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Exercise 7.1 · Q8

Q.If A=(100010ab−1)A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1\end{pmatrix}, show that A2A^2 is a unit matrix.

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We multiply A⋅AA\cdot A row by column and show every entry of the result matches the identity matrix I3I_3.

Step 1. Set up A2=A⋅AA^2=A\cdot A with A=(100010ab−1)A=\begin{pmatrix}1&0&0\\0&1&0\\a&b&-1\end{pmatrix}.

Step 2. Compute the first two rows. Since rows 11 and 22 of AA are exactly the first two rows of I3I_3, multiplying them into any matrix just reproduces that matrix's first two rows. So rows 1,21,2 of A2A^2 equal rows 1,21,2 of AA itself: (1,0,0)(1,0,0) and (0,1,0)(0,1,0).

Step 3. Compute row 3 of A2A^2. Row 3 of AA is (a, b, −1)(a,\,b,\,-1).

(3,1): a(1)+b(0)+(−1)(a)=a−a=0(3,1):\ a(1)+b(0)+(-1)(a)=a-a=0

(3,2): a(0)+b(1)+(−1)(b)=b−b=0(3,2):\ a(0)+b(1)+(-1)(b)=b-b=0 …

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