |M|$ for an $n \times n$ matrix $M$.

Now let's apply these to the given condition.

1. Set up the given equation. We have $|5 \cdot \text{Adj} A| = 5$. Here $n = 3$, so the scalar $5$ multiplies the entire $3 \times 3$ matrix $\text{Adj} A$. Using the scalar multiplication rule:

So the equation becomes

2. Express $|\text{Adj} A|$ in terms of $|A|$. …

We start with the given: $A$ is $3\times 3$, $\det(A) = -1$, and

We need $\det((\det A)B)$. Since $\det A = -1$, this is $\det((-1)B) = \det(-B)$. For a $3\times 3$ matrix, $\det(-B) = (-1)^3 \det B = -\det B$. So we really need $\det B$.

Concept and Intuition

The adjugate satisfies $\operatorname{Adj}(M) = \det(M) \, M^{-1}$ for invertible $M$. This lets us turn adjugates into powers of the determinant times inverses. Since $A$ is invertible ($\det A \neq 0$), we can simplify the nested adjugate step by step. The final result will be a scalar multiple of $A$, making $\det B$ easy.

Step-by-step solution

1. Simplify $\operatorname{Adj}(A^2)$ For any invertible matrix $M$, $\operatorname{Adj}(M) = \det(M) \, M^{-1}$. Here $M = A^2$, so $\det(A^2) = (\det A)^2 = (-1)^2 = 1$. Thus

2. Simplify the inner product $A \cdot \operatorname{Adj}(A^2)$

3. Now take the adjugate of that result We have

Using $\operatorname{Adj}(M) = \det(M) M^{-1}$ again with $M = A^{-1}$:

so

Hence

4. Find $B$ Invert both sides: $$$$

We are given a square matrix $A$ of order $3$. The problem asks for $|\operatorname{Adj}(\operatorname{Adj} A^2)|$. The key is to recall the relationship between a matrix and its adjugate: for any $n \times n$ matrix $M$,

and taking determinants gives

Here $n=3$, so $|\operatorname{Adj}(M)| = |M|^2$. We apply this twice, carefully handling the square on $A$.

1. First, find $|A^2|$. Since $|A^2| = |A|^2$ (determinant of a product is the product of determinants), we have

2. Apply the adjugate determinant formula to $A^2$. For $M = A^2$ and $n=3$,

3. Now consider $\operatorname{Adj}(\operatorname{Adj} A^2)$. Let $B = \operatorname{Adj}(A^2)$. Then we need $|\operatorname{Adj}(B)|$. Again using the formula with $n=3$,

4. Substitute $|B|$ from step 2. We have $|B| = |\operatorname{Adj}(A^2)| = |A|^4$. Therefore, $$|\operatorname{Adj}(B)| = (|A|^4)^2 = |A|^8. …$$

We evaluate

Step 1 — Clear the denominators.

Multiply $R_1$ by $c$, $R_2$ by $a$, $R_3$ by $b$. This scales the determinant by $abc$, so

Step 2 — Reduce.

Write $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$ and expand; the symmetric structure evaluates to …

Concept and Intuition

A key determinant property: replacing a row/column by itself plus a multiple of another row/column leaves the determinant unchanged. Options built this way (like adding row 3, an all-1s or all-(-1)s row, into rows 1/2, or subtracting one column from an adjacent one) preserve the value $D$. But summing every pair of the original three columns (as in option D) is a genuinely different linear combination whose determinant scales by a fixed factor of 2, via the identity $\det[u+v,\,v+w,\,w+u]=2\det[u,v,w]$.

Step-by-Step Solution

1. Let rows of the original matrix be $R_1=(a,b,c)$, $R_2=(a^2,b^2,c^2)$, $R_3=(1,1,1)$, so $D=\det(R_1,R_2,R_3)$.
2. (A): rows are $R_1+R_3$, $R_2+R_3$, $R_3$ — obtained by $R_1\to R_1+R_3$, $R_2\to R_2+R_3$, both determinant-preserving row operations, so this equals $D$.
3. (B): viewing by columns $C_a,C_b,C_c$ (with $C_a=(a,a^2,1)^T$ etc.), the new columns are $C_a-C_b$, $C_b-C_c$, $C_c$ — again column operations of the "add multiple of one column to another" type, preserving the determinant, so this equals $D$ (verified directly by multilinear expansion too).
4. (C): rows are $R_1+R_2$, $R_1+R_3$, $-R_3$; expanding by multilinearity, $\det(R_1+R_2,R_1+R_3,-R_3)=-\det(R_1+R_2,R_1+R_3,R_3)$. Since $\det(R_1+R_2,R_1+R_3,R_3)=\det(R_2,R_1,R_3)=-D$ (the $R_1,R_3$ terms vanish by repeated rows), the whole expression is $-(-D)=D$. So (C) equals $D$ too. …

Concept and Intuition

Rotating axes about a fixed origin is an orthogonal (congruence) transformation of the conic's coefficient matrix $M$: $M'=R^{T}MR$ with $R$ orthogonal ($\det R=\pm1$). Since $\det(M')=\det(R)^2\det(M)=\det(M)$, this determinant is unchanged by rotation — so instead of transforming the whole equation, we can just compute the determinant from the original coefficients.

Step-by-Step Solution

1. Original equation: $x^2+y^2-2x-4y-20=0$, so in the general conic form $ax^2+2hxy+by^2+2gx+2fy+c=0$: $a=1$, $h=0$, $b=1$, $2g=-2\Rightarrow g=-1$, $2f=-4\Rightarrow f=-2$, $c=-20$.
2. This determinant is invariant under rotation of axes about the same origin (translation is not involved here, only rotation), so we can evaluate it directly using these original coefficients rather than the rotated ones. …

We are given a $3\times 3$ determinant with entries that look like ratios of $a,b,c$. The key idea is to factor out common factors from each row and each column to simplify the matrix into something more manageable. This works because factoring a constant from a row multiplies the determinant by that constant; factoring from a column does the same. By doing this cleverly, we can reduce the determinant to one with simple integer entries.

Let’s denote the determinant by $\Delta$:

1. Factor out common factors from each row.

   • Row 1: each term has a factor $\frac{1}{a}$. Actually, look: $-\frac{bc}{a^2} = \frac{1}{a}\cdot\left(-\frac{bc}{a}\right)$, $\frac{c}{a} = \frac{1}{a}\cdot c$, $\frac{b}{a} = \frac{1}{a}\cdot b$. So factor $\frac{1}{a}$ from row 1.
   • Row 2: factor $\frac{1}{b}$ (since $\frac{c}{b} = \frac{1}{b}\cdot c$, etc.)
   • Row 3: factor $\frac{1}{c}$.

   Thus:

2. Now factor out common factors from each column.

   • Column 1: entries are $-\frac{bc}{a},\; c,\; b$. Factor $\frac{1}{a}$? Actually, notice $-\frac{bc}{a} = \frac{1}{a}(-bc)$, $c = \frac{1}{a}(ac)$, $b = \frac{1}{a}(ab)$. So factor $\frac{1}{a}$ from column 1.
   • Column 2: entries are $c,\; -\frac{ac}{b},\; a$. Factor $\frac{1}{b}$? Check: $c = \frac{1}{b}(bc)$, $-\frac{ac}{b} = \frac{1}{b}(-ac)$, $a = \frac{1}{b}(ab)$. Yes, factor $\frac{1}{b}$.
   • Column 3: entries are $b,\; a,\; -\frac{ab}{c}$. Factor $\frac{1}{c}$: $b = \frac{1}{c}(bc)$, $a = \frac{1}{c}(ac)$, $-\frac{ab}{c} = \frac{1}{c}(-ab)$.

   So:

The product of the six factors is $\frac{1}{a^2 b^2 c^2}$.

3. Factor out common factors from each row again (the new matrix).

   • Row 1: each term has factor $bc$.
   • Row 2: each term has factor $ac$.
   • Row 3: each term has factor $ab$.

   So:

   $\Delta = \frac{1}{a^2 b^2 c^2} \cdot (bc)(ac)(ab) \begin{vmatrix} -1 & 1 & 1 \\[6pt] 1 & -1 & 1 \\[6pt] 1 & 1 & -1 \end{vmatrix}$ …

Cofactor expansion along the first row

Each $2\times2$ minor:

So the determinant equals

$(b^2c^3-b^3c^2)-(a^2c^3-a^3c^2)+(a^2b^3-a^3b^2).$ …