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Exercise 3.3 · Q10

Q.Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(i) [351−1]\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}
(ii) [6−22−23−12−13]\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}
(iii) [33−1−2−21−4−52]\begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}
(iv) [15−12]\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}
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Every square matrix AA can be uniquely written as A=P+QA = P + Q where P=12(A+AT)P = \frac{1}{2}(A + A^T) is symmetric and Q=12(A−AT)Q = \frac{1}{2}(A - A^T) is skew-symmetric. We apply this decomposition to each given matrix.

The idea is beautiful in its simplicity. Any square matrix can be split into two parts: one that is symmetric (equal to its own transpose) and one that is skew-symmetric (equal to the negative of its transpose). The trick is to use the transpose itself to manufacture these parts.

If you take any matrix AA, then A+ATA + A^T is always symmetric — because transposing it gives itself back. Similarly, A−ATA - A^T is always skew-symmetric — because transposing it flips its sign. Halving each gives the exact decomposition.

For any square matrix AA:

P=12(A+AT)(symmetric)P = \frac{1}{2}(A + A^T) \quad \text{(symmetric)}

Q=12(A−AT)(skew-symmetric)Q = \frac{1}{2}(A - A^T) \quad \text{(skew-symmetric)}

and A=P+QA = P + Q.

Let's apply this to each matrix.


(i) A=[351−1]A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}

Step 1: Find ATA^T.

Transpose means swap rows and columns:

AT=[315−1]A^T = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}

Step 2: Compute P=12(A+AT)P = \frac{1}{2}(A + A^T).

Add element-wise:

A+AT=[3+35+11+5−1+(−1)]=[666−2]A + A^T = \begin{bmatrix} 3+3 & 5+1 \\ 1+5 & -1+(-1) \end{bmatrix} = \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix}

Now halve:

P=12[666−2]=[333−1]P = \frac{1}{2} \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}

Check: PP is symmetric — P12=P21=3P_{12} = P_{21} = 3.

Step 3: Compute Q=12(A−AT)Q = \frac{1}{2}(A - A^T).

Subtract:

A−AT=[3−35−11−5−1−(−1)]=[04−40]A - A^T = \begin{bmatrix} 3-3 & 5-1 \\ 1-5 & -1-(-1) \end{bmatrix} = \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix}

Halve:

Q=12[04−40]=[02−20]Q = \frac{1}{2} \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}

Check: QQ is skew-symmetric — diagonal entries are zero, and Q12=−Q21Q_{12} = -Q_{21}.

Step 4: Verify. …

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