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NCERT Exemplar · Q50

Q.Express the matrix [2311−12412]\begin{bmatrix} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{bmatrix} as the sum of a symmetric and a skew symmetric matrix.

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Every square matrix can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix using the formulas A=12(A+AT)+12(A−AT)A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T). For the given matrix, the symmetric part is 12(A+AT)\frac{1}{2}(A + A^T) and the skew-symmetric part is 12(A−AT)\frac{1}{2}(A - A^T).

Why This Works: The Symmetric + Skew-Symmetric Decomposition

Any square matrix AA can be split into two special parts. The symmetric part is a matrix that equals its own transpose — it's "mirrored" across the main diagonal. The skew-symmetric part is a matrix whose transpose equals its negative — the off-diagonal entries are opposite in sign, and the diagonal entries are always zero.

The beauty is that you don't need to guess. There's a direct formula:

A=12(A+AT)⏟symmetric+12(A−AT)⏟skew-symmetricA = \underbrace{\frac{1}{2}(A + A^T)}_{\text{symmetric}} + \underbrace{\frac{1}{2}(A - A^T)}_{\text{skew-symmetric}}

Why does this work? If you take the transpose of 12(A+AT)\frac{1}{2}(A + A^T), you get 12(AT+A)\frac{1}{2}(A^T + A), which is the same matrix — so it's symmetric. And the transpose of 12(A−AT)\frac{1}{2}(A - A^T) is 12(AT−A)=−12(A−AT)\frac{1}{2}(A^T - A) = -\frac{1}{2}(A - A^T), which is exactly the condition for skew-symmetry.

Let's apply this to the given matrix.


Step 1: Write down the given matrix and find its transpose

We have:

A=[2311−12412]A = \begin{bmatrix} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{bmatrix}

The transpose ATA^T is obtained by swapping rows and columns:

AT=[2143−11122]A^T = \begin{bmatrix} 2 & 1 & 4 \\ 3 & -1 & 1 \\ 1 & 2 & 2 \end{bmatrix}


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

Add AA and ATA^T entry by entry:

A+AT=[2+23+11+41+3−1+(−1)2+14+11+22+2]=[4454−23534]A + A^T = \begin{bmatrix} 2+2 & 3+1 & 1+4 \\ 1+3 & -1+(-1) & 2+1 \\ 4+1 & 1+2 & 2+2 \end{bmatrix} = \begin{bmatrix} 4 & 4 & 5 \\ 4 & -2 & 3 \\ 5 & 3 & 4 \end{bmatrix}

Now multiply each entry by 12\frac{1}{2}:

P=12(A+AT)=[222.52−11.52.51.52]P = \frac{1}{2}(A + A^T) = \begin{bmatrix} 2 & 2 & 2.5 \\ 2 & -1 & 1.5 \\ 2.5 & 1.5 & 2 \end{bmatrix}

Notice that PP is symmetric — the entry at (i,j)(i,j) equals the entry at (j,i)(j,i). For example, P12=2P_{12} = 2 and P21=2P_{21} = 2.


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

Subtract ATA^T from AA:

A−AT=[2−23−11−41−3−1−(−1)2−14−11−22−2]=[02−3−2013−10]A - A^T = \begin{bmatrix} 2-2 & 3-1 & 1-4 \\ 1-3 & -1-(-1) & 2-1 \\ 4-1 & 1-2 & 2-2 \end{bmatrix} = \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix}

Now multiply by 12\frac{1}{2}:

Q=12(A−AT)=[01−1.5−100.51.5−0.50]Q = \frac{1}{2}(A - A^T) = \begin{bmatrix} 0 & 1 & -1.5 \\ -1 & 0 & 0.5 \\ 1.5 & -0.5 & 0 \end{bmatrix} …

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