Let M and N be two 3 x 3 matrices such that MN = NM. Further, if M ≠ N² and M² = N⁴, then (a) determ

Q: Wait, why can't we just say $M + N^2 = 0$ since $(M - N^2)(M + N^2) = 0$?

Ah, the classic trap! In real numbers, $ab=0$ implies $a=0$ or $b=0$. But matrices are different. A matrix can be non-zero and still have a determinant of zero. We must use the determinant property: $\det(AB) = \det(A)\det(B) = 0$.

Q: Why does $\det(M + N^2) = 0$ force the determinant of $(M^2 + MN^2)$ to be zero?

Because $M^2 + MN^2$ factors into $M(M + N^2)$. The determinant of a product is the product of determinants. Since $\det(M + N^2) = 0$, the whole product becomes zero.

Q: How can we be sure a non-zero matrix $U$ exists such that $(M^2 + MN^2)U = 0$?

If a matrix $A$ has $\det(A) = 0$, it is singular. This means there exists a non-zero vector $X$ such that $AX = 0$. We can simply construct $U$ by placing $X$ in one column and zeros elsewhere.

Question

Let $M$ and $N$ be two $3 	imes 3$ matrices such that $MN = NM$. Further, if $M 
eq N^2$ and $M^2 = N^4$, then

Options:

(A) determinant of $(M^2 + MN^2)$ is $0$

(B) there is $3 	imes 3$ non-zero matrix $U$ such that $(M^2 + MN^2)U$ is the zero matrix

(C) determinant of $(M^2 + MN^2) \geq 1$

(D) for a $3 	imes 3$ matrix $U$, if $(M^2 + MN^2)U$ equals the zero matrix then $U$ is the zero matrix

Sigma Percentile · Accepted Answer

### Analyzing the Commutativity Condition

We are given two $3 	imes 3$ matrices, $M$ and $N$, such that $MN = NM$. This commutativity is our most powerful tool, allowing us to treat these matrices with the same algebraic grace we apply to scalar variables.

We begin with the given condition:
$$M^2 = N^4$$

This can be rewritten as:
$$M^2 - (N^2)^2 = 0$$

### Factoring the Matrix Expression

Because $M$ and $N$ commute, it follows that $M$ and $N^2$ also commute. This allows us to factor the expression using the difference of squares identity:
$$(M - N^2)(M + N^2) = 0$$

We are given the constraint that $M 
eq N^2$. Therefore, the matrix $(M - N^2)$ is not the zero matrix.

### Evaluating the Determinant

In matrix algebra, the product of two non-zero matrices can result in the zero matrix if the matrices are singular. This implies that the determinant of the second factor must be zero:
$$\det(M + N^2) = 0$$

Now, consider the expression $M^2 + MN^2$. We can factor this as:
$$M^2 + MN^2 = M(M + N^2)$$

Taking the determinant of this product, we apply the property $\det(AB) = \det(A) \cdot \det(B)$:
$$\det(M^2 + MN^2) = \det(M) \cdot \det(M + N^2)$$

Since $\det(M + N^2) = 0$, the entire product is zero:
$$\det(M^2 + MN^2) = 0$$

### Conclusion

This result confirms that the matrix $M^2 + MN^2$ is singular. Consequently, there exists a non-zero vector $X$ such that:
$$(M^2 + MN^2)X = 0$$

This confirms that the matrix is not invertible, which is the core requirement for **Option A**. The existence of such a vector $X$ further allows for the construction of a matrix $U$ to satisfy the conditions described in **Option B**.

Animated Solution for Mathematics - Matrices and Determinants: Let $M$ and $N$ be two $3 \times 3$ matrices such that $M N = N M$ . Further, if $M \neq = N^{2}$ and $M^{2} = N^{4}$ , then

Select Answer:

Visualized Solution

Given Conditions and Commutativity

Rearranging the Main Equation

The Power of Commutativity

Factorizing the Difference of Squares

Analyzing the Factors

The Non-Invertibility Argument

Concluding the Determinant is Zero

Setting Up the Target Expression

Applying the Determinant Property

Evaluating the Final Determinant

Exploring the Singular Matrix Property

Constructing the Non-Zero Matrix $U$

Final Conclusion

The Sigma Insight: Properties of Determinants

Analyzing the Commutativity Condition

Factoring the Matrix Expression

Evaluating the Determinant

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 M and N be two 3×3 matrices such that MN=NM. Further, if M=N2 and M2=N4, then