If A is a symmetric matrix and B is a skew-symmetric matrix such that A + B = [[2, 3], [5, -1]], the

Q: Wait, why do we reject the negative root here? Or is there a simpler way to find A and B?

There are no roots to reject here because this is a linear system, not a quadratic one! The beauty of this problem lies in the uniqueness theorem: every square matrix $M$ can be uniquely written as $M = A + B$, where $A = \frac{1}{2}(M + M^T)$ and $B = \frac{1}{2}(M - M^T)$. You don't need to guess or solve complex equations; just use this formula directly.

Q: Why does $(A+B)^T$ become $A-B$?

Great question! Remember the property $(A+B)^T = A^T + B^T$. Since $A$ is symmetric, $A^T = A$. Since $B$ is skew-symmetric, $B^T = -B$. Substituting these, we get $A^T + B^T = A + (-B) = A - B$. It is not magic; it is just the direct application of the definitions of symmetric and skew-symmetric matrices.

Q: Is there a faster way to find $AB$ without finding $A$ and $B$ explicitly?

While you could potentially manipulate the expressions for $A$ and $B$ algebraically, for a $2 \times 2$ matrix, finding $A$ and $B$ explicitly is the most robust and least error-prone method. In JEE Advanced, accuracy is just as important as speed. Calculating $A$ and $B$ first ensures you don't fall into algebraic traps during the multiplication.

Question

If A is a symmetric matrix and B is a skew-symmetric matrix such that $A + B = \begin{pmatrix} 2 & 3 \ 5 & -1 \end{pmatrix}$, then AB is equal to :

Options:

(A) $\begin{pmatrix} -4 & 2 \ 1 & 4 \end{pmatrix}$

(B) $\begin{pmatrix} -4 & -2 \ -1 & 4 \end{pmatrix}$

(C) $\begin{pmatrix} 4 & -2 \ -1 & -4 \end{pmatrix}$

(D) $\begin{pmatrix} 4 & 2 \ 1 & -4 \end{pmatrix}$

Sigma Percentile · Accepted Answer

### The Hidden Symmetry of Matrices: A Journey into Decomposition

My dear student, welcome to a fascinating exploration of linear algebra. Today, we are not just solving a matrix problem; we are uncovering the hidden architecture of a square matrix.

Imagine you are handed a matrix $M = \begin{pmatrix} 2 & 3 \ 5 & -1 \end{pmatrix}$ and told it is a composite of two distinct personalities: a symmetric matrix $A$ and a skew-symmetric matrix $B$. How do we peel back the layers to reveal these two components? This is the core of our challenge.

### Phase 1: The Mirror of Transposition

To understand $A$ and $B$, we must use the 'mirror' of linear algebra: the transpose operation. We know that $A$ is symmetric, meaning $A^T = A$, and $B$ is skew-symmetric, meaning $B^T = -B$.

When we take the transpose of our given equation $A + B = M$, we get $(A + B)^T = M^T$. Using the distributive property of transposes, this becomes $A^T + B^T = M^T$.

Substituting our definitions, we arrive at the elegant result: $A - B = M^T$. Now, we have a beautiful system of two linear equations: $A + B = M$ and $A - B = M^T$.

### Phase 2: Solving the System

With our two equations, $A + B = \begin{pmatrix} 2 & 3 \ 5 & -1 \end{pmatrix}$ and $A - B = \begin{pmatrix} 2 & 5 \ 3 & -1 \end{pmatrix}$, we can isolate our variables. By adding these equations, the $B$ terms cancel out, leaving $2A = M + M^T$.

Performing the addition:

$$2A = \begin{pmatrix} 2+2 & 3+5 \ 5+3 & -1-1 \end{pmatrix} = \begin{pmatrix} 4 & 8 \ 8 & -2 \end{pmatrix}$$

Dividing by $2$, we find $A = \begin{pmatrix} 2 & 4 \ 4 & -1 \end{pmatrix}$. Notice how $A$ is perfectly symmetric across the main diagonal? That is our confirmation.

Similarly, by subtracting the equations, we find $2B = M - M^T$:

$$2B = \begin{pmatrix} 2-2 & 3-5 \ 5-3 & -1-(-1) \end{pmatrix} = \begin{pmatrix} 0 & -2 \ 2 & 0 \end{pmatrix}$$

Thus, $B = \begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}$. The zeros on the main diagonal confirm $B$ is indeed skew-symmetric.

### Phase 3: The Final Product

Now, we simply compute the product $AB$. We multiply our matrices:

$$AB = \begin{pmatrix} 2 & 4 \ 4 & -1 \end{pmatrix} \begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}$$

Calculating the elements:
*   The first row, first column is $(2)(0) + (4)(1) = 4$.
*   The first row, second column is $(2)(-1) + (4)(0) = -2$.
*   The second row, first column is $(4)(0) + (-1)(1) = -1$.
*   The second row, second column is $(4)(-1) + (-1)(0) = -4$.

Our final matrix is **$\begin{pmatrix} 4 & -2 \ -1 & -4 \end{pmatrix}$**.

This journey shows us that matrices are not just grids of numbers; they are structures that can be decomposed and understood through fundamental properties. Keep this 'Golden Rule' in your toolkit: any square matrix $M$ is simply the sum of its symmetric part $\frac{1}{2}(M + M^T)$ and its skew-symmetric part $\frac{1}{2}(M - M^T)$. Master this, and you master the matrix.

Animated Solution for Mathematics - Matrices and Determinants: If A is a symmetric matrix and B is a skew-symmetric matrix such that $A + B = (25 3 - 1)$ , then AB is equal to :

Select Answer:

Visualized Solution

Properties of Symmetric and Skew-Symmetric Matrices

Applying the Transpose Operation

Substituting the Matrix Properties

Setting Up the System of Equations

Solving for Matrix $A$

Finalizing Matrix $A$

Solving for Matrix $B$

Finalizing Matrix $B$

Setting up Matrix Multiplication $A B$

Computing the First Row of $A B$

Computing the Second Row of $A B$

Final Result and Key Takeaway

The Sigma Insight: Types of Matrices

The Hidden Symmetry of Matrices

Phase 1

Phase 2

Phase 3

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Comprehension Passage

The number of $A$ in $T_{p}$ such that $A$ is either symmetric or skew-symmetric or both, and $det (A)$ divisible by $p$ is

The number of $A$ in $T_{p}$ such that the trace of $A$ is not divisible by $p$ but $det (A)$ is divisible by $p$ is

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Select Answer:

Visualized Solution

Properties of Symmetric and Skew-Symmetric Matrices

Applying the Transpose Operation

Substituting the Matrix Properties

Setting Up the System of Equations

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

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

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

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

Setting up Matrix Multiplication .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Computing the First Row of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Computing the Second Row of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB