Let A = [1 -1 1; 2 1 -3; 1 1 1] and 10B = [4 2 2; -5 0 α; 1 -2 3]. If B is the inverse of matrix A,

Q: Wait, why can't I just calculate $A^{-1}$ using the adjoint method and then compare it to $B$?

You absolutely could, but in a JEE Advanced exam, time is your most precious currency. Calculating the adjoint, the determinant, and then dividing every element by the determinant is a long, error-prone process. The 'Sniper Approach'—using the property $A(10B) = 10I$—allows you to extract the specific information you need without doing the heavy lifting of finding the entire inverse matrix. Always look for the shortcut that uses the properties of the system rather than brute force.

Q: Why did we multiply by $10$ on both sides? Is that allowed?

Yes, it is perfectly valid! Since $B = A^{-1}$, we know $AB = I$. Multiplying both sides by a scalar constant like $10$ is a linear operation. So, $10(AB) = 10I$, which is the same as $A(10B) = 10I$. We do this specifically to clear the fractions that would appear if we tried to work with $B$ directly, making our arithmetic much cleaner and less prone to silly mistakes.

Q: How did I know to multiply the first row of $A$ by the third column of $10B$?

Great question! We know that $A(10B) = 10I$. The matrix $10I$ is a diagonal matrix where all off-diagonal elements are $0$. By choosing the first row of $A$ and the third column of $10B$, we are calculating the element at position $(1,3)$ of the resulting matrix. Since the $(1,3)$ element of $10I$ is $0$, this creates a simple linear equation involving $\alpha$ that we can solve instantly. It is all about choosing the path of least resistance.

Question

Let $A = \begin{bmatrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{bmatrix}$ and $10B = \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & \alpha \ 1 & -2 & 3 \end{bmatrix}$. If $B$ is the inverse of matrix $A$, then $\alpha$ is

Options:

(A) $5$

(B) $-1$

(C) $2$

(D) $-2$

Sigma Percentile · Accepted Answer

### The Matrix Mystery: Why Brute Force is Your Enemy

Welcome, future engineers. Today, we are going to dismantle a classic matrix problem that often trips up even the brightest students.

When you see a matrix $A$ and a matrix $10B$ where $B$ is the inverse of $A$, your first instinct might be to jump into the deep end: calculating the inverse of $A$. Stop. Take a breath.

In the world of JEE Advanced, the most elegant solution is rarely the one that requires the most writing. Let us embark on a journey to solve this using the 'Sniper Approach'.

### The Fundamental Truth

We are given:

$$A = \begin{bmatrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{bmatrix}, \quad 10B = \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & \alpha \ 1 & -2 & 3 \end{bmatrix}$$

We are told $B = A^{-1}$. The definition of an inverse is simple yet profound: $AB = I$.

This is the bedrock of our solution. However, working with $B$ directly would introduce fractions, and fractions are where calculation errors love to hide.

Instead, let us multiply both sides of $AB = I$ by $10$. This gives us:

$$A(10B) = 10I$$

Now, we are working with the exact matrix provided in the question. No fractions, no mess, just pure linear algebra.

### The Sniper Approach

Now, look at the target: $10I$. This is the identity matrix scaled by $10$, which looks like this:

$$10I = \begin{bmatrix} 10 & 0 & 0 \ 0 & 10 & 0 \ 0 & 0 & 10 \end{bmatrix}$$

Notice the zeros in the off-diagonal positions? This is a goldmine.

We need to find $\alpha$, which sits in the second row, third column of $10B$. But wait—if we multiply the first row of $A$ by the third column of $10B$, we get the element at position $(1,3)$ of the product matrix.

Looking at our target $10I$, the element at $(1,3)$ is $0$. This is our key! By focusing only on this specific multiplication, we bypass the need to calculate the entire matrix product.

### The Calculation

Let us execute this with precision. We take the first row of $A$, which is $\begin{bmatrix} 1 & -1 & 1 \end{bmatrix}$, and the third column of $10B$, which is $\begin{bmatrix} 2 \ \alpha \ 3 \end{bmatrix}$.

The dot product is:

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

Simplifying this, we get:

$$2 - \alpha + 3 = 0$$

Combining the constants, we have $5 - \alpha = 0$, which leads us directly to **$\alpha = 5$**.

### The Takeaway

Did you see how we turned a potentially tedious problem into a simple arithmetic equation? The beauty of this problem lies not in the complexity of the matrices, but in the strategy of the approach.

Always look for the property that simplifies the system. When you are in the exam hall, remember: don't just calculate—think. You have the tools, you have the logic, and now you have the strategy. Go forth and conquer.

Animated Solution for Mathematics - Matrices and Determinants: 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

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Visualized Solution

Introduction to Matrices $A$ and $10 B$

The Fundamental Property of Inverse

Defining the Target Matrix $10 I$

Strategy for Finding $α$

Equating to the Target Element

Raw Setup (Substitution)

Atomic Compute (Execution)

Final Conclusion

The Sigma Insight: Adjoint and Inverse of a Matrix

The Matrix Mystery

The Fundamental Truth

The Sniper Approach

The Calculation

The Takeaway

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Visualized Solution

Introduction to Matrices .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 10B

The Fundamental Property of Inverse

Defining the Target Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 10I

Strategy for Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α