Let M be a 2 x 2 symmetric matrix with integer entries. Then M is invertible if (a) The first column

Q: Wait, why do we reject the negative root here? Or rather, why do we say the determinant must be non-zero?

A matrix is invertible if and only if it has a non-zero determinant. If the determinant is zero, the matrix is singular, meaning it maps space into a lower dimension (like a line or a point), losing information. We aren't rejecting a root; we are enforcing the condition $\det(M) \neq 0$ to ensure the matrix can be reversed.

Q: Why can't we just use energy conservation or some other physics principle?

This is a pure linear algebra problem! In physics, we often use energy conservation to solve for motion, but here, the 'energy' is the determinant. The determinant is the fundamental scalar that tells us if a transformation is 'collapsing' the space. If the determinant is zero, the transformation is irreversible, just like a system that loses all its degrees of freedom.

Q: Is option D really always true? What if $ac$ is a perfect square but $ac \neq b^2$?

That is a brilliant catch! Let's look closer. Option D states the product $ac$ is not the square of an integer. If $ac$ is not a perfect square, it can never equal $b^2$ because $b$ is an integer, and the square of any integer is a perfect square. Thus, $ac - b^2$ can never be zero. It is a robust guarantee of invertibility!

Question

Let $M$ be a $2 	imes 2$ symmetric matrix with integer entries. Then $M$ is invertible if

Options:

(A) The first column of $M$ is the transpose of the second row of $M$

(B) The second row of $M$ is the transpose of the first column of $M$

(C) $M$ is a diagonal matrix with non-zero entries in the main diagonal

(D) The product of entries in the main diagonal of $M$ is not the square of an integer

Sigma Percentile · Accepted Answer

### The Geometry of Invertibility: A Matrix Odyssey

Imagine you are standing in a two-dimensional plane. You have a transformation, a machine that takes any point $(x, y)$ and maps it to a new location.

This machine is represented by our matrix:
$$M = \begin{bmatrix} a & b \ b & c \end{bmatrix}$$

Because $M$ is symmetric, it has a beautiful, balanced structure. The off-diagonal elements are identical, reflecting a kind of equilibrium. But the real question is: can we reverse this machine? This is the essence of invertibility.

### The Gatekeeper: The Determinant

To know if we can reverse the machine, we must look at the determinant:
$$\det(M) = ac - b^2$$

Think of the determinant as the 'scaling factor' of the area. If the determinant is zero, the machine crushes our 2D plane into a 1D line or a 0D point.

Once crushed, we cannot recover the original coordinates. Therefore, for the matrix to be invertible, we absolutely require $\det(M) 
eq 0$, or:
$$ac - b^2 
eq 0$$

This is our golden rule.

### Testing the Options: The Detective Work

Let us investigate the options provided. Options A and B suggest that the first column is the transpose of the second row.

The first column is $\begin{bmatrix} a \ b \end{bmatrix}$, and the second row is $\begin{bmatrix} b & c \end{bmatrix}$. Its transpose is $\begin{bmatrix} b \ c \end{bmatrix}$.

If these are equal, then $a = b$ and $b = c$. This forces our matrix to be:
$$M = \begin{bmatrix} a & a \ a & a \end{bmatrix}$$

Now, calculate the determinant:
$$\det(M) = a \cdot a - a \cdot a = a^2 - a^2 = 0$$

The machine has crushed our space! These options are definitely not the answer.

### The Elegance of Diagonal Matrices

Now, consider Option C: $M$ is a diagonal matrix with non-zero entries. This means $b = 0$.

Our determinant simplifies beautifully to:
$$\det(M) = ac - 0^2 = ac$$

Since the problem states the diagonal entries $a$ and $c$ are non-zero, their product $ac$ must also be non-zero. The machine is perfectly functional and invertible. Option C is a winner.

### The Number Theory Twist

Finally, let us tackle Option D. It claims that the product $ac$ is not the square of an integer.

We know $b$ is an integer, so $b^2$ is a perfect square. If $ac$ is strictly not a perfect square, then $ac$ can never equal $b^2$.

Consequently, $ac - b^2$ can never be zero. The determinant is guaranteed to be non-zero. This is a powerful, elegant result.

Both Options C and D provide the conditions we need to ensure our matrix remains a valid, reversible transformation. You have successfully navigated the matrix, proving that even in the abstract world of integers, geometry and algebra dance together in perfect harmony.

Animated Solution for Mathematics - Matrices and Determinants: Let $M$ be a $2 \times 2$ symmetric matrix with integer entries. Then $M$ is invertible if

Select Answer:

Visualized Solution

Defining the Symmetric Matrix $M$

Condition for Invertibility

The Determinant Equation

Checking Options A & B

Determinant for Options A & B

Checking Option C (Diagonal Matrix)

Determinant for Option C

Checking Option D

Determinant for Option D

Final Conclusion

The Sigma Insight: Adjoint and Inverse of a Matrix

The Geometry of Invertibility

The Gatekeeper

Testing the Options

The Elegance of Diagonal Matrices

The Number Theory Twist

Similar Questions

For $3 \times 3$ matrices $M$ and $N$ , which of the following statement(s) is (are) NOT correct?

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

If $A = e^{t} e^{t} e^{t} e^{- t} cos t - e^{- t} cos t - e^{- t} sin t 2 e^{- t} sin t e^{- t} sin t - e^{- t} sin t + e^{- t} cos t - 2 e^{- t} cos t$ Then A is-

Let $M$ be a $3 \times 3$ invertible matrix with real entries and let $I$ denote the $3 \times 3$ identity matrix. If $M^{- 1} = adj (adj M)$ , then which of the following statement is/are ALWAYS TRUE?

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 :

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

Let $A = [1 - 1 24]$ . 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 $2 \times 2$ 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 $A$ be a $2 \times 2$ matrix. \\ Statement-1: $a d j (a d j A) = A$ \\ Statement-2: $∣ a d j A ∣ = ∣ A ∣$

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 M be a 2×2 symmetric matrix with integer entries. Then M is invertible if