Let I be the identity matrix of order 3 × 3 and for the matrix A = [[λ, 2, 3], [4, 5, 6], [7, -1, 2]

Q: Wait, why can't I just calculate $A^2$ and then find the adjoint? It seems safer.

While technically possible, it is a time-sink that will cost you precious minutes in the exam hall. JEE Advanced rewards those who use properties. The identity $\text{adj}(A^2) = (\text{adj } A)^2$ is a standard tool that saves you from tedious arithmetic. Always look for the property before you start the calculation!

Q: Why does the determinant of $A^{-1}$ equal $-1$?

Remember the fundamental property $|A| \cdot |A^{-1}| = |I| = 1$. Since we are given $|A| = -1$, it follows that $-1 \cdot |A^{-1}| = 1$, which means $|A^{-1}| = -1$. It is a simple but vital relationship to keep in mind!

Q: Is it always true that $\text{adj}(kA) = k^{n-1} \text{adj } A$?

Yes! For an $n \times n$ matrix, pulling a scalar out of an adjoint operation requires raising it to the power of $n-1$. Here $n=3$, so it becomes $k^2$. Don't forget this power, or your scalar will be off by a factor of $k$!

Question

Let $I$ be the identity matrix of order $3 	imes 3$ and for the matrix $A = \begin{bmatrix} \lambda & 2 & 3 \ 4 & 5 & 6 \ 7 & -1 & 2 \end{bmatrix}$, $|A| = -1$. Let $B$ be the inverse of the matrix $	ext{adj}(A 	ext{ adj}(A^2))$. Then $|(\lambda B + I)|$ is equal to _____

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you first look at the matrix 
$$A = \begin{bmatrix} \lambda & 2 & 3 \ 4 & 5 & 6 \ 7 & -1 & 2 \end{bmatrix}$$
you might feel a shiver. There is an unknown $\lambda$ lurking in the first row.

However, in the world of JEE Advanced, every constraint is a gift. We are told $|A| = -1$. This is our anchor.

### Finding the Hidden Variable

We start by expanding the determinant along the first row:
$$|A| = \lambda(10 - (-6)) - 2(8 - 42) + 3(-4 - 35)$$

Simplifying the arithmetic, we obtain:
$$16\lambda - 2(-34) + 3(-39) = -1$$
$$16\lambda + 68 - 117 = -1$$

This yields the linear equation:
$$16\lambda - 49 = -1$$

Solving for $\lambda$:
$$16\lambda = 48 \Rightarrow \lambda = 3$$
We have successfully unlocked the first part of the puzzle.

### The Adjoint Labyrinth

The problem introduces a matrix $B$, defined as the inverse of the adjoint of $A \cdot 	ext{adj}(A^2)$. We use the property $	ext{adj}(A^2) = (	ext{adj } A)^2$ to rewrite the expression:
$$A \cdot 	ext{adj}(A^2) = A \cdot (	ext{adj } A) \cdot (	ext{adj } A)$$

Using the fundamental property $A \cdot 	ext{adj } A = |A| I$, the expression collapses:
$$|A| I \cdot 	ext{adj } A = |A| 	ext{adj } A$$

We now find the adjoint of this term. For a $3 	imes 3$ matrix, pulling a scalar $k$ out of an adjoint results in $k^{n-1} = k^2$:
$$	ext{adj}(|A| 	ext{adj } A) = |A|^2 	ext{adj}(	ext{adj } A)$$

Applying the property $	ext{adj}(	ext{adj } A) = |A|^{n-2} A = |A| A$, we get:
$$|A|^2 (|A| A) = |A|^3 A$$

Given $|A| = -1$, this simplifies to $-A$. Therefore, $B = (-A)^{-1} = -A^{-1}$.

### The Final Transformation

We now tackle the final expression $|\lambda B + I|$. Substituting $\lambda = 3$ and $B = -A^{-1}$:
$$|3(-A^{-1}) + I| = |I - 3A^{-1}|$$

We factor out $A^{-1}$ from the determinant:
$$|(A - 3I)A^{-1}| = |A - 3I| \cdot |A^{-1}|$$

Since $|A| = -1$, then $|A^{-1}| = -1$. Our target is $-|A - 3I|$. We construct $A - 3I$:
$$A - 3I = \begin{bmatrix} 3-3 & 2 & 3 \ 4 & 5-3 & 6 \ 7 & -1 & 2-3 \end{bmatrix} = \begin{bmatrix} 0 & 2 & 3 \ 4 & 2 & 6 \ 7 & -1 & -1 \end{bmatrix}$$

Calculating the determinant:
$$|A - 3I| = 0 - 2(-4 - 42) + 3(-4 - 14)$$
$$|A - 3I| = -2(-46) + 3(-18) = 92 - 54 = 38$$

Since our expression was $-|A - 3I|$, the final result is **-38**.

Animated Solution for Mathematics - Matrices and Determinants: Let $I$ be the identity matrix of order $3 \times 3$ and for the matrix $A = λ 47 25 - 1 362$ , $∣ A ∣ = - 1$ . Let $B$ be the inverse of the matrix $adj (A adj (A^{2}))$ . Then $∣ (λ B + I) ∣$ is equal to _____

Enter Numerical Value:

Visualized Solution

Analyze the Matrix $A$

Expand Determinant to find $λ$

Calculate $λ$

Define Matrix $B$

Simplify the Inner Term

Adjoint of a Scalar Multiple

Adjoint of Adjoint Property

Final Expression for $B$

Setup the Required Determinant

Factor out $A^{- 1}$

Construct Matrix $A - 3 I$

Calculate $∣ A - 3 I ∣$

Final Conclusion

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

Finding the Hidden Variable

The Adjoint Labyrinth

The Final Transformation

Similar Questions

For a $3 \times 3$ matrix M, let trace(M) denote the sum of all the diagonal elements of M. Let A be a $3 \times 3$ matrix such that $∣ A ∣ = \frac{1}{2}$ and trace $(A) = 3$ . If $B = a d j (a d j (2 A))$ , then the value of $∣ B ∣ +$ trace $(B)$ equals :

Let $A$ be a $3 \times 3$ matrix such that $∣ adj (adj (adj A)) ∣ = 1 2^{4}$ . Then $∣ A^{- 1} adj A ∣$ is equal to

Let $A$ be a $2 \times 2$ symmetric matrix such that $A [11] = [37]$ and the determinant of $A$ be 1. If $A^{- 1} = α A + β I$ , where $I$ is an identity matrix of order $2 \times 2$ , then $α + β$ equals _______

Let $a \in R$ and $A$ be a matrix of order $3 \times 3$ such that $det (A) = - 4$ and $A + I = 12 a a 11 102$ , where $I$ is the identity matrix of order $3 \times 3$ . If $det ((a + 1) adj ((a - 1) A))$ is $2^{m} 3^{n}$ , $m, n \in {0, 1, 2, \dots, 20}$ , then $m + n$ is equal to :

If A, B, and $(a d j (A^{- 1}) + a d j (B^{- 1}))$ are non-singular matrices of same order, then the inverse of $A (a d j (A^{- 1}) + a d j (B^{- 1}))^{- 1} B$ , is equal to

Let A be a $3 \times 3$ matrix such that $A + A^{T} = O$ . If $A 1 - 1 0 = 332, A^{2} 1 - 1 0 = - 3 19 - 24$ and $det (adj (2 adj (A + I))) = (2)^{α} \cdot (3)^{β} \cdot (11)^{γ}, α, β, γ$ are non-negative integers, then $α + β + γ$ is equal to .........

Let $A = [1 - 1 24]$ . If $A^{- 1} = α I + β A, α, β \in R, I$ is a $2 \times 2$ identity matrix, then $4 (α - β)$ is equal to :

Let $A$ be a matrix of order $3 \times 3$ and $∣ A ∣ = 5$ . If $∣2 adj (3 A adj (2 A)) ∣ = 2^{α} \cdot 3^{β} \cdot 5^{γ}, α, β, γ \in N$ then $α + β + γ$ is equal to

If $B = 50 α 223 11 - 1$ is the inverse of a $3 \times 3$ matrix A, then the sum of all values of $α$ for which $det (A) + 1 = 0$ , is :

Let $B = 11 α 32 α α 34$ , $α > 2$ be the adjoint of a matrix $A$ and $∣ A ∣ = 2$ . Then $[α - 2 α α] B α - 2 α α$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Expand Determinant to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Define Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B