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જો $A=\left[\begin{array}{rrr}1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1\end{array}\right],$ હોય, તો સાબિત કરો કે $A^{3}-23 A-40I=0$
Solution
We have ${A^2} = A \cdot A = $ $\left[ {\begin{array}{*{20}{r}}
1&2&3 \\
3&{ – 2}&1 \\
4&2&1
\end{array}} \right]$ $\left[ {\begin{array}{*{20}{r}}
1&2&3 \\
3&{ – 2}&1 \\
4&2&1
\end{array}} \right] = $ $\left[ {\begin{array}{*{20}{c}}
{19}&4&8 \\
1&{12}&8 \\
{14}&6&{15}
\end{array}} \right]$
So ${A^3} = A{A^2} = $ $\left[ {\begin{array}{*{20}{r}}
1&2&3 \\
3&{ – 2}&1 \\
4&2&1
\end{array}} \right]$$\left[ {\begin{array}{*{20}{r}}
{19}&4&8 \\
1&{12}&8 \\
{14}&6&{15}
\end{array}} \right] = $ $\left[ {\begin{array}{*{20}{r}}
{63}&{46}&{69} \\
{69}&{ – 6}&{23} \\
{92}&{46}&{63}
\end{array}} \right]$
Now
$A^{3}-23 A-40 I=$ $\left[ {\begin{array}{*{20}{r}}
{63}&{46}&{69} \\
{69}&{ – 6}&{23} \\
{92}&{46}&{63}
\end{array}} \right]$ $-23\left[\begin{array}{rrr}1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1\end{array}\right]-40\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$ = \left[ {\begin{array}{*{20}{c}}
{63}&{46}&{69} \\
{69}&{ – 6}&{23} \\
{92}&{46}&{63}
\end{array}} \right] + $ $\left[ {\begin{array}{*{20}{c}}
{ – 23}&{ – 46}&{ – 69} \\
{ – 69}&{46}&{ – 23} \\
{ – 92}&{ – 46}&{ – 23}
\end{array}} \right] + $ $\left[ {\begin{array}{*{20}{c}}
{ – 40}&0&0 \\
0&{ – 40}&0 \\
0&0&{ – 40}
\end{array}} \right]$
$=\left[\begin{array}{lll}63-23-40 & 46-46+0 & 69-69+0 \\ 69-69+0 & -6+46-40 & 23-23+0 \\ 92-92+0 & 46-46+0 & 63-23-40\end{array}\right]$
$=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=0$
