Question

Please help me in solving this question. ​

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:A = \begin{gathered}\sf \left[\begin{array}{ccc}1&-1&2\\3&0&-2\\1&0&3\end{array}\right]\end{gathered}$

Calculation of | A |

$\rm :\longmapsto\: |A| = \rm \: = \: \: \: \begin{gathered}\sf \left | \begin{array}{ccc}1&\ - 1&2\\3&0& - 2\\1& 0& 3\end{array}\right | \end{gathered}$

★ Expanding along Second column, we get

$\: \: \rm= \: \:1(9 + 2)$

$\: \: \rm= \: \:11$

$\bf\implies \: |A| = 11$

Calculation of Adjoint of A :-

$\rm :\longmapsto\:c_{11} = {( - 1)}^{1 + 1}\begin{array}{|cc|}\sf 0 &\sf - 2 \\ \sf 0 &\sf 3 \\\end{array} = 0$

$\rm :\longmapsto\:c_{12} = {( - 1)}^{1 + 2}\begin{array}{|cc|}\sf 3 &\sf - 2 \\ \sf 1 &\sf 3 \\\end{array} = - 11$

$\rm :\longmapsto\:c_{13} = {( - 1)}^{1 + 3}\begin{array}{|cc|}\sf 3 &\sf 0 \\ \sf 1 &\sf 0 \\\end{array} = 0$

$\rm :\longmapsto\:c_{21} = {( - 1)}^{2+ 1}\begin{array}{|cc|}\sf - 1 &\sf 2 \\ \sf 0 &\sf 3 \\\end{array} = 3$

$\rm :\longmapsto\:c_{22} = {( - 1)}^{2+ 2}\begin{array}{|cc|}\sf 1 &\sf 2 \\ \sf 1 &\sf 3 \\\end{array} = 1$

$\rm :\longmapsto\:c_{23} = {( - 1)}^{2+ 3}\begin{array}{|cc|}\sf 1 &\sf - 1 \\ \sf 1 &\sf 0 \\\end{array} = - 1$

$\rm :\longmapsto\:c_{31} = {( - 1)}^{3+ 1}\begin{array}{|cc|}\sf - 1 &\sf 2 \\ \sf 0 &\sf - 2 \\\end{array} = 2$

$\rm :\longmapsto\:c_{32} = {( - 1)}^{3+ 2}\begin{array}{|cc|}\sf 1 &\sf 2 \\ \sf 3&\sf - 2 \\\end{array} = 8$

$\rm :\longmapsto\:c_{33} = {( - 1)}^{3+ 3}\begin{array}{|cc|}\sf 1 &\sf - 1 \\ \sf 3&\sf 0 \\\end{array} = 3$

Hence,

$\rm :\longmapsto\:adjA = \begin{gathered}\sf \left[\begin{array}{ccc}0& 3&2\\ - 11&1&8\\0& - 1&3\end{array}\right]\end{gathered}$

Now,

Consider,

$\rm :\longmapsto\:AadjA$

$\: \: \rm= \: \:\begin{gathered}\sf \left[\begin{array}{ccc}1&-1&2\\3&0&-2\\1&0&3\end{array}\right]\end{gathered}\begin{gathered}\sf \left[\begin{array}{ccc}0& 3&2\\ - 11&1&8\\0& - 1&3\end{array}\right]\end{gathered}$

$\: \: \rm= \: \:\begin{gathered}\sf \left[\begin{array}{ccc}11&0&0\\0&11&0\\0&0&11\end{array}\right]\end{gathered}$

$\: \: \rm= \: 11\:\begin{gathered}\sf \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\end{gathered}$

$\: \: \rm= \: \: |A| I$

$\bf\implies \:A(adjA) = |A|I - - (1)$

Now,

Consider,

$\rm :\longmapsto\:(adjA)A$

$\: \: \rm= \: \:\begin{gathered}\sf \left[\begin{array}{ccc}0& 3&2\\ - 11&1&8\\0& - 1&3\end{array}\right]\end{gathered}\begin{gathered}\sf \left[\begin{array}{ccc}1&-1&2\\3&0&-2\\1&0&3\end{array}\right]\end{gathered}$

$\: \: \rm= \: \:\begin{gathered}\sf \left[\begin{array}{ccc}11&0&0\\0&11&0\\0&0&11\end{array}\right]\end{gathered}$

$\: \: \rm= \: 11\:\begin{gathered}\sf \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\end{gathered}$

$\: \: \rm= \: \: |A| I$

$\bf\implies \:(adjA) A= |A|I - - (2)$

★ From equation (1) and equation (2), we get

$\bf\implies \:(adjA) A= A(adjA) = |A|I$