AB = \(\left(\begin{array}{rr} 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right) \left[\begin{array}{rr} 0 & 1 \\ 2 & -1 \\ 0 & 2 \end{array}\right] \)
\(=\left[\begin{array}{rr} 1+4+0 \ \ 0-2 + 6 \\ 3+4+0 \ 0-2+2 \end{array}\right] =\left[\begin{array}{rr} 5 & 4 \\ 7 & 0 \end{array}\right] \)
\(\begin{gathered} A B+C=\left[\begin{array}{ll} 5 & 4 \\ 7 & 0 \end{array}\right]+\left(\begin{array}{rr} 0 & 1 \\ -2 & 5 \end{array}\right)=\left[\begin{array}{ll} 5 & 5 \\ 5 & 5 \end{array}\right] \\ \end{gathered}\)
இது சரி 
(ii) \(\begin{gathered} B C=\left[\begin{array}{rr} 1 & 0\\ 2 & -1 \\ 0 & 2 \end{array}\right] \end{gathered}\left(\begin{array}{rr} 0 & 1 \\ -2 & 5 \end{array}\right)\)
\(=\left[\begin{array}{rr} 0+0 \ 1+0 \\ 0+2 \ 2-5 \\ 0-4 \ 0+7 \end{array}\right] =\left[\begin{array}{rr} 0 & 1 \\ 2 & -3 \\ -4 & 7 \end{array}\right] \)
இது சரி 
(iii) BA \(=\left[\begin{array}{rr} 1 & 0\\ 2 & -1 \\ 0 & 2 \end{array}\right] \left(\begin{array}{rr} 1&2&3 \\ 3&2&1 \end{array}\right)\)
\(=\left[\begin{array}{rr} 1+0 & 2+0 & 3+0 \\2-3 & 4-2 &6-1 \\ 0+6 & 0+4 & 0+2 \end{array}\right] \)
\(=\left[\begin{array}{rr} 1 & 2& 3\\ -1 & 2 & -5 \\ 6&4&2 \end{array}\right]\)
BA + C காண இயலாது 
(AB)C = \(=\left[\begin{array}{ll} 5 & 4 \\ 7 & 0 \end{array}\right]+\left(\begin{array}{rr} 0 & 1 \\ -2 & 5 \end{array}\right)=\left[\begin{array}{ll} 0-8 & 5+20 \\ 0+0 & 7+0 \end{array}\right] =\left[\begin{array}{ll} -8 & 25 \\ 0 & 7 \end{array}\right] \)
இது சரியல்ல எனவே (i) மற்றும் (ii) மட்டுமே சரியானது 

கொடுக்கப்பட்டது \(\begin{aligned} 2 x &+\left[\begin{array}{ll} 1 & 3 \\ 5 & 7 \end{array}\right]=\left[\begin{array}{ll} 5 & 7 \\ 9 & 5 \end{array}\right] \\ \end{aligned}\)
\(\begin{aligned} 2 x &=\left[\begin{array}{ll} 5 & 7 \\ 9 & 5 \end{array}\right]-\left[\begin{array}{ll} 1 & 3 \\ 5 & 7 \end{array}\right] \\ \end{aligned}\)
\(\begin{aligned} &=\left[\begin{array}{cc} 4 & 4 \\ 4 & -2 \end{array}\right] \\ \end{aligned}\)
\(\begin{aligned} x &=\frac{1}{2}\left[\begin{array}{cc} 4 & 4 \\ 4 & -2 \end{array}\right] \\ \end{aligned}\)
\(\begin{aligned} &=\left[\begin{array}{cc} 2 & 2 \\ 2 & -1 \end{array}\right] \end{aligned}\)

\(\begin{aligned} &A=\left(\begin{array}{rrrr} 1 & 3 & 5 & 7 \\ 2 & 4 & 6 & 8 \\ 9 & 11 & 13 & 15 \end{array}\right), \end{aligned}\) A ன் வரிசை = 3 x 4 
\(\begin{aligned} &\mathrm{~A}^{\mathrm{T}}=\left(\begin{array}{rrr} 1 & 2 & 9 \\ 3 & 4 & 11 \\ 5 & 6 & 13 \\ 7 & 8 & 15 \end{array}\right), \end{aligned}\) A^T ன் வரிசை = 4 x 3

\(\sqrt{4 x^4-24 x^3+76 x^2+a x+b}\)

கொடுக்கப்பட்ட பல்லுறுப்புக் கோவை ஒரு முழு வர்க்கமாதலால்,
a + 120 = 0, b - 100 = 0
∴ a = -120, b = 100

\(x^4+64+16 x^2 =x^4+16 x^2+64 \)
\(=\left(x^2\right)^2+2(8)\left(x^2\right)+(8)^2 \)
\(=\left(x^2+8\right)^2\)

\(\frac{x}{x^2-25}-\frac{8}{x^2+6 x+5} =\frac{x}{(x+5)(x-5)}-\frac{8}{(x+5)(x+1)}  \)
\(=\frac{x(x+1)-8(x-5)}{(x+5)(x-5)(x+1)}  \)
\(=\frac{x^2+x-8 x+40}{(x+5)(x-5)(x+1)}  \)
\(=\frac{x^2-7 x+40}{\left(x^2-25\right)(x+1)}\)

\(\frac{3 y-3}{y} \div \frac{7 y-7}{3 y^2}=\frac{3(y-1)}{y} \times \frac{3 y^2}{7(y-1)}=\frac{9 y}{7}\)