Maths Solve the Following Question.(2 Marks)

$\begin{aligned} A B^T & =\left[\begin{array}{ccc}2 & 1 & -3 \\ 0 & 2 & 6\end{array}\right]\left[\begin{array}{cc}1 & 3 \\ 0 & -1 \\ -2 & 4\end{array}\right] \\ & =\left[\begin{array}{cc}2+0+6 & 6-1-12 \\ 0+0-12 & 0-2+24\end{array}\right] \\ & =\left[\begin{array}{cc}8 & -7 \\ -12 & 22\end{array}\right]\end{aligned}$

and $A^{\mathrm{T}} B=\left[\begin{array}{cc}2 & 0 \\ 1 & 2 \\ -3 & 6\end{array}\right]\left[\begin{array}{ccc}1 & 0 & -2 \\ 3 & -1 & 4\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{ccc}2+0 & 0+0 & -4+0 \\ 1+6 & 0-2 & -2+8 \\ -3+18 & 0-6 & 6+24\end{array}\right] \\ & =\left[\begin{array}{ccc}2 & 0 & -4 \\ 7 & -2 & 6 \\ 15 & -6 & 30\end{array}\right]\end{aligned}$

$(A+B) \cdot(A-B) \neq A^2-B^2$

i.e, to prove that $A(A-B)+B(A-B) \neq A^2-B^2$

i.e, to prove that $A^2-A B+B A-B^2 \neq A^2-B^2$

i.e, to prove that $A B \neq B A$.

$\begin{aligned} A B & =\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right] \\ & =\left[\begin{array}{cc}0+1 & 0+0 \\ 0+0 & -1+0\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]\end{aligned}$

$\begin{aligned} \mathbf{B A} & =\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \\ & =\left[\begin{array}{cc}0-1 & 0-0 \\ 0+0 & 1+0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & 1\end{array}\right]\end{aligned}$

∴ AB ≠ BA

$\begin{aligned} & =\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] \\ & =0\end{aligned}$

$\begin{aligned} & =\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]-5\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]-14\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] \\ & =\left[\begin{array}{cc}9+20 & -15-10 \\ -12-8 & 20+4\end{array}\right]-\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]-\left[\begin{array}{cc}14 & 0 \\ 0 & 14\end{array}\right] \\ & =\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]-\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]-\left[\begin{array}{cc}14 & 0 \\ 0 & 14\end{array}\right] \\ & =\left[\begin{array}{cc}29-15-14 & -25+25-0 \\ -20+20-0 & 24-10-14\end{array}\right]\end{aligned}$

$\begin{aligned} & =\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] \\ & =0\end{aligned}$

$\begin{aligned} & \omega^3=1 \\ & \omega^3-1=0 \\ & (\omega-1)\left(\omega^2+\omega+1\right)=0 \\ & \omega=1 \text { or } \omega^2+\omega+1=0\end{aligned}$

But, $\omega$ is a complex number.

$1+w+w^2=0 \ldots \ldots$ (i)

AB + BA + A – 2B

$=\left[\begin{array}{cc}1 & \omega \\ \omega^2 & 1\end{array}\right]\left[\begin{array}{cc}\omega^2 & 1 \\ 1 & \omega\end{array}\right]+\left[\begin{array}{cc}\omega^2 & 1 \\ 1 & \omega\end{array}\right]\left[\begin{array}{cc}1 & \omega \\ \omega^2 & 1\end{array}\right]$

$+\left[\begin{array}{cc}1 & \omega \\ \omega^2 & 1\end{array}\right]-2\left[\begin{array}{cc}\omega^2 & 1 \\ 1 & \omega\end{array}\right]$

$=\left[\begin{array}{cc}\omega^2+\omega & 1+\omega^2 \\ \omega^4+1 & \omega^2+\omega\end{array}\right]+\left[\begin{array}{cc}\omega^2+\omega^2 & \omega^3+1 \\ 1+\omega^3 & \omega+\omega\end{array}\right]$

$+\left[\begin{array}{cc}1 & \omega \\ \omega^2 & 1\end{array}\right]-\left[\begin{array}{cc}2 \omega^2 & 2 \\ 2 & 2 \omega\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{ll}\omega^2+\omega+2 \omega^2+1-2 \omega^2 & 1+\omega^2+\omega^3+1+\omega-2 \\ \omega^4+1+1+\omega^3+\omega^2-2 & \omega^2+\omega+2 \omega+1-2 \omega\end{array}\right] \\ & =\left[\begin{array}{ll}\omega^2+\omega+1 & \omega^2+\omega+1 \\ \omega^2+\omega+1 & \omega^2+\omega+1\end{array}\right]\end{aligned}$

$\because\left[\because \omega^3=1 \therefore \omega^4=\omega\right]$

$ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] $ $\ldots[$ From (i)]

which is a null matrix.

$=\left[\begin{array}{ccc}1+8 & 3-2 & 2-6 \\ 3+8 & 9-2 & 6-6 \\ -1+0 & -3+0 & -2+0\end{array}\right]$

$=\left[\begin{array}{ccc}9 & 1 & -4 \\ 11 & 7 & 0 \\ -1 & -3 & -2\end{array}\right]$

$|A B|=\left|\begin{array}{ccc}9 & 1 & -4 \\ 11 & 7 & 0 \\ -1 & -3 & -2\end{array}\right|$

$\begin{aligned} & =9(-14+0)-1(-22+0)-4(-33+7) \\ & =-126+22+104\end{aligned}$

$=0$

$\mathrm{AB}$ is a singular matrix.

$=\left[\begin{array}{cc}6 & -9 \\ 9 & -6 \\ -3 & 12\end{array}\right]-\left[\begin{array}{cc}-15 & 10 \\ 20 & -5 \\ 5 & -15\end{array}\right]$

$\therefore \quad 3 A-5 B^{\top}=\left[\begin{array}{cc}21 & -19 \\ -11 & -1 \\ -8 & 27\end{array}\right]$

$\therefore \quad\left(3 A-5 B^{\mathrm{T}}\right)^{\mathrm{T}}=\left[\begin{array}{ccc}21 & -11 & -8 \\ -19 & -1 & 27\end{array}\right]$

$\ldots$..i)

$3 A^T-5 B=3\left[\begin{array}{ccc}2 & 3 & -1 \\ -3 & -2 & 4\end{array}\right]-5\left[\begin{array}{ccc}-3 & 4 & 1 \\ 2 & -1 & -3\end{array}\right]$

$=\left[\begin{array}{ccc}6 & 9 & -3 \\ -9 & -6 & 12\end{array}\right]-\left[\begin{array}{ccc}-15 & 20 & 5 \\ 10 & -5 & -15\end{array}\right]$

$=\left[\begin{array}{ccc}21 & -11 & -8 \\ -19 & -1 & 27\end{array}\right]$

...(ii)

From (i) and (ii), we get

$\left(3 A-5 B^T\right)^T=3 A^T-5 B$

$\therefore \quad \mathbf{A}^{\mathrm{T}}=\left[\begin{array}{ccc}2 & 3 & -1 \\ -3 & -2 & 4\end{array}\right]$ and $B^{\mathrm{T}}=\left[\begin{array}{cc}-3 & 2 \\ 4 & -1 \\ 1 & -3\end{array}\right]$

$\begin{aligned} \therefore \quad A+2 B^T & =\left[\begin{array}{cc}2 & -3 \\ 3 & -2 \\ -1 & 4\end{array}\right]+2\left[\begin{array}{cc}-3 & 2 \\ 4 & -1 \\ 1 & -3\end{array}\right] \\ & =\left[\begin{array}{cc}2 & -3 \\ 3 & -2 \\ -1 & 4\end{array}\right]+\left[\begin{array}{cc}-6 & 4 \\ 8 & -2 \\ 2 & -6\end{array}\right]\end{aligned}$

$\therefore \quad A+2 B^T=\left[\begin{array}{cc}-4 & 1 \\ 11 & -4 \\ 1 & -2\end{array}\right]$

$\therefore \quad\left(A+2 B^{\mathrm{T}}\right)^{\mathrm{T}}=\left[\begin{array}{ccc}-4 & 11 & 1 \\ 1 & -4 & -2\end{array}\right]$

$\ldots$...(i)

$\begin{aligned} A^{\mathrm{T}}+2 B & =\left[\begin{array}{ccc}2 & 3 & -1 \\ -3 & -2 & 4\end{array}\right]+2\left[\begin{array}{ccc}-3 & 4 & 1 \\ 2 & -1 & -3\end{array}\right] \\ & =\left[\begin{array}{ccc}2 & 3 & -1 \\ -3 & -2 & 4\end{array}\right]+\left[\begin{array}{ccc}-6 & 8 & 2 \\ 4 & -2 & -6\end{array}\right] \\ & =\left[\begin{array}{ccc}-4 & 11 & 1 \\ 1 & -4 & -2\end{array}\right] \quad \ldots \text { (ii) }\end{aligned}$

From (i) and (ii), we get

$\left(A+2 B^{\mathrm{T}}\right)^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}}+2 \mathrm{~B}$

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

$\ldots$...(i)

and $A+5 B=\left[\begin{array}{ccc}0 & 0 & 1 \\ -1 & 0 & 0\end{array}\right]$

$\ldots(i i)$

By (i) $\times 5+$ (ii), we get

$\begin{aligned} 16 \mathrm{~A} & =5\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 0 & 5\end{array}\right]+\left[\begin{array}{ccc}0 & 0 & 1 \\ -1 & 0 & 0\end{array}\right] \\ & =\left[\begin{array}{ccc}-5 & 10 & 5 \\ 5 & 0 & 25\end{array}\right]+\left[\begin{array}{ccc}0 & 0 & 1 \\ -1 & 0 & 0\end{array}\right]\end{aligned}$

$\begin{array}{ll}\therefore & 16 A=\left[\begin{array}{ccc}-5 & 10 & 6 \\ 4 & 0 & 25\end{array}\right] \\ \therefore & A=\frac{1}{16}\left[\begin{array}{ccc}-5 & 10 & 6 \\ 4 & 0 & 25\end{array}\right]\end{array}$

By (i) (ii) $\times 3$, we get

$\begin{aligned} & -16 B=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 0 & 5\end{array}\right]-3\left[\begin{array}{ccc}0 & 0 & 1 \\ -1 & 0 & 0\end{array}\right] \\ & =\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 0 & 5\end{array}\right]-\left[\begin{array}{ccc}0 & 0 & 3 \\ -3 & 0 & 0\end{array}\right] \\ & \therefore \quad-16 B=\left[\begin{array}{ccc}-1 & 2 & -2 \\ 4 & 0 & 5\end{array}\right] \\ & \therefore \quad B=\frac{-1}{16}\left[\begin{array}{ccc}-1 & 2 & -2 \\ 4 & 0 & 5\end{array}\right] \\ & \therefore \quad B=\frac{1}{16}\left[\begin{array}{ccc}1 & -2 & 2 \\ -4 & 0 & -5\end{array}\right] \\ & \end{aligned}$

$\begin{aligned} & 2 A-B=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] \ldots \ldots(i) \\ & \text { and } A+3 B=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] \\ & \text { By (i) } \times 3+\text { (ii), we get } \\ & 7 A=3\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]+\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] \\ & \therefore \quad 7 \mathrm{~A}=\left[\begin{array}{cc}3 & -3 \\ 0 & 3\end{array}\right]+\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] \\ & \therefore \quad 7 \mathrm{~A}=\left[\begin{array}{cc}4 & -4 \\ 0 & 4\end{array}\right] \\ & \therefore \quad \mathrm{A}=\frac{1}{7}\left[\begin{array}{cc}4 & -4 \\ 0 & 4\end{array}\right] \\ & \end{aligned}$

By (i) - (ii) $\times 2$, we get

$\begin{aligned}-7 B & =\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]-2\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] \\ & =\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]-\left[\begin{array}{cc}2 & -2 \\ 0 & 2\end{array}\right] \\ \therefore \quad-7 B & =\left[\begin{array}{cc}-1 & 1 \\ 0 & -1\end{array}\right] \\ \therefore \quad B & =\frac{1}{7}\left[\begin{array}{cc}-1 & 1 \\ 0 & -1\end{array}\right]\end{aligned}$

$\therefore \quad B=\frac{1}{7}\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]$

$\begin{aligned} & Q\left(x_2, y_2\right)=Q(1,-1) \\ & R\left(x_2, y_2\right)=R(11,4)\end{aligned}$

If A(ΔPQR) = 0, then the points P, Q, R are collinear.

$\mathrm{A}(\Delta \mathrm{PQR})=\frac{1}{2}\left|\begin{array}{ccc}5 & 1 & 1 \\ 1 & -1 & 1 \\ 11 & 4 & 1\end{array}\right|$

$=\frac{1}{2}[5(-1-4)-1(1-11)$

$+1(4+11)]$

$\begin{aligned} & =\frac{1}{2}[5(-5)-1(-10)+1(15)] \\ & =\frac{1}{2}(-25+10+15)=0\end{aligned}$

$\begin{aligned} & \mathrm{M}\left(\mathrm{x}_2, \mathrm{y}_2\right)=\mathrm{M}(5,7) \\ & \mathrm{N}\left(\mathrm{x}_3 \mathrm{y}_3\right)=\mathrm{N}(8,9)\end{aligned}$

If A(ΔLMN) = 0, then the points L, M, N are collinear.

$A(\Delta L \mathrm{LN})=\frac{1}{2}\left|\begin{array}{lll}2 & 5 & 1 \\ 5 & 7 & 1 \\ 8 & 9 & 1\end{array}\right|$

$=\frac{1}{2}[2(7-9)-5(5-8)+1(45-56)]$

$=\frac{1}{2}[2(-2)-5(-3)+1(-11)]$

$=\frac{1}{2}(-4+15-11)=0$

∴ The points L, M, N are collinear.

$\mathrm{A}(\Delta \mathrm{LMN})=\frac{33}{2}$ sq. units

Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$

$\pm \frac{33}{2}=\frac{1}{2}\left|\begin{array}{ccc}3 & -5 & 1 \\ -2 & k & 1 \\ 1 & 4 & 1\end{array}\right|$

$\Rightarrow \pm \frac{33}{2}=\frac{1}{2}[3(k-4)-(-5)(-2-1)+1(-8-k)]$

$\begin{aligned} & \Rightarrow \pm 33=3 k-12-15-8-k \\ & \Rightarrow \pm 33=2 k-35 \\ & \Rightarrow 2 k-35=33 \text { or } 2 k-35=-33 \\ & \Rightarrow 2 k=68 \text { or } 2 k=2 \\ & \Rightarrow k=34 \text { or } k=1\end{aligned}$

$\begin{aligned} & Q\left(x_2, y_2\right)=Q(4,0) \\ & R\left(x_3, y_3\right)=R(0,2)\end{aligned}$

A(ΔPQR) = 4 sq.units

Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$

$\therefore \quad \pm 4=\frac{1}{2}\left|\begin{array}{lll}k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1\end{array}\right|$

$\therefore \quad \pm 4=\frac{1}{2}[k(0-2)-0+1(8-0)]$

$\therefore \quad \pm 8=-2 k+8$

$\therefore \quad 8=-2 k+8 \quad$ or $\quad-8=-2 k+8$

$\therefore \quad-2 \mathrm{k}=0 \quad$ or $\quad 2 \mathrm{k}=16$

$\therefore \quad k=0 \quad$ or $\quad k=8$

$\begin{aligned} & M\left(x_2, y_2\right)=M(-2,2) \\ & N\left(x_3, y_3\right)=N(5,4)\end{aligned}$

Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$

$\mathrm{A}(\Delta \mathrm{LMN})=\frac{1}{2}\left|\begin{array}{rrr}1 & 1 & 1 \\ -2 & 2 & 1 \\ 5 & 4 & 1\end{array}\right|$

$=\frac{1}{2}[1(2-4)-1(-2-5)+1(-8-10)]$

$=\frac{1}{2}[1(-2)-1(-7)+1(-18)]$

$=\frac{1}{2}(-2+7-18)$

$=-\frac{13}{3}$

Since area cannot be negative,

$\mathrm{A}(\Delta \mathrm{LMN})=\frac{13}{2}$ sq.units

$\begin{aligned} & Q\left(x_2, y_2\right)=Q(-1,3) \\ & R\left(x_3, y_3\right)=R(2,-1)\end{aligned}$

Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$

$\mathrm{A}(\Delta \mathrm{P} Q \mathrm{R})=\frac{1}{2}\left|\begin{array}{ccc}3 & 6 & 1 \\ -1 & 3 & 1 \\ 2 & -1 & 1\end{array}\right|$

$=\frac{1}{2}[3(3+1)-6(-1-2)+1(1-6)]$

$=\frac{1}{2}[3(4)-6(-3)+1(-5)]$

$=\frac{1}{2}(12+18-5)$

$A(\triangle P Q R)=\frac{25}{2}$ sq.units

$\begin{aligned} & B\left(x_2, y_2\right)=B(2,4) \\ & C\left(x_3, y_3\right)=C(0,0)\end{aligned}$

Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$

$\mathrm{A}(\Delta \mathrm{ABC})=\frac{1}{2}\left|\begin{array}{ccc}-1 & 2 & 1 \\ 2 & 4 & 1 \\ 0 & 0 & 1\end{array}\right|$

$=\frac{1}{2}[-1(4-0)-2(2-0)+1(0-0)]$

$=\frac{1}{2}(-4-4)=\frac{1}{2}(-8)=-4$

Since area cannot be negative, A(ΔABC) = 4 sq.units

$\left|\begin{array}{ccc}k-2 & k-1 & -17 \\ k-1 & k-2 & -18 \\ 1 & 1 & -5\end{array}\right|=0$

Applying $R_1 \rightarrow R_1-R_2$, we get

$\left|\begin{array}{ccc}-1 & 1 & 1 \\ k-1 & k-2 & -18 \\ 1 & 1 & -5\end{array}\right|=0$

⇒ -1(-5k + 10 + 18) – 1(-5k + 5 + 18) + 1(k – 1 – k + 2) = 0 ⇒ -1(-5k + 28) – 1(-5k + 23) + 1(1) = 0 ⇒ 5k – 28 + 5k – 23 + 1 = 0 ⇒ 10k – 50 = 0 ⇒ k = 5

$\left|\begin{array}{rrr}3 & 1 & -2 \\ k & 2 & -3 \\ 2 & -1 & -3\end{array}\right|=0$

⇒ 3(-6 – 3) – 1(-3k + 6) – 2(-k – 4) = 0 ⇒ 3(-9) – 1(-3k + 6) – 2(-k – 4) = 0 ⇒ -27 + 3k – 6 + 2k + 8 = 0 ⇒ 5k – 25 = 0 ⇒ k = 5

(k + 1)x + (k – 1)y + (k – 1) = 0 (k – 1)x + (k + 1)y + (k – 1) = 0 (k – 1)x + (k – 1)y + (k + 1) = 0 Since these equations are consistent,

$\left|\begin{array}{lll}k+1 & k-1 & k-1 \\ k-1 & k+1 & k-1 \\ k-1 & k-1 & k+1\end{array}\right|=0$

Applying $C_1 \rightarrow C_1-C_2$, we get

$\left|\begin{array}{ccc}2 & k-1 & k-1 \\ -2 & k+1 & k-1 \\ 0 & k-1 & k+1\end{array}\right|=0$

Applying $C_2 \rightarrow C_2-C_3$, we get

$\left|\begin{array}{ccc}2 & 0 & k-1 \\ -2 & 2 & k-1 \\ 0 & -2 & k+1\end{array}\right|=0$

⇒ 2(2k + 2 + 2k – 2) – 0 + (k – 1) (4 – 0) = 0 ⇒ 2(4k) + (k – 1)4 = 0 ⇒ 8k + 4k – 4 = 0 ⇒ 12k – 4 = 0

$\Rightarrow k=\frac{4}{12}=\frac{1}{3}$

Applying $R_1 \rightarrow R_1+R_2$, we get

L.H.S. $=\left|\begin{array}{ccc}x+y+z & x+y+z & x+y+z \\ z & x & y \\ 1 & 1 & 1\end{array}\right|$

Taking $(x+y+z)$ common from $\mathrm{R}_1$, we get

L.H.S. $=(x+y+z)\left|\begin{array}{lll}1 & 1 & 1 \\ z & x & y \\ 1 & 1 & 1\end{array}\right|$

$=(x+y+z)(0)$

$\ldots\left[\because R_1\right.$ and $R_3$ are identical $]$

$=0=$ R.H.S.

$\begin{aligned} & \Rightarrow 1(4-16)-2 x(1-16)+4 x(1-4)=0 \\ & \Rightarrow 1(-12)-2 x(-15)+4 x(-3)=0 \\ & \Rightarrow-12+30 x-12 x=0 \\ & \Rightarrow 18 x=12 \\ & \Rightarrow x=\frac{12}{18}=\frac{2}{3}\end{aligned}$

$\begin{aligned} & \Rightarrow 1\left(-10 x^2+10 x\right)-4\left(5 x^2+5\right)+20(2 x+2)=0 \\ & \Rightarrow-10 x^2+10 x-20 x^2-20+40 x+40=0 \\ & \Rightarrow-30 x^2+50 x+20=0\end{aligned}$

$\Rightarrow 3 x^2-5 x-2=0 \ldots$....[Dividing throughout by $(-10)$ ]

$\begin{aligned} & \Rightarrow 3 x^2-6 x+x-2=0 \\ & \Rightarrow 3 x(x-2)+1(x-2)=0 \\ & \Rightarrow(x-2)(3 x+1)=0 \\ & \Rightarrow x-2=0 \text { or } 3 x+1=0 \\ & \Rightarrow x=2 \text { or } x=-\frac{1}{3}\end{aligned}$

$\begin{aligned} & =\left[\begin{array}{ccc}5-6+2 & -5-9+6 & 30+3+4 \\ 0-12+8 & 10-21+12 & -25-15-2\end{array}\right] \\ & =\left[\begin{array}{ccc}1 & -8 & 37 \\ -4 & 1 & -42\end{array}\right] \\ & \therefore X=\frac{1}{4}\left[\begin{array}{ccc}1 & -8 & 37 \\ -4 & 1 & -42\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{4} & -2 & \frac{37}{4} \\ -1 & \frac{1}{4} & -\frac{21}{2}\end{array}\right] \\ & \end{aligned}$

$=100\left|\begin{array}{ccc}1 & 2 & 3 \\ 505 & 606 & 707 \\ 1 & 2 & 3\end{array}\right|$ by using property
$=100 \times 0\left(\mathrm{R}_1\right.$ and $\mathrm{R}_3$ are identical $)$
$=0$

$\begin{aligned} & \therefore \quad A^{\mathrm{T}} \mathrm{A}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right] \\ & =\left[\begin{array}{cc}\cos ^2 \alpha+\sin ^2 \alpha & \cos \alpha \sin \alpha-\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha-\cos \alpha \sin \alpha & \sin ^2 \alpha+\cos ^2 \alpha\end{array}\right]\end{aligned}$

$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\therefore A^{\top} A=I$, where $I$ is the unit matrix of order 2 .

$\begin{aligned} \therefore \quad C & =\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]+2\left[\begin{array}{cc}-1 & 3 \\ 4 & -1\end{array}\right]-3\left[\begin{array}{cc}5 & 4 \\ -2 & 3\end{array}\right] \\ & =\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]+\left[\begin{array}{cc}-2 & 6 \\ 8 & -2\end{array}\right]-\left[\begin{array}{cc}15 & 12 \\ -6 & 9\end{array}\right] \\ & =\left[\begin{array}{cc}1-2-15 & 0+6-12 \\ 0+8+6 & 1-2-9\end{array}\right]\end{aligned}$

$\begin{array}{ll}\therefore & C=\left[\begin{array}{cc}-16 & -6 \\ 14 & -10\end{array}\right] \\ \therefore & C^{\mathrm{T}}=\left[\begin{array}{cc}-16 & 14 \\ -6 & -10\end{array}\right]\end{array}$

$\therefore \quad 3 A=3\left[\begin{array}{ccc}1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9\end{array}\right]=\left[\begin{array}{ccc}3 & 6 & -15 \\ 6 & -9 & 12 \\ -15 & 12 & 27\end{array}\right]$

$\therefore \quad(3 \mathrm{~A})^{\mathrm{T}}=\left[\begin{array}{ccc}3 & 6 & -15 \\ 6 & -9 & 12 \\ -15 & 12 & 27\end{array}\right]$

$\ldots$..(i)

$A^T=\left[\begin{array}{ccc}1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9\end{array}\right]$

$\therefore \quad 3 A^{\mathrm{T}}=3\left[\begin{array}{ccc}1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9\end{array}\right]=\left[\begin{array}{ccc}3 & 6 & -15 \\ 6 & -9 & 12 \\ -15 & 12 & 27\end{array}\right]$

...(ii)

From (i) and (ii), we get

$(3 \mathrm{~A})^{\top}=3 \mathrm{~A}^{\top}$

$\therefore \quad 2 A=2\left[\begin{array}{cc}5 & -3 \\ 4 & -3 \\ -2 & 1\end{array}\right]=\left[\begin{array}{cc}10 & -6 \\ 8 & -6 \\ -4 & 2\end{array}\right]$

$\therefore \quad(2 \mathrm{~A})^{\mathrm{T}}=\left[\begin{array}{ccc}10 & 8 & -4 \\ -6 & -6 & 2\end{array}\right]$

$\ldots$ (i)

$A^T=\left[\begin{array}{ccc}5 & 4 & -2 \\ -3 & -3 & 1\end{array}\right]$

$\therefore \quad 2 A^{\top}=2\left[\begin{array}{ccc}5 & 4 & -2 \\ -3 & -3 & 1\end{array}\right]$

$=\left[\begin{array}{rrr}10 & 8 & -4 \\ -6 & -6 & 2\end{array}\right]$

$\ldots$ (ii)

From (i) and (ii), we get

$(2 \mathrm{~A})^{\top}=2 \mathrm{~A}^{\top}$

$\begin{aligned} & =\left[\begin{array}{cc}\cos ^2 \alpha-\sin ^2 \alpha & \cos \alpha \sin \alpha+\cos \alpha \sin \alpha \\ -\cos \alpha \sin \alpha-\cos \alpha \sin \alpha & -\sin ^2 \alpha+\cos ^2 \alpha\end{array}\right] \\ & =\left[\begin{array}{cc}\cos ^2 \alpha-\sin ^2 \alpha & 2 \sin \alpha \cos \alpha \\ -2 \sin \alpha \cos \alpha & \cos ^2 \alpha-\sin ^2 \alpha\end{array}\right] \\ & =\left[\begin{array}{cc}\cos 2 \alpha & \sin 2 \alpha \\ -\sin 2 \alpha & \cos 2 \alpha\end{array}\right]\end{aligned}$

∴ [6 + 12x + 14] =[0] ∴ By equality of matrices, we get ∴ 12x + 20 = 0 ∴ 12x =-20

$\therefore x=\frac{-5}{3}$

∴ By equality of matrices, we get 3k = 3 ∴ k = 1

$\begin{aligned} & =\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]-5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]+7\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\ & =\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]-\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]+\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right] \\ & =\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]+\left[\begin{array}{cc}7 & 0 \\ 0 & 7\end{array}\right] \\ & =\left[\begin{array}{cc}8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7\end{array}\right]\end{aligned}$

$=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0$

i.e, to prove $A^2+A B+B A+B^2=A^2+A B+B^2$,

i.e., to prove BA = 0.

$\begin{aligned} & B A=\left[\begin{array}{cc}5 & -4 \\ 10 & -8\end{array}\right]\left[\begin{array}{cc}8 & 4 \\ 10 & 5\end{array}\right] \\ & {\left[\begin{array}{ll}40-40 & 20-20 \\ 80-80 & 40-40\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]}\end{aligned}$

$\begin{aligned} & =\left[\begin{array}{lll}1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1\end{array}\right]-\left[\begin{array}{lll}4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4\end{array}\right] \\ & =\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]-\left[\begin{array}{ccc}4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4\end{array}\right] \\ & =\left[\begin{array}{lll}9-4 & 8-8 & 8-8 \\ 8-8 & 9-4 & 8-8 \\ 8-8 & 8-8 & 9-4\end{array}\right]\end{aligned}$

$=\left[\begin{array}{lll}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right]$

which is a scalar matrix.