- Home
- Standard 12
- Mathematics
જો $\mathrm{A}=\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right],$ હોય, તો સાબિત કરો કે $A^{3}-6 A^{2}+7 A+2 I=0$
Solution
${A^2} = AA = $ $\left[ {\begin{array}{*{20}{l}}
1&0&2 \\
0&2&1 \\
2&0&3
\end{array}} \right]\left[ {\begin{array}{*{20}{l}}
1&0&2 \\
0&2&1 \\
2&0&3
\end{array}} \right]$
$=\left[\begin{array}{lll}1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9\end{array}\right]$ $=\left[\begin{array}{lll}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]$
Now, $A^{3}=A^{2} A$
$=\left[\begin{array}{lll}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]$
$=\left[\begin{array}{lll}5+0+16 & 0+0+0 & 10+0+24 \\ 2+0+10 & 0+8+0 & 4+4+15 \\ 8+0+26 & 0+0+0 & 16+0+39\end{array}\right]$
$=\left[\begin{array}{lll}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{array}\right]$
Substituting the matrices in the given equation $A^{3}-6 A^{2}+7 A+2 I$
$ = \left[ {\begin{array}{*{20}{l}}
{21}&0&{34} \\
{12}&8&{23} \\
{34}&0&{55}
\end{array}} \right]$ $ – 6\left[ {\begin{array}{*{20}{l}}
5&0&8 \\
2&4&5 \\
8&0&{13}
\end{array}} \right]$ $ + 7\left[ {\begin{array}{*{20}{l}}
1&0&2 \\
0&2&1 \\
2&0&3
\end{array}} \right]$ $ + 2\left[ {\begin{array}{*{20}{l}}
1&0&0 \\
0&1&0 \\
0&0&1
\end{array}} \right]$
$=\left[\begin{array}{lll}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{array}\right]-\left[\begin{array}{ccc}30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78\end{array}\right]$ $+\left[\begin{array}{ccc}7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21\end{array}\right]+\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$
$=\left[\begin{array}{ccc}21+7+2 & 0+0+0 & 34+14+0 \\ 12+0+0 & 8+14+2 & 23+7+0 \\ 34+14+0 & 0+0+0 & 55+21+2\end{array}\right]$ $-\left[\begin{array}{ccc}30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78\end{array}\right]$
$=\left[\begin{array}{ccc}30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78\end{array}\right]-\left[\begin{array}{ccc}30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78\end{array}\right]$
$=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=0$
$\therefore A^{3}-6 A^{2}+7 A+2 I=0$
