If A = [[2, 2], [9, 4]] and I = [[1, 0], [0, 1]], then 10A^-1 is equal to: (A) 6I - A (B) A - 6I (C)

Q: Why did we keep the $\frac{1}{-10}$ outside the matrix instead of dividing every term immediately?

Great question! In competitive exams, we value efficiency. By keeping the scalar outside, we saw that multiplying by 10 would cancel the denominator perfectly. It's about working smarter, not harder. Dividing early often introduces fractions that make addition and subtraction much more prone to silly mistakes.

Q: Is there a faster way to solve this without finding the inverse explicitly?

Yes! You could use the Cayley-Hamilton Theorem, which states that every square matrix satisfies its own characteristic equation: $A^2 - \text{tr}(A)A + |A|I = 0$. Multiplying by $A^{-1}$ gives $A - \text{tr}(A)I + |A|A^{-1} = 0$. This is a powerful shortcut for advanced students to relate $A^{-1}$ directly to $A$ and $I$ without manual inversion!

Q: What if the determinant was zero? Would the inverse still exist?

If $|A| = 0$, the matrix is singular, and the inverse does not exist. Always check the determinant first—it's the gatekeeper of invertibility. If you ever find a determinant of zero in a problem asking for an inverse, stop and re-check your calculations, as the problem might be testing your knowledge of singular matrices.

Question

If $A = \begin{bmatrix} 2 & 2 \ 9 & 4 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$, then $10A^{-1}$ is equal to:

Options:

(A) $6I - A$

(B) $A - 6I$

(C) $4I - A$

(D) $A - 4I$

Sigma Percentile · Accepted Answer

Welcome, future engineers! Today, we are going to peel back the layers of a classic matrix problem. Matrices are not just grids of numbers; they are the language of transformations, and mastering them is a rite of passage for every JEE aspirant.

Let's look at our matrix $A = \begin{bmatrix} 2 & 2 \ 9 & 4 \end{bmatrix}$. Our mission is to find $10A^{-1}$.

### The Strategy of the Mastermind

Before we rush into calculations, let's pause. In the JEE Advanced, time is your most precious currency.

We could calculate $A^{-1}$ and then multiply by $10$, but we should also keep an eye on the options. The options are expressed in terms of $A$ and $I$. This suggests that our final answer will be a linear combination of these two.

### The Inverse Toolset

To find the inverse of any $2 	imes 2$ matrix, we rely on the formula:

$$A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$$

This is our roadmap. First, we need the determinant, $|A|$. For our matrix, this is:

$$|A| = (2 	imes 4) - (2 	imes 9) = 8 - 18 = -10$$

The determinant is $-10$. Since it is non-zero, we are safe—the inverse exists.

Now, for the adjoint. For a $2 	imes 2$ matrix, we don't need to calculate cofactors. We simply swap the main diagonal elements ($2$ and $4$) and negate the off-diagonal elements ($2$ and $9$).

Thus, the adjoint is:

$$	ext{adj}(A) = \begin{bmatrix} 4 & -2 \ -9 & 2 \end{bmatrix}$$

### The Art of Strategic Laziness

Now, we combine these into our formula:

$$A^{-1} = \frac{1}{-10} \begin{bmatrix} 4 & -2 \ -9 & 2 \end{bmatrix}$$

Here is where many students make a mistake. They immediately divide every element by $-10$, creating messy decimals or fractions like $-0.4$ or $0.2$. Don't do that!

Look at the goal: $10A^{-1}$. If we multiply our expression by $10$, the $10$ in the numerator and the $-10$ in the denominator will cancel out perfectly. This is the beauty of mathematical foresight.

$$10A^{-1} = 10 	imes \left( \frac{1}{-10} \begin{bmatrix} 4 & -2 \ -9 & 2 \end{bmatrix} ight) = -1 	imes \begin{bmatrix} 4 & -2 \ -9 & 2 \end{bmatrix}$$

Distributing the negative sign gives us:

$$10A^{-1} = \begin{bmatrix} -4 & 2 \ 9 & -2 \end{bmatrix}$$

### The Final Verification

We have our result. Now, we compare it with the options. Let's test the expression $A - 6I$:

$$A - 6I = \begin{bmatrix} 2 & 2 \ 9 & 4 \end{bmatrix} - 6 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \ 9 & 4 \end{bmatrix} - \begin{bmatrix} 6 & 0 \ 0 & 6 \end{bmatrix}$$

Performing the subtraction element-wise:

$$A - 6I = \begin{bmatrix} 2-6 & 2-0 \ 9-0 & 4-6 \end{bmatrix} = \begin{bmatrix} -4 & 2 \ 9 & -2 \end{bmatrix}$$

It matches perfectly! We have arrived at the solution. Remember, math is not about brute force; it is about finding the most efficient path to the truth. Keep practicing, stay curious, and you will conquer these problems with ease.

Animated Solution for Mathematics - Matrices and Determinants: If $A = [2924]$ and $I = [1001]$ , then $10 A^{- 1}$ is equal to:

Select Answer:

Visualized Solution

Identify Matrix $A$

The Inverse Formula

Calculate Determinant $∣ A ∣$

Simplify the Determinant

Find Adjoint of $A$

Substitute into Inverse Formula

Calculate $10 A^{- 1}$

Distribute the Minus Sign

Evaluate the Options

Setup $A - 6 I$

Perform Matrix Subtraction

Final Conclusion

The Sigma Insight: Adjoint and Inverse of a Matrix

The Strategy of the Mastermind

The Inverse Toolset

The Art of Strategic Laziness

The Final Verification

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A=[29​24​] and I=[10​01​], then 10A−1 is equal to:

Select Answer:

Visualized Solution

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

The Inverse Formula

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

Simplify the Determinant

Find Adjoint of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Substitute into Inverse Formula

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 10A−1