class 12 maths matrices

Compute the following s

(i) $\left[ {\begin{array}{cccccccccccccccccccc}a&b\\{ - b}&a\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}a&b\\b&a\end{array}} \right]$

(ii) $\left[ {\begin{array}{cccccccccccccccccccc}{{a^2} + {b^2}}&{{b^2} + {c^2}}\\{{a^2} + {c^2}}&{{a^2} + {b^2}}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{2ab}&{2bc}\\{ - 2ac}&{ - 2ab}\end{array}} \right]$

(iii) $\left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&4&{ - 6}\\8&5&{16}\\2&8&5\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{12}&7&6\\8&0&5\\3&2&4\end{array}} \right]$

(iv) $\left[ {\begin{array}{cccccccccccccccccccc}{{{\cos }^2}x}&{{{\sin }^2}x}\\{{{\sin }^2}x}&{{{\cos }^2}x}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{{{\sin }^2}x}&{{{\cos }^2}x}\\{{{\cos }^2}x}&{{{\sin }^2}x}\end{array}} \right]$

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📘 Matrices NCERT,Ex.3.2,Q.No.2,Page.80 SA

Compute the following s

(i) $\left[ {\begin{array}{cccccccccccccccccccc}a&b\\{ - b}&a\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}a&b\\b&a\end{array}} \right]$

(ii) $\left[ {\begin{array}{cccccccccccccccccccc}{{a^2} + {b^2}}&{{b^2} + {c^2}}\\{{a^2} + {c^2}}&{{a^2} + {b^2}}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{2ab}&{2bc}\\{ - 2ac}&{ - 2ab}\end{array}} \right]$

(iii) $\left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&4&{ - 6}\\8&5&{16}\\2&8&5\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{12}&7&6\\8&0&5\\3&2&4\end{array}} \right]$

(iv) $\left[ {\begin{array}{cccccccccccccccccccc}{{{\cos }^2}x}&{{{\sin }^2}x}\\{{{\sin }^2}x}&{{{\cos }^2}x}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{{{\sin }^2}x}&{{{\cos }^2}x}\\{{{\cos }^2}x}&{{{\sin }^2}x}\end{array}} \right]$

Official Solution

VVidaara Team ✓ Verified solution NCERT & Exemplar

.:

(i)

We have,
$\left[ {\begin{array}{cccccccccccccccccccc}a&b\\{ - b}&a\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}a&b\\b&a\end{array}} \right]$ $= \left[ {\begin{array}{cccccccccccccccccccc}{a + a}&{b + b}\\{ - b + b}&{a + a}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{2a}&{2b}\\0&{2a}\end{array}} \right]$

(ii)

We have, $\left[ {\begin{array}{cccccccccccccccccccc}{{a^2} + {b^2}}&{{b^2} + {c^2}}\\{{a^2} + {c^2}}&{{a^2} + {b^2}}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{2ab}&{2bc}\\{ - 2ac}&{ - 2ab}\end{array}} \right]$
$= \left[ {\begin{array}{cccccccccccccccccccc}{{a^2} + {b^2} + 2ab}&{{b^2} + {c^2} + 2bc}\\{{a^2} + {c^2} - 2ac}&{{a^2} + {b^2} - 2ab}\end{array}} \right]$
$= \left[ {\begin{array}{cccccccccccccccccccc}{{{(a + b)}^2}}&{{{(b + c)}^2}}\\{{{(a - c)}^2}}&{{{(a - b)}^2}}\end{array}} \right]$

(iii)

We have, $\left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&4&{ - 6}\\8&5&{16}\\2&8&5\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{12}&7&6\\8&0&5\\3&2&4\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 1 + 12}&{4 + 7}&{ - 6 + 6}\\{8 + 8}&{5 + 0}&{16 + 5}\\{2 + 3}&{8 + 2}&{5 + 4}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{11}&{11}&0\\{16}&5&{21}\\5&{10}&9\end{array}} \right]$

(iv)

We have, $\left[ {\begin{array}{cccccccccccccccccccc}{{{\cos }^2}x}&{{{\sin }^2}x}\\{{{\sin }^2}x}&{{{\cos }^2}x}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{{{\sin }^2}x}&{{{\cos }^2}x}\\{{{\cos }^2}x}&{{{\sin }^2}x}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{{{\cos }^2}x + {{\sin }^2}x}&{{{\sin }^2}x + {{\cos }^2}x}\\{{{\sin }^2}x + {{\cos }^2}x}&{{{\cos }^2}x + {{\sin }^2}x}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&1\\1&1\end{array}} \right]$

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