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Q.Express the matrix (15−12)\begin{pmatrix}1 & 5\\-1 & 2\end{pmatrix} as the sum of a symmetric matrix and a skew-symmetric matrix. OR If A=(31−12)A=\begin{pmatrix}3 & 1\\-1 & 2\end{pmatrix}, then show that A2−5A+7I=0A^2-5A+7I=0. Hence find A−1A^{-1}.

Tripura TbseHigher Secondary (+2 Stage) Examination 2025Subjective· 4mImportance★★★★★
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Any square matrix AA splits as A=A+AT2+A−AT2A=\dfrac{A+A^T}{2}+\dfrac{A-A^T}{2}, where the first part is symmetric and the second is skew-symmetric.

Let A=(15−12)A=\begin{pmatrix}1&5\\-1&2\end{pmatrix}, so AT=(1−152)A^T=\begin{pmatrix}1&-1\\5&2\end{pmatrix}.

Symmetric part P=A+AT2P=\dfrac{A+A^T}{2}:

A+AT=(1+15−1−1+52+2)=(2444) ⇒ P=(1222)A+A^T=\begin{pmatrix}1+1&5-1\\-1+5&2+2\end{pmatrix}=\begin{pmatrix}2&4\\4&4\end{pmatrix}\ \Rightarrow\ P=\begin{pmatrix}1&2\\2&2\end{pmatrix}

(indeed symmetric: PT=PP^T=P).

Skew-symmetric part Q=A−AT2Q=\dfrac{A-A^T}{2}:

A−AT=(1−15−(−1)−1−52−2)=(06−60) ⇒ Q=(03−30)A-A^T=\begin{pmatrix}1-1&5-(-1)\\-1-5&2-2\end{pmatrix}=\begin{pmatrix}0&6\\-6&0\end{pmatrix}\ \Rightarrow\ Q=\begin{pmatrix}0&3\\-3&0\end{pmatrix}

(indeed skew-symmetric: QT=−QQ^T=-Q).

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