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Q.Express [15−12]\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix} as the sum of a symmetric and a skew-symmetric matrix.

Karnataka PUCKarnataka II PUC Board 2025Subjective· 3mImportance★★★★★
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Any square matrix A=12(A+A′)+12(A−A′)A = \frac12(A+A') + \frac12(A-A'); the first part is symmetric, the second skew-symmetric.

Let

A=[15−12].A = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}.

Step 1 — write the transpose:

A′=[1−152].A' = \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}.

Step 2 — symmetric part P=12(A+A′)P = \dfrac{1}{2}(A + A'):

A+A′=[1+15−1−1+52+2]=[2444],A + A' = \begin{bmatrix} 1+1 & 5-1 \\ -1+5 & 2+2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 4 & 4 \end{bmatrix},

P=12[2444]=[1222].P = \frac{1}{2}\begin{bmatrix} 2 & 4 \\ 4 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix}.

Here P′=PP' = P, so PP is symmetric.

Step 3 — skew-symmetric part Q=12(A−A′)Q = \dfrac{1}{2}(A - A'):

A−A′=[1−15+1−1−52−2]=[06−60],A - A' = \begin{bmatrix} 1-1 & 5+1 \\ -1-5 & 2-2 \end{bmatrix} = \begin{bmatrix} 0 & 6 \\ -6 & 0 \end{bmatrix},

Q=12[06−60]=[03−30].Q = \frac{1}{2}\begin{bmatrix} 0 & 6 \\ -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}. …

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