Determinants and Matrices — Maths STD 11 Science — Question

$\begin{aligned} A-I & =\left[\begin{array}{ccc}1 & 2 & 0 \\ 5 & 4 & 2 \\ 0 & 7 & -3\end{array}\right]-\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \\ & =\left[\begin{array}{ccc}1-1 & 2-0 & 0-0 \\ 5-0 & 4-1 & 2-0 \\ 0-0 & 7-0 & -3-1\end{array}\right]=\left[\begin{array}{ccc}0 & 2 & 0 \\ 5 & 3 & 2 \\ 0 & 7 & -4\end{array}\right]\end{aligned}$

$\begin{aligned} \therefore \quad & (A+I)(A-I) \\ \quad & {\left[\begin{array}{ccc}2 & 2 & 0 \\ 5 & 5 & 2 \\ 0 & 7 & -2\end{array}\right]\left[\begin{array}{lll}0 & 2 & 0 \\ 5 & 3 & 2 \\ 0 & 7 & -4\end{array}\right] }\end{aligned}$

$=\left[\begin{array}{ccc}0+10+0 & 4+6+0 & 0+4-0 \\ 0+25+0 & 10+15+14 & 0+10-8 \\ 0+35+0 & 0+21-14 & 0+14+8\end{array}\right]$

$=\left[\begin{array}{ccc}10 & 10 & 4 \\ 25 & 39 & 2 \\ 35 & 7 & 22\end{array}\right]$

[Note: Answer given in the textbook is $\left[\begin{array}{ccc}9 & 6 & 4 \\ 15 & 32 & -2 \\ 35 & -7 & 29\end{array}\right]$

However, as per our calculation it is $\left[\begin{array}{ccc}10 & 10 & 4 \\ 25 & 39 & 2 \\ 35 & 7 & 22\end{array}\right]$.]