If B = [[5, 2, 1], [0, 2, 1], [α, 3, -1]] is the inverse of a 3x3 matrix A, then the sum of all valu

Q: Why is $\det(B) = 1/\det(A)$? Can't I just invert the matrix?

That is a fantastic question! While you could theoretically invert the matrix, it would be a nightmare of algebraic complexity. The property $\det(A^{-1}) = 1/\det(A)$ is a powerful shortcut derived from $\det(AA^{-1}) = \det(I) = 1$. Using properties is the hallmark of a JEE topper—work smarter, not harder!

Q: Do I really need to solve for $\alpha$ individually?

Not at all! The question specifically asks for the 'sum of all values of $\alpha$'. Whenever you see a quadratic equation and a request for the sum of roots, your mind should immediately jump to Vieta's formulas. It saves time and minimizes the chance of calculation errors.

Q: What if $\det(A) = 0$?

If $\det(A) = 0$, the matrix $A$ is singular and does not have an inverse. Since the problem explicitly states that $B$ is the inverse of $A$, we are guaranteed that $\det(A) \neq 0$. This is a subtle constraint that ensures the problem is well-defined.

Question

If $B = \begin{pmatrix} 5 & 2 & 1 \ 0 & 2 & 1 \ \alpha & 3 & -1 \end{pmatrix}$ is the inverse of a $3 	imes 3$ matrix A, then the sum of all values of $\alpha$ for which $\det(A) + 1 = 0$, is :

Options:

(A) $0$

(B) $2$

(C) $1$

(D) $-1$

Sigma Percentile · Accepted Answer

### The Elegance of Inverse Matrices

Welcome, future engineer! Today, we are going to unravel a problem that might look like a daunting wall of numbers, but is actually a beautiful dance of matrix properties.

We are given a matrix $B$, which is the inverse of some matrix $A$, and we are tasked with finding the sum of all possible values of $\alpha$ that satisfy the condition $\det(A) + 1 = 0$. Let us embark on this journey together.

### Phase 1: The Bridge

First, we must establish the connection between $A$ and $B$. We know that $B = A^{-1}$.

In the realm of linear algebra, the determinant of an inverse matrix is simply the reciprocal of the determinant of the original matrix. Mathematically, this is expressed as:

$$\det(B) = \frac{1}{\det(A)}$$

This is our bridge! It allows us to translate the condition $\det(A) + 1 = 0$ into something useful for matrix $B$.

Since $\det(A) = -1$, it follows immediately that:

$$\det(B) = \frac{1}{-1} = -1$$

We have already simplified the problem significantly without touching a single element of matrix $B$.

### Phase 2: The Expansion

Now, let us look at the matrix $B$ itself:

$$B = \begin{pmatrix} 5 & 2 & \alpha \ 0 & 2 & 1 \ \alpha & 3 & -1 \end{pmatrix}$$

To find $\alpha$, we must calculate its determinant. We will expand along the first row, keeping the sign convention $(+, -, +)$ in mind.

The expansion looks like this:

$$\det(B) = 5(2(-1) - 3(1)) - 2(0(-1) - \alpha(1)) + \alpha(0(3) - 2(\alpha))$$

Let us break this down carefully. The first term is $5(-2 - 3) = 5(-5) = -25$.

The second term is $-2(0 - \alpha) = 2\alpha$. The third term is $\alpha(0 - 2\alpha) = -2\alpha^2$.

Combining these, we get the expression:

$$\det(B) = -2\alpha^2 + 2\alpha - 25$$

### Phase 3: The Quadratic Equation

We have two expressions for $\det(B)$: one from the property of inverses, which is $-1$, and one from our expansion, which is $-2\alpha^2 + 2\alpha - 25$.

Equating them gives us:

$$-2\alpha^2 + 2\alpha - 25 = -1$$

Rearranging this, we get $-2\alpha^2 + 2\alpha - 24 = 0$. Dividing by $-2$ simplifies this to:

$$\alpha^2 - \alpha - 12 = 0$$

This is a standard quadratic equation. While we could solve for $\alpha$ using the quadratic formula, we are asked for the **sum** of all values of $\alpha$.

### Phase 4: The Elegant Shortcut

This is where we apply the wisdom of Vieta's formulas. For any quadratic equation $a\alpha^2 + b\alpha + c = 0$, the sum of the roots is given by $-\frac{b}{a}$.

In our equation $\alpha^2 - \alpha - 12 = 0$, we have $a = 1$ and $b = -1$.

Therefore, the sum of the values of $\alpha$ is:

$$-\frac{-1}{1} = 1$$

And there we have it! The problem that seemed complex at first glance dissolves into a simple application of properties and Vieta's formulas.

Keep practicing this mindset—looking for the elegant path rather than the brute-force one—and you will master JEE Advanced mathematics in no time. The final answer is **1**.

Animated Solution for Mathematics - Matrices and Determinants: 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 :

Select Answer:

Visualized Solution

The Given Matrix and Condition

Determinant Property of Inverse Matrices

Finding $det (A)$

Evaluating $det (B)$

Expanding $det (B)$

The Determinant Expansion

Simplifying the Terms

The Determinant Expression

Equating to Find $α$

Forming the Quadratic Equation

Sum of Roots of a Quadratic

Final Calculation

The Sigma Insight: Adjoint and Inverse of a Matrix

The Elegance of Inverse Matrices

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let $A = 121 - 1 11 1 - 3 1$ and $10 B = 4 - 5 1 20 - 2 2 α 3$ . If $B$ is the inverse of matrix $A$ , then $α$ is

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 $A = [1 - 1 24]$ . 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 $A = [20 - 1 2]$ . If $B = I -^{5} C_{1} (adj A) +^{5} C_{2} (adj A)^{2} - \dots -^{5} C_{5} (adj A)^{5}$ , then the sum of all elements of the matrix $B$ is

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 _____

If $A = [1 λ 510]$ , $A^{- 1} = α A + β I$ and $α + β = - 2$ , then $4 α^{2} + β^{2} + λ^{2}$ 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 $A = {a_{ij}}$ be a $3 \times 3$ matrix, where $a_{ij} = ⎩ ⎨ ⎧ (- 1)^{j - i} 2 (- 1)^{i + j} if i < j if i = j if i > j$ then $det (3 Adj (2 A^{- 1}))$ is equal to

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