Let A be a square matrix all of whose entries are integers. Then which one of the following is true?

Q: Wait, why do we reject the negative root here? Or is it that the determinant can be any integer?

Great question! The determinant of an integer matrix is always an integer. However, the condition $\det(A) = \pm 1$ is special because it ensures that the inverse matrix $A^{-1}$ also consists entirely of integers. If the determinant were, say, $2$, then the inverse would involve dividing by $2$, which would likely introduce fractions. So, we aren't rejecting anything; we are simply identifying the specific case where the inverse remains within the set of integers.

Q: Why can't we just use energy conservation? Wait, this is linear algebra, not physics!

Haha, I love the enthusiasm! You are absolutely right—this is pure linear algebra. In physics, we often look for conserved quantities, but here, our 'conserved' property is the integrity of the integer entries. We are essentially 'conserving' the integer nature of the matrix elements through the operations of addition, multiplication, and division by $\pm 1$.

Q: Is it possible for the inverse to exist if the determinant is zero?

No, it is not. If the determinant of a matrix is zero, the matrix is singular, meaning it does not have an inverse. The formula $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$ would involve division by zero, which is undefined. So, for the inverse to exist, the determinant must be non-zero.

Question

Let $A$ be a square matrix all of whose entries are integers. Then which one of the following is true?

Options:

(A) If $det A = \pm 1$, then $A^{-1}$ exists but all its entries are not necessarily integers

(B) If $det A 
eq \pm 1$, then $A^{-1}$ exists and all its entries are non integers

(C) If $det A = \pm 1$, then $A^{-1}$ exists but all its entries are integers

(D) If $det A = \pm 1$, then $A^{-1}$ need not exist

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a vast, perfectly ordered grid of numbers. Every single entry in this square matrix $A$ is an integer.

There are no messy fractions, no irrational decimals—just the clean, solid bedrock of the integers $\mathbb{Z}$. This is our starting point, and it is a world of beautiful, rigid structure.

Today, we are going to explore what happens when we try to invert this structure.

### The Map: The Inverse Formula

To find the inverse of our matrix $A$, we rely on the classic, powerful formula:

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

This formula is our map. It tells us that the inverse is composed of two distinct entities: the determinant, $\det(A)$, which acts as a scalar gatekeeper, and the adjoint matrix, $	ext{adj}(A)$, which is the engine of the transformation.

Our goal is to understand how these two components preserve the 'integer-ness' of our matrix.

### The Engine: The Adjoint Matrix

Let us look closely at the adjoint matrix, $	ext{adj}(A)$. By definition, it is the transpose of the cofactor matrix.

This means every entry in the adjoint matrix is a cofactor $C_{ji}$ of the original matrix $A$. Now, pause and think: how is a cofactor calculated?

It is the determinant of a smaller submatrix, multiplied by a sign. And what is a determinant? It is a sum of products of the matrix entries.

Since our original matrix $A$ consists only of integers, and integers are closed under addition, subtraction, and multiplication, every single cofactor must also be an integer!

This is a profound realization. It means that the entire adjoint matrix $	ext{adj}(A)$ is a matrix of pure integers. The transpose operation just rearranges these integers, so the adjoint remains a matrix of integers.

### The Gatekeeper: The Determinant

Now, we return to our formula: $A^{-1} = \frac{1}{\det(A)} 	ext{adj}(A)$. We have an adjoint matrix filled with integers.

To find the inverse, we must multiply this entire matrix by the scalar $\frac{1}{\det(A)}$. This is equivalent to dividing every single integer entry in the adjoint matrix by the determinant.

If we want the resulting inverse matrix $A^{-1}$ to also consist only of integers, this division must not create any fractions. This is where the condition $\det(A) = \pm 1$ becomes the hero of our story.

If $\det(A) = 1$, the multiplier is $\frac{1}{1} = 1$. If $\det(A) = -1$, the multiplier is $\frac{1}{-1} = -1$.

In both scenarios, we are simply multiplying our integer-filled adjoint matrix by $1$ or $-1$.

### The Conclusion: A Perfect Result

When you multiply an integer by $1$ or $-1$, the result is always an integer. No fractions are introduced.

The integrity of the integer entries is preserved perfectly. Thus, if **$\det(A) = \pm 1$**, the inverse matrix $A^{-1}$ is guaranteed to have only integer entries.

We have navigated the complexities of the inverse formula, understood the integer nature of the cofactors, and seen how the determinant acts as the final arbiter of the inverse's form. It is a beautiful, logical journey that reveals the hidden harmony within matrix algebra.

Animated Solution for Mathematics - Matrices and Determinants: Let $A$ be a square matrix all of whose entries are integers. Then which one of the following is true?

Select Answer:

Visualized Solution

Defining the Integer Matrix $A$

The Matrix Inverse Formula

Analyzing the Adjoint Matrix

Integer Nature of Cofactors

Entries of the Adjoint Matrix

The Role of the Determinant

Applying the Condition $det (A) = \pm 1$

Final Conclusion

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Map

The Engine

The Gatekeeper

The Conclusion

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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; } Let A be a square matrix all of whose entries are integers. Then which one of the following is true?