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Worked Examples · Example 22

Q.Express the matrix B=[2−2−4−1341−2−3]B = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} as the sum of a symmetric and a skew symmetric matrix.

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Any square matrix can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix. For BB, the symmetric part is 12(B+BT)\frac{1}{2}(B+B^T) and the skew-symmetric part is 12(B−BT)\frac{1}{2}(B-B^T). The result is B=[2−32−32−3231−321−3]+[0−12−52120352−30]B = \begin{bmatrix} 2 & -\frac{3}{2} & -\frac{3}{2} \\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3 \end{bmatrix} + \begin{bmatrix} 0 & -\frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0 \end{bmatrix}.

The Core Idea: Every Square Matrix Has a Built-in Mirror

This problem is about matrix decomposition — breaking a matrix into two special pieces that reveal hidden structure. Every square matrix AA can be written as:

A=Symmetric+Skew-symmetricA = \text{Symmetric} + \text{Skew-symmetric}

Why does this always work? Because any matrix AA can be "averaged" with its own transpose. The symmetric part is the average of AA and ATA^T; the skew-symmetric part is half their difference. This is analogous to writing any function as the sum of an even and an odd function — a deep mathematical symmetry.

For any square matrix AA:

Symmetric part: P=12(A+AT)\text{Symmetric part: } P = \frac{1}{2}(A + A^T)

Skew-symmetric part: Q=12(A−AT)\text{Skew-symmetric part: } Q = \frac{1}{2}(A - A^T)

Then A=P+QA = P + Q, PT=PP^T = P, and QT=−QQ^T = -Q.

Step-by-Step Construction

1. Find the transpose BTB^T

The transpose swaps rows and columns. For B=[2−2−4−1341−2−3]B = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}, we get:

BT=[2−11−23−2−44−3]B^T = \begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}

Notice how the diagonal stays the same (2, 3, -3) — this is always true for the transpose.

2. Compute the symmetric part P=12(B+BT)P = \frac{1}{2}(B + B^T)

Add corresponding entries:

B+BT=[2+2−2+(−1)−4+1−1+(−2)3+34+(−2)1+(−4)−2+4−3+(−3)]=[4−3−3−362−32−6]B + B^T = \begin{bmatrix} 2+2 & -2+(-1) & -4+1 \\ -1+(-2) & 3+3 & 4+(-2) \\ 1+(-4) & -2+4 & -3+(-3) \end{bmatrix} = \begin{bmatrix} 4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{bmatrix}

Now divide by 2:

P=12[4−3−3−362−32−6]=[2−32−32−3231−321−3]P = \frac{1}{2} \begin{bmatrix} 4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{bmatrix} = \begin{bmatrix} 2 & -\frac{3}{2} & -\frac{3}{2} \\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3 \end{bmatrix}

Check: PP is symmetric — P12=P21=−32P_{12} = P_{21} = -\frac{3}{2}, P13=P31=−32P_{13} = P_{31} = -\frac{3}{2}, P23=P32=1P_{23} = P_{32} = 1. Good.

3. Compute the skew-symmetric part Q=12(B−BT)Q = \frac{1}{2}(B - B^T)

Subtract entry by entry:

B−BT=[2−2−2−(−1)−4−1−1−(−2)3−34−(−2)1−(−4)−2−4−3−(−3)]=[0−1−51065−60]B - B^T = \begin{bmatrix} 2-2 & -2-(-1) & -4-1 \\ -1-(-2) & 3-3 & 4-(-2) \\ 1-(-4) & -2-4 & -3-(-3) \end{bmatrix} = \begin{bmatrix} 0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0 \end{bmatrix}

Divide by 2:

Q=12[0−1−51065−60]=[0−12−52120352−30]Q = \frac{1}{2} \begin{bmatrix} 0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -\frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0 \end{bmatrix} …

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