If M is a 3 x 3 matrix, where det M = 1 and MMᵀ = I, where 'I' is an identity matrix, prove that det

Q: Wait, why can't we just say $\det(M - I) = \det(M) - \det(I)$?

This is a classic trap! Determinants are not linear operators. In general, $\det(A + B) \neq \det(A) + \det(B)$. You must treat the matrix expression as a single entity. Always manipulate the matrix expression first using algebraic properties (like factoring) before applying the determinant operator.

Q: Why does the scalar $-1$ become $(-1)^3$ when pulled out of the determinant?

Great catch! Remember that when you factor a scalar $k$ out of a matrix $A$ of order $n \times n$, you are effectively multiplying every row by $k$. Since there are $n$ rows, you pull out $k$ exactly $n$ times. Thus, $\det(kA) = k^n \det(A)$. Since our matrix is $3 \times 3$, $n=3$, so $(-1)^3 = -1$.

Q: Is this result true for any matrix where $MM^T = I$?

Not necessarily! The proof relies on the fact that the dimension is odd ($n=3$). If the dimension were even, say $n=2$, then $(-1)^2 = 1$, and the equation would become $\det(M-I) = \det(M-I)$, which doesn't force the determinant to be zero. The odd dimension is the secret ingredient that forces the sign change!

Question

If $M$ is a $3 	imes 3$ matrix, where $\det M = 1$ and $MM^T = I$, where '$I$' is an identity matrix, prove that $\det(M - I) = 0$.

Sigma Percentile · Accepted Answer

### The Geometry of Orthogonality: A Journey into Matrix Symmetry

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a matrix problem; we are uncovering a deep, structural truth about orthogonal matrices.

Imagine you are standing in a three-dimensional space. An orthogonal matrix $M$ represents a rotation or a reflection—a transformation that preserves the very fabric of distance and angles.

When we are given that $M$ is a $3 	imes 3$ matrix with $\det(M) = 1$ and $MM^T = I$, we are looking at a pure rotation in 3D space. Our goal is to prove that $\det(M - I) = 0$. This isn't just algebra; it's a statement about the existence of a fixed axis of rotation.

### Phase 1: The Power of Substitution

We begin with the expression $\det(M - I)$. We look to our given condition: $MM^T = I$.

We can replace the identity matrix $I$ in our expression with the product $MM^T$. This introduces the matrix $M$ into both terms, creating a common factor. Our expression transforms into $\det(M - MM^T)$.

Since $M$ is on the left of both terms, we factor it out to the left:
$$\det(M(I - M^T))$$

### Phase 2: The Determinant's Elegance

Now, we invoke the multiplicative property of determinants: $\det(AB) = \det(A)\det(B)$. This allows us to split our expression into $\det(M) \cdot \det(I - M^T)$.

We are given that $\det(M) = 1$. This is a beautiful simplification, leaving us with:
$$\det(I - M^T)$$

### Phase 3: The Transpose and the Scalar Trap

We need to reach $\det(M - I)$, but we are currently at $\det(I - M^T)$. We use the property that the determinant of a matrix is identical to the determinant of its transpose: $\det(A) = \det(A^T)$.

So, $\det(I - M^T) = \det((I - M^T)^T)$. Using the properties of transposes, $(I - M^T)^T = I^T - (M^T)^T = I - M$. Thus, we have $\det(I - M)$.

Now, we factor out a $-1$. Remember, when we factor out a scalar $k$ from a matrix of order $n$, it comes out as $k^n$. Since $n=3$, we pull out $(-1)^3 = -1$:
$$\det(I - M) = -\det(M - I)$$

### The Final Revelation

We have shown that $\det(M - I) = -\det(M - I)$. Adding $\det(M - I)$ to both sides gives:
$$2\det(M - I) = 0$$

This forces $\det(M - I) = 0$. We have arrived!

We have proven that for any $3 	imes 3$ rotation matrix, the determinant of $M - I$ must be zero. This is the mathematical signature of a rotation axis—a line of vectors that remain unchanged by the transformation. The final result is **$\det(M - I) = 0$**.

Animated Solution for Mathematics - Matrices and Determinants: If $M$ is a $3 \times 3$ matrix, where $det M = 1$ and $M M^{T} = I$ , where ' $I$ ' is an identity matrix, prove that $det (M - I) = 0$ .

Visualized Solution

Given Conditions for $M$

Substituting $I = M M^{T}$

Factoring the Matrix $M$

Applying $det (A B) = det (A) det (B)$

Using $det (A) = det (A^{T})$

Factoring Out $- 1$

Final Conclusion: $det (M - I) = 0$

The Sigma Insight: Properties of Determinants

The Geometry of Orthogonality

Phase 1

Phase 2

Phase 3

The Final Revelation

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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 M is a 3×3 matrix, where detM=1 and MMT=I, where 'I' is an identity matrix, prove that det(M−I)=0.

Visualized Solution

Given Conditions for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I=MMT

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

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(AB)=det(A)det(B)

Using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(A)=det(AT)

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

Final Conclusion: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(M−I)=0