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| XI Basic Mathematics Solution Exercise 5.3 |
Exercise 5.3
1) Find the adjoints and inverse of the following matrices, if possible.
(i)$\left( {\begin{array}{*{20}{c}}2&1\\4&6\end{array}} \right)$
Solution:
Let A = $\left( {\begin{array}{*{20}{c}}2&1\\4&6\end{array}} \right)$
A11 = 6, A12 = –4, A21 = –1, A22 = 2
Matrix of cofactors = $\left( {\begin{array}{*{20}{c}}{{{\rm{A}}_{11}}}&{{{\rm{A}}_{12}}}\\{{{\rm{A}}_{21}}}&{{{\rm{A}}_{22}}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6&{ - 4}\\{ - 1}&2\end{array}} \right)$
Adj A = $\left( {\begin{array}{*{20}{c}}6&{ - 1}\\{ - 4}&2\end{array}} \right)$
|A| = $\left| {\begin{array}{*{20}{c}}2&1\\4&6\end{array}} \right|$ = 12 – 4 = 8 ≠ 0.
A–1 = $\frac{{{\rm{adj}}.{\rm{A}}}}{{\left| {\rm{A}} \right|}}$ = $\frac{{\left( {\begin{array}{*{20}{c}}6&{ - 1}\\{ - 4}&2\end{array}} \right)}}{8}$ = $\left( {\begin{array}{*{20}{c}}{\frac{3}{4}}&{ - \frac{1}{8}}\\{ - \frac{1}{2}}&{\frac{1}{4}}\end{array}} \right)$.
(ii)$\left( {\begin{array}{*{20}{c}}2&4\\1&3\end{array}} \right)$
Solution:
Let A = $\left( {\begin{array}{*{20}{c}}2&4\\1&3\end{array}} \right)$
A11 = 3, A12 = –1, A21 = –4, A22 = 2
Matrix of cofactors = $\left( {\begin{array}{*{20}{c}}{{{\rm{A}}_{11}}}&{{{\rm{A}}_{12}}}\\{{{\rm{A}}_{21}}}&{{{\rm{A}}_{22}}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&{ - 1}\\{ - 4}&2\end{array}} \right)$
Adj A = $\left( {\begin{array}{*{20}{c}}3&{ - 4}\\{ - 1}&2\end{array}} \right)$
|A| = $\left| {\begin{array}{*{20}{c}}2&4\\1&3\end{array}} \right|$ = 6 – 4 = 2 ≠ 0.
So, A–1exists.
A–1 = $\frac{{{\rm{adj}}.{\rm{A}}}}{{\left| {\rm{A}} \right|}}$ = $\frac{{\left( {\begin{array}{*{20}{c}}3&{ - 4}\\{ - 1}&2\end{array}} \right)}}{2}$ = $\left( {\begin{array}{*{20}{c}}{\frac{3}{2}}&{ - 2}\\{ - \frac{1}{2}}&1\end{array}} \right)$.
(iii)$\left( {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right)$
Solution:
Let A = $\left( {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right)$
A11 = 1, A12 = –1, A21 = –2, A22 = 3
Matrix of cofactors = $\left( {\begin{array}{*{20}{c}}{{{\rm{A}}_{11}}}&{{{\rm{A}}_{12}}}\\{{{\rm{A}}_{21}}}&{{{\rm{A}}_{22}}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 2}&3\end{array}} \right)$
Adj A = $\left( {\begin{array}{*{20}{c}}1&{ - 2}\\{ - 1}&3\end{array}} \right)$
|A| = $\left| {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right|$ = 3 – 2 = 1 ≠ 0.
So, A–1exists.
A–1 = $\frac{{{\rm{adj}}.{\rm{A}}}}{{\left| {\rm{A}} \right|}}$ = $\frac{{\left( {\begin{array}{*{20}{c}}1&{ - 2}\\{ - 1}&3\end{array}} \right)}}{1}$ = $\left( {\begin{array}{*{20}{c}}1&{ - 2}\\{ - 1}&3\end{array}} \right).$
(iv)$\left( {\begin{array}{*{20}{c}}1&0&1\\0&1&1\\1&1&0\end{array}} \right)$
Solution:
A11 = cofactor of 1 = $\left| {\begin{array}{*{20}{c}}1&1\\1&0\end{array}} \right|$ = 0 – 1 = –1
A12 = cofactor of 0 = $ - \left| {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right|$ = –(0 – 1) = 1
A13 = cofactor of 1 = $\left| {\begin{array}{*{20}{c}}0&1\\1&1\end{array}} \right|$ = 0 – 1 = –1
A21 = cofactor of 0 = $ - \left| {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right|$ = –(0 – 1) = 1
A22 = cofactor of 1 = $\left| {\begin{array}{*{20}{c}}1&1\\1&0\end{array}} \right|$ = 0 – 1 = –1
A23 = cofactor of 1 = $ - \left| {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right|$ = –(1 – 0) = 1
A31 = cofactor of 1 = $\left| {\begin{array}{*{20}{c}}0&1\\1&1\end{array}} \right|$ = 0 – 1 = –1
A32 = cofactor of 0 = $ - \left| {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right|$ = –(1 – 0)= 1
Matrix of cofactors = $\left( {\begin{array}{*{20}{c}}{{{\rm{A}}_{11}}}&{{{\rm{A}}_{12}}}&{{{\rm{A}}_{13}}}\\{{{\rm{A}}_{21}}}&{{{\rm{A}}_{22}}}&{{{\rm{A}}_{23}}}\\{{{\rm{A}}_{31}}}&{{{\rm{A}}_{32}}}&{{{\rm{A}}_{33}}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 1}&1&{ - 1}\\1&{ - 1}&{ - 1}\\{ - 1}&{ - 1}&1\end{array}} \right)$
Adj. A = $\left( {\begin{array}{*{20}{c}}{ - 1}&1&{ - 1}\\1&{ - 1}&{ - 1}\\{ - 1}&{ - 1}&1\end{array}} \right)$
|A| = $\left| {\begin{array}{*{20}{c}}1&0&1\\0&1&1\\1&1&0\end{array}} \right|$
= 1$\left| {\begin{array}{*{20}{c}}1&1\\1&0\end{array}} \right| - 0\left| {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right| + 1\left| {\begin{array}{*{20}{c}}0&1\\1&1\end{array}} \right|$
= 1(0 – 1) – 0 + 1(0 – 1) = –2 ≠ 0.
So, A–1 exists.
A–1 = $\frac{{{\rm{Adj}},{\rm{A}}}}{{\left| {\rm{A}} \right|}}$ = $\frac{{\left( {\begin{array}{*{20}{c}}{ - 1}&1&{ - 1}\\1&{ - 1}&{ - 1}\\{ - 1}&{ - 1}&1\end{array}} \right)}}{{ - 2}}$ = $ - \frac{1}{2}\left( {\begin{array}{*{20}{c}}{ - 1}&1&{ - 1}\\1&{ - 1}&{ - 1}\\{ - 1}&{ - 1}&1\end{array}} \right)$
(v)$\left( {\begin{array}{*{20}{c}}1&2&4\\0&1&6\\1&3&2\end{array}} \right)$
Solution:
A11 = $\left| {\begin{array}{*{20}{c}}1&6\\3&2\end{array}} \right|$ = 2 – 18 = –6;
A12 = $ - \left| {\begin{array}{*{20}{c}}0&6\\1&2\end{array}} \right|$ = –(0 – 6) = 6;
A13 = $\left| {\begin{array}{*{20}{c}}0&1\\1&3\end{array}} \right|$ = 0 – 1 = –1;
A21 = $\left| {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right|$ = –(4 – 12) = 8;
A22 = $\left| {\begin{array}{*{20}{c}}1&4\\1&2\end{array}} \right|$ = 2 – 4 = –2;
A23 = $\left| {\begin{array}{*{20}{c}}1&2\\1&3\end{array}} \right|$ = –(3 – 2) = –1;
A31 = $\left| {\begin{array}{*{20}{c}}2&4\\1&4\end{array}} \right|$ = 12 – 4 = 8;
A32 = $ - \left| {\begin{array}{*{20}{c}}1&4\\0&6\end{array}} \right|$ = –(6 – 0) = –6;
A33 = $\left| {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right|$ = 1 – 0 = 1;
Matrix of cofactors = $\left( {\begin{array}{*{20}{c}}{{{\rm{A}}_{11}}}&{{{\rm{A}}_{12}}}&{{{\rm{A}}_{13}}}\\{{{\rm{A}}_{21}}}&{{{\rm{A}}_{22}}}&{{{\rm{A}}_{23}}}\\{{{\rm{A}}_{31}}}&{{{\rm{A}}_{32}}}&{{{\rm{A}}_{33}}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 16}&6&{ - 1}\\8&{ - 2}&{ - 1}\\8&{ - 6}&1\end{array}} \right)$
Adj. A = $\left( {\begin{array}{*{20}{c}}{ - 16}&8&8\\6&{ - 2}&{ - 6}\\{ - 1}&{ - 1}&1\end{array}} \right)$
|A| = $\left| {\begin{array}{*{20}{c}}1&2&4\\0&1&6\\1&3&2\end{array}} \right|$
= 1$\left| {\begin{array}{*{20}{c}}1&6\\3&2\end{array}} \right| + 1\left| {\begin{array}{*{20}{c}}2&4\\1&6\end{array}} \right|$
= 1(2 – 18) + 1(12 – 4) = – 8 ≠ 0.
So, A–1 exists.
A–1 = $\frac{{{\rm{Adj}},{\rm{A}}}}{{\left| {\rm{A}} \right|}}$ = $\frac{{\left( {\begin{array}{*{20}{c}}{ - 16}&8&8\\6&{ - 2}&{ - 6}\\{ - 1}&{ - 1}&1\end{array}} \right)}}{{ - 8}}$ = $ - \frac{1}{2}\left( {\begin{array}{*{20}{c}}2&{ - 1}&{ - 1}\\{ - \frac{3}{4}}&{\frac{1}{4}}&{\frac{3}{4}}\\{\frac{1}{8}}&{\frac{1}{8}}&{ - \frac{1}{8}}\end{array}} \right)$
(vi)$\left( {\begin{array}{*{20}{c}}1&2&{ - 1}\\2&0&1\\0&3&{ - 1}\end{array}} \right)$
Solution:
Let A = $\left( {\begin{array}{*{20}{c}}1&{ - 1}&1\\2&1&{ - 3}\\1&1&1\end{array}} \right)$
A11 = $\left| {\begin{array}{*{20}{c}}1&{ - 3}\\1&1\end{array}} \right|$ = 1 + 3 = 4;
A12 = $ - \left| {\begin{array}{*{20}{c}}2&{ - 3}\\1&2\end{array}} \right|$ = –(2 + 3) = –5;
A13 = $\left| {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right|$ = 2 – 1 = 1;
A21 = $\left| {\begin{array}{*{20}{c}}{ - 1}&1\\1&1\end{array}} \right|$ = –(–1 – 1) = 2;
A22 = $\left| {\begin{array}{*{20}{c}}1&1\\1&1\end{array}} \right|$ = 0;
A23 = $\left| {\begin{array}{*{20}{c}}1&{ - 1}\\1&1\end{array}} \right|$ = –(1 + 1) = –2;
A31 = $\left| {\begin{array}{*{20}{c}}{ - 1}&1\\1&{ - 3}\end{array}} \right|$ = 3 – 1 = 2;
A32 = $ - \left| {\begin{array}{*{20}{c}}1&1\\2&3\end{array}} \right|$ = –(–3 – 2) = 5;
A33 = $\left| {\begin{array}{*{20}{c}}1&{ - 1}\\2&1\end{array}} \right|$ = 1 + 2 = 3;
Matrix of cofactors = $\left( {\begin{array}{*{20}{c}}{{{\rm{A}}_{11}}}&{{{\rm{A}}_{12}}}&{{{\rm{A}}_{13}}}\\{{{\rm{A}}_{21}}}&{{{\rm{A}}_{22}}}&{{{\rm{A}}_{23}}}\\{{{\rm{A}}_{31}}}&{{{\rm{A}}_{32}}}&{{{\rm{A}}_{33}}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}4&{ - 5}&1\\2&0&{ - 2}\\2&5&3\end{array}} \right)$Adj. A = $\left( {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&5\\1&{ - 2}&3\end{array}} \right)$
|A| = $\left| {\begin{array}{*{20}{c}}1&{ - 1}&1\\2&1&{ - 3}\\1&1&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}0&{ - 2}&0\\2&1&{ - 3}\\1&1&1\end{array}} \right|$(R1à R1 – R3).
= –(–2) $\left| {\begin{array}{*{20}{c}}2&{ - 3}\\1&1\end{array}} \right|$
= 2(2 + 3) = 10 ≠ 0.
So, A–1 exists.
A–1 = $\frac{{{\rm{Adj}},{\rm{A}}}}{{\left| {\rm{A}} \right|}}$ = $\frac{{\left( {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&5\\1&{ - 2}&3\end{array}} \right)}}{{10}}$ = $\frac{1}{{10}}\left( {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&5\\1&{ - 2}&3\end{array}} \right)$.
2) If A = $\left( {\begin{array}{*{20}{c}}7&{ - 3}\\6&2\end{array}} \right)$, prove that A–1 = $\frac{{{\rm{Adj}}.{\rm{A}}}}{{\left| {\rm{A}} \right|}}$ = $\frac{1}{{32}}\left( {\begin{array}{*{20}{c}}2&3\\{ - 6}&7\end{array}} \right)$.
Soln:
A = $\left( {\begin{array}{*{20}{c}}7&{ - 3}\\6&2\end{array}} \right)$;
|A| = $\left| {\begin{array}{*{20}{c}}7&{ - 3}\\6&2\end{array}} \right|$ = 14 – 6 * (–3) = 32.
A11 = 2, A12 = –6, A21 = 3, A22 = 7.
Matrix of cofactors = $\left( {\begin{array}{*{20}{c}}2&{ - 6}\\3&7\end{array}} \right)$;
Adj A. = $\left( {\begin{array}{*{20}{c}}2&3\\{ - 6}&7\end{array}} \right)$
So, A–1 = $\frac{{{\rm{Adj}}.{\rm{A}}}}{{\left| {\rm{A}} \right|}}$ = $\frac{1}{{32}}\left( {\begin{array}{*{20}{c}}2&3\\{ - 6}&7\end{array}} \right)$
3) Prove that two matrices $\left( {\begin{array}{*{20}{c}}{ - 3}&{ - 2}\\5&3\end{array}} and \right)\left( {\begin{array}{*{20}{c}}3&2\\{ - 5}&{ - 3}\end{array}} \right)$ are the inverses of each other.
Solution:
Or, $\left( {\begin{array}{*{20}{c}}{ - 3}&{ - 2}\\5&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&2\\{ - 5}&{ - 3}\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}{ - 9 + 10}&{ - 6 + 6}\\{15 - 15}&{10 - 9}\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)$Also, $\left( {\begin{array}{*{20}{c}}3&2\\{ - 5}&{ - 3}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 3}&{ - 2}\\5&3\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}{ - 9 + 10}&{ - 6 + 6}\\{15 - 15}&{10 - 9}\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)$
So, $\left( {\begin{array}{*{20}{c}}{ - 3}&{ - 2}\\5&3\end{array}} \right)$ and $\left( {\begin{array}{*{20}{c}}3&2\\{ - 5}&{ - 3}\end{array}} \right)$ are inverse of each other.
4) If A = $\left( {\begin{array}{*{20}{c}}5&3\\4&2\end{array}} \right)$ verify that A(Adj.A) = (Adj.A)A = |A|I.
Solution:
A = $\left( {\begin{array}{*{20}{c}}5&3\\4&2\end{array}} \right)$. Then |A| = $\left| {\begin{array}{*{20}{c}}5&3\\4&2\end{array}} \right|$ = 10 – 12 = – 2
A11 = 2, A13 = –4, A21 = –3,A22 = 5.
Matrix of cofactors = $\left( {\begin{array}{*{20}{c}}2&{ - 4}\\{ - 3}&5\end{array}} \right)$;
Adj. A = $\left( {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 4}&5\end{array}} \right)$;
A(Adj. A) –$\left( {\begin{array}{*{20}{c}}5&3\\4&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 4}&5\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}{10 - 12}&{ - 15 + 15}\\{8 - 8}&{ - 12 + 10}\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}{ - 2}&0\\0&{ - 2}\end{array}} \right)$
(Adj. A)A = $\left( {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 4}&5\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&3\\4&2\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}{10 - 12}&{6 - 6}\\{ - 20 + 20}&{ - 12 + 10}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 2}&0\\0&{ - 2}\end{array}} \right).$
And |A|I = –2$\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}{ - 2}&0\\0&{ - 2}\end{array}} \right)$
So, A(Adj.A) = (Adj.A)A = |A|I.
5) If A=$\left( {\begin{array}{*{20}{c}}2&3\\4&5\end{array}} and B= \right)\left( {\begin{array}{*{20}{c}}6&0\\{ - 5}&2\end{array}} \right)$ verify that (AB)–1 = B–1.A–1.
Solution:
AB = $\left( {\begin{array}{*{20}{c}}2&3\\4&5\end{array}} \right)\left( {\begin{array}{*{20}{c}}6&0\\{ - 5}&2\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}{12 - 15}&{0 + 6}\\{24 - 25}&{0 + 10}\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}{ - 3}&6\\{ - 1}&{10}\end{array}} \right)$
|AB| = $\left| {\begin{array}{*{20}{c}}{ - 3}&6\\{ - 1}&{10}\end{array}} \right|$ = –30 + 6 = –24
The cofactors of –3 = 10. The cofactors of 6 = 1.
The cofactors of –1 = –6. The cofactors of 10 = –3.
Matrix of cofactors = $\left( {\begin{array}{*{20}{c}}{10}&1\\{ - 6}&{ - 3}\end{array}} \right);$
Adj.(AB) = $\left( {\begin{array}{*{20}{c}}{10}&{ - 6}\\1&{ - 3}\end{array}} \right)$
(AB)–1 = $\frac{{{\rm{Adj}}.\left( {{\rm{AB}}} \right)}}{{\left| {{\rm{AB}}} \right|}}$ = $\frac{{\left( {\begin{array}{*{20}{c}}{10}&{ - 6}\\1&{ - 3}\end{array}} \right)}}{{ - 24}}$ = $\frac{1}{{24}}\left( {\begin{array}{*{20}{c}}{10}&{ - 6}\\1&{ - 3}\end{array}} \right)$ …(i)
A = $\left( {\begin{array}{*{20}{c}}2&3\\4&5\end{array}} \right)$;
|A| = $\left| {\begin{array}{*{20}{c}}2&3\\4&5\end{array}} \right|$ = 10 – 12 = –2
A11 = 5, A12 = –4, A21= –3, A22= 2.
Matrix of cofactors = $\left( {\begin{array}{*{20}{c}}5&{ - 3}\\{ - 4}&2\end{array}} \right)$;
Adj. A = $\left( {\begin{array}{*{20}{c}}5&{ - 3}\\{ - 4}&2\end{array}} \right)$
Or, A–1 = $\frac{{{\rm{Adj}}.{\rm{A}}}}{{\left| {\rm{A}} \right|}}$ = $\frac{{\left( {\begin{array}{*{20}{c}}5&{ - 3}\\{ - 4}&2\end{array}} \right)}}{{ - 2}}$
B = $\left( {\begin{array}{*{20}{c}}6&0\\{ - 5}&2\end{array}} \right)$;
|B| = $\left| {\begin{array}{*{20}{c}}6&0\\{ - 5}&2\end{array}} \right|$ = 12 – 0 = 12
B11 = 2, B12 = 5, B21 = 0, B22 = 6
Matrix of cofactors = $\left( {\begin{array}{*{20}{c}}2&5\\0&6\end{array}} \right)$;
Adj. B = $\left( {\begin{array}{*{20}{c}}2&0\\5&6\end{array}} \right)$
Or, B–1 = $\frac{{{\rm{Adj}}.{\rm{B}}}}{{\left| {\rm{B}} \right|}}$ = $\frac{{\left( {\begin{array}{*{20}{c}}2&0\\5&6\end{array}} \right)}}{{12}}$
B–1.A–1 = $\frac{1}{{12}}\left( {\begin{array}{*{20}{c}}2&0\\5&6\end{array}} \right).\frac{1}{{ - 2}}\left( {\begin{array}{*{20}{c}}5&{ - 3}\\{ - 4}&2\end{array}} \right)$ = $ - \frac{1}{{24}}\left( {\begin{array}{*{20}{c}}{10 - 0}&{ - 6 + 0}\\{25 - 24}&{ - 15 + 12}\end{array}} \right)$
= $ - \frac{1}{{24}}\left( {\begin{array}{*{20}{c}}{10}&{ - 6}\\1&{ - 3}\end{array}} \right)$ = $\frac{1}{{24}}\left( {\begin{array}{*{20}{c}}{10}&{ - 6}\\{ - 1}&3\end{array}} \right)$ …(ii)
From (i) and (ii), (AB)–1 = B–1.A–1.
6) Given matrix $\left( {\begin{array}{*{20}{c}}3&{ - 1}\\5&{ - 2}\end{array}} , \right)$, find a matrix $\left( {\begin{array}{*{20}{c}}{\rm{p}}&{\rm{q}}\\{\rm{r}}&{\rm{s}}\end{array}} \right)$ such that they are inverse if each other.
Solution:
Or, $\left( {\begin{array}{*{20}{c}}3&{ - 1}\\5&{ - 2}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\rm{p}}&{\rm{q}}\\{\rm{r}}&{\rm{s}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)$
Or, $\left( {\begin{array}{*{20}{c}}{3{\rm{p}} - {\rm{r}}}&{3{\rm{q}} - {\rm{s}}}\\{5{\rm{p}} - 2{\rm{r}}}&{5{\rm{q}} - 2{\rm{s}}}\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)$
3p – r = 1 …(i)
3q – s = 0 …(ii)
5p – 2r = 0 …(iii)
5q – 2s = 1 …(iv)
So, $\left( {\begin{array}{*{20}{c}}{\rm{p}}&{\rm{q}}\\{\rm{r}}&{\rm{s}}\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}2&{ - 1}\\5&{ - 3}\end{array}} \right)$
Also, $\left( {\begin{array}{*{20}{c}}{\rm{p}}&{\rm{q}}\\{\rm{r}}&{\rm{s}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&{ - 1}\\5&{ - 2}\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}2&{ - 1}\\5&{ - 3}\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&{ - 1}\\5&{ - 2}\end{array}} \right)$
= $\left( {\begin{array}{*{20}{c}}{6 - 5}&{ - 2 + 2}\\{15 - 15}&{ - 5 + 6}\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)$.
7) If A = $\left( {\begin{array}{*{20}{c}}1&2&{ - 2}\\{ - 1}&3&0\\0&{ - 2}&1\end{array}} \right)$, find A–1 and verify that A A–1 =I.
Solution:
A = $\left( {\begin{array}{*{20}{c}}1&2&{ - 2}\\{ - 1}&3&0\\0&{ - 2}&1\end{array}} \right)$
Or, |A| = $\left| {\begin{array}{*{20}{c}}1&2&{ - 2}\\{ - 1}&3&0\\0&{ - 2}&1\end{array}} \right| = 1\left| {\begin{array}{*{20}{c}}3&0\\{ - 2}&1\end{array}} \right| - 2\left| {\begin{array}{*{20}{c}}{ - 1}&0\\0&1\end{array}} \right| - 2\left| {\begin{array}{*{20}{c}}{ - 1}&3\\0&{ - 2}\end{array}} \right|$
= 1(3 – 0) – 2(–1 – 0) – 2(2 – 0) = 3 + 2 – 4 = 1.
A11 = $\left| {\begin{array}{*{20}{c}}3&0\\{ - 2}&1\end{array}} \right|$ = 3;
A12 = $ - \left| {\begin{array}{*{20}{c}}{ - 1}&0\\0&1\end{array}} \right|$ = 1;
A13 = $\left| {\begin{array}{*{20}{c}}{ - 1}&3\\0&{ - 2}\end{array}} \right|$ = 2;
A21 = $ - \left| {\begin{array}{*{20}{c}}2&{ - 2}\\{ - 2}&1\end{array}} \right|$ = 2;
A22 = $\left| {\begin{array}{*{20}{c}}1&{ - 2}\\0&1\end{array}} \right|$ = 1;
A23 = $ - \left| {\begin{array}{*{20}{c}}1&2\\{ - 1}&{ - 2}\end{array}} \right|$ = 2;
A31 = $\left| {\begin{array}{*{20}{c}}2&{ - 2}\\3&0\end{array}} \right|$ = 6;
A32 = $ - \left| {\begin{array}{*{20}{c}}1&{ - 2}\\{ - 1}&0\end{array}} \right|$ = 2;
A33 = $\left| {\begin{array}{*{20}{c}}1&2\\{ - 1}&3\end{array}} \right|$ = 5;
Matrix of cofactors = $\left( {\begin{array}{*{20}{c}}3&1&2\\2&1&2\\6&2&5\end{array}} \right)$;
Adj. A = $\left( {\begin{array}{*{20}{c}}3&2&6\\1&1&2\\2&2&5\end{array}} \right)$
Or, A–1 = $\frac{{{\rm{adj}}.{\rm{A}}}}{{\left| {\rm{A}} \right|}}$ = $\left( {\begin{array}{*{20}{c}}3&2&6\\1&1&2\\2&2&5\end{array}} \right)$ (|A| = 1)
A.A–1 = $\left( {\begin{array}{*{20}{c}}1&2&{ - 2}\\{ - 1}&3&0\\0&{ - 2}&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&2&6\\1&1&2\\2&2&5\end{array}} \right)$
= $\left( {\begin{array}{*{20}{c}}{3 + 2 - 4}&{2 + 2 - 4}&{6 + 4 - 10}\\{ - 3 + 3 + 0}&{2 - 3 + 0}&{ - 6 + 6 + 0}\\{0 - 2 + 2}&{0 - 2 + 2}&{0 - 4 + 5}\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)$ = I.