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.

$\begin{aligned} & =\left[\begin{array}{cc}3-6 & -1-14 \\ 15+18 & -5+42\end{array}\right]-\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right] \\ & =\left[\begin{array}{cc}-3 & -15 \\ 33 & 37\end{array}\right]-\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right] \\ & =\left[\begin{array}{cc}-3-2 & -15-0 \\ 33-0 & 37-2\end{array}\right] \\ & =\left[\begin{array}{cc}-5 & -15 \\ 33 & 35\end{array}\right]\end{aligned}$

$=\left[\begin{array}{cc}4-9 & 1+6 \\ 16+6 & 4-4\end{array}\right]$

$=\left[\begin{array}{cc}-5 & 7 \\ 22 & 0\end{array}\right]$

...(i)

$\mathrm{BA}=\left[\begin{array}{rr}4 & 1 \\ 3 & -2\end{array}\right]\left[\begin{array}{cc}1 & -3 \\ 4 & 2\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{cc}4+4 & -12+2 \\ 3-8 & -9-4\end{array}\right] \\ & =\left[\begin{array}{cc}8 & -10 \\ -5 & -13\end{array}\right]\end{aligned}$

...(ii)

From (i) and (ii), we get AB ≠ BA

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

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

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

$\begin{aligned} & =-\left[\begin{array}{ccc}1 & 2 & -3 \\ -3 & 7 & -8 \\ 0 & -6 & 1\end{array}\right]+\left[\begin{array}{ccc}9 & -1 & 2 \\ -4 & 2 & 5 \\ 4 & 0 & -3\end{array}\right] \\ & =-\left[\begin{array}{ccc}1+9 & 2-1 & -3+2 \\ -3-4 & 7+2 & -8+5 \\ 0+4 & -6+0 & 1-3\end{array}\right] \\ & =-\left[\begin{array}{ccc}10 & 1 & -1 \\ -7 & 9 & -3 \\ 4 & -6 & -2\end{array}\right] \\ & =\left[\begin{array}{ccc}-10 & -1 & 1 \\ 7 & -9 & 3 \\ -4 & 6 & 2\end{array}\right]\end{aligned}$

$\begin{aligned} & =\left[\begin{array}{rr}1 & -2 \\ 5 & 3\end{array}\right]-\left[\begin{array}{cc}2 & -6 \\ 8 & -14\end{array}\right]+\left[\begin{array}{ll}6 & 0 \\ 0 & 6\end{array}\right] \\ & =\left[\begin{array}{cc}1-2+6 & -2+6+0 \\ 5-8+0 & 3+14+6\end{array}\right] \\ & =\left[\begin{array}{cc}5 & 4 \\ -3 & 23\end{array}\right]\end{aligned}$

$\therefore A=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$

Given, $a_{i j}=\mathrm{i}-\mathrm{j}$

$\begin{aligned} & a_{11}=1-1=0, a_{12}=1-2=-1, a_{13}=1-3=-2 \\ & a_{21}-2-1=1, a_{22}=2-2=0, a_{23}=2-3=-1, \\ & a_{31}=3-1=2, a_{32}=3-2=1, a_{33}=3-3=0\end{aligned}$

$\therefore \quad A=\left[\begin{array}{ccc}0 & -1 & -2 \\ 1 & 0 & -1 \\ 2 & 1 & 0\end{array}\right]$

$\therefore \quad A^{\mathrm{T}}=\left[\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & 1 \\ -2 & -1 & 0\end{array}\right]=-\left[\begin{array}{ccc}0 & -1 & -2 \\ 1 & 0 & -1 \\ 2 & 1 & 0\end{array}\right]$

$\therefore \mathrm{A}^{\top}=-\mathrm{A}$, i.e $_v \mathrm{~A}=-\mathrm{A}^{\top}$

∴ A is a skew-symmetric matrix.

$\therefore \quad A^{\mathrm{T}}=\left[\begin{array}{ccc}0 & y & \frac{3}{2} \\ -5 i & 0 & -\sqrt{2} \\ x & z & 0\end{array}\right]$

Since $\mathrm{A}$ is a skew-symmetric matrix,

$\mathrm{A}=-\mathrm{A}^{\mathrm{T}}$

$\therefore \quad\left[\begin{array}{ccc}0 & -5 i & x \\ y & 0 & \mathrm{z} \\ \frac{3}{2} & -\sqrt{2} & 0\end{array}\right]=-\left[\begin{array}{ccc}0 & y & \frac{3}{2} \\ -5 i & 0 & -\sqrt{2} \\ x & z & 0\end{array}\right]$

$\therefore \quad\left[\begin{array}{ccc}0 & -5 \mathrm{i} & x \\ y & 0 & z \\ \frac{3}{2} & -\sqrt{2} & 0\end{array}\right]=\left[\begin{array}{ccc}0 & -y & \frac{-3}{2} \\ 5 \mathrm{i} & 0 & \sqrt{2} \\ -x & -z & 0\end{array}\right]$

$\therefore \quad$ By equality of matrices, we get

$x=\frac{-3}{2}, y=5 i, z=\sqrt{2}$

$\therefore \quad A^{\mathrm{T}}=\left[\begin{array}{ccc}1 & \mathrm{~b} & -4 \\ \frac{3}{5} & -5 & \mathrm{c} \\ \mathrm{a} & -7 & 0\end{array}\right]$

Since A is a symmetric matrix,

$A=A^T$

$\therefore \quad\left[\begin{array}{ccc}1 & \frac{3}{5} & a \\ b & -5 & -7 \\ -4 & c & 0\end{array}\right]=\left[\begin{array}{ccc}1 & b & -4 \\ \frac{3}{5} & -5 & c \\ a & -7 & 0\end{array}\right]$

$\therefore \quad$ By equality of matrices, we get

$a=-4, b=\frac{3}{5}, c=-7$

$\therefore|A|=\left[\begin{array}{cc}7 & 5 \\ -4 & 7\end{array}\right]=49+20=69 \neq 0$

$\therefore|A|=\left|\begin{array}{ccc}3 & 5 & 7 \\ -2 & 1 & 4 \\ 3 & 2 & 5\end{array}\right|$

= 3(5 – 8) – 5(-10 – 12) + 7(-4 – 3) = -9 + 110 – 49 = 52 ≠ 0 ∴ A is a non-singular matrix.

$\therefore|A|=\left|\begin{array}{ccc}5 & 0 & 5 \\ 1 & 99 & 100 \\ 6 & 99 & 105\end{array}\right|$

Applying $\mathrm{C}_2 \rightarrow \mathrm{C}_2+\mathrm{C}_1$

$|A|=\left|\begin{array}{ccc}5 & 5 & 5 \\ 1 & 100 & 100 \\ 6 & 105 & 105\end{array}\right|$

$=0 \ldots\left[\because C_2\right.$ and $C_3$ are identical $]$

∴ A is a singular matrix.

$\therefore \quad|A|=\left|\begin{array}{ccc}a & b & c \\ p & q & r \\ 2 a-p & 2 b-q & 2 c-r\end{array}\right|$

Applying $R_3 \rightarrow R_3+R_2$, wé get$|A|=\left|\begin{array}{ccc}a & b & c \\ p & q & r \\ 2 a & 2 b & 2 c\end{array}\right|$

Taking 2 common from $R_3$, we get

$|A|=2\left|\begin{array}{lll}a & b & c \\ p & q & r \\ a & b & c\end{array}\right|$

$=2(0) \quad \ldots\left[\because R_1\right.$ and $R_3$ are identical $]$

$=0$

$\therefore \quad$ A is a singular matrix.

$\therefore \quad a_{11}=\frac{(1+1)^3}{5}=\frac{2^3}{5}=\frac{8}{5}, a_{12}=\frac{(1+2)^3}{5}=\frac{3^3}{5}=\frac{27}{5}$

$\begin{aligned} & a_{21}=\frac{(2+1)^3}{5}=\frac{3^3}{5}=\frac{27}{5}, a_{22}=\frac{(2+2)^3}{5}=\frac{4^3}{5}=\frac{64}{5} \\ & a_{31}=\frac{(3+1)^3}{5}=\frac{4^3}{5}=\frac{64}{5}, a_{32}=\frac{(3+2)^3}{5}=\frac{5^3}{5}=\frac{125}{5}\end{aligned}$

$\therefore \quad A=\left[\begin{array}{cc}\frac{8}{5} & \frac{27}{5} \\ \frac{27}{5} & \frac{64}{5} \\ \frac{64}{5} & \frac{125}{5}\end{array}\right]=\frac{1}{5}\left[\begin{array}{cc}8 & 27 \\ 27 & 64 \\ 64 & 125\end{array}\right]$

$\begin{aligned} & \therefore a_{11}=1-3(1)=1-3=-2 \\ & a_{12}=1-3(2)=1-6=-5 \\ & a_{21}=2-3(1)=2-3=-1 \\ & a_{22}=2-3(2)=2-6=-4 \\ & a_{31}=3-3(1)=3-3=0 \\ & a_{32}=3-3(2)=3-6=-3\end{aligned}$

$\therefore A=\left[\begin{array}{cc}-2 & -5 \\ -1 & -4 \\ 0 & -3\end{array}\right]$

$\therefore \quad A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{52}\end{array}\right]$

$a_{i j}=\frac{(i-j)^2}{5-i}$

$\therefore \quad a_{11}=\frac{(1-1)^2}{5-1}=0, a_{12}=\frac{(1-2)^2}{5-1}=\frac{(-1)^2}{4}=\frac{1}{4}$,

$a_{21}=\frac{(2-1)^2}{5-2}=\frac{1}{3}, a_{22}=\frac{(2-2)^2}{5-2}=0$

$a_{31}=\frac{(3-1)^2}{5-3}=\frac{2^2}{2}=2, a_{32}=\frac{(3-2)^2}{5-3}=\frac{1}{2}$

$\therefore \quad A=\left[\begin{array}{ll}0 & \frac{1}{4} \\ \frac{1}{3} & 0 \\ 2 & \frac{1}{2}\end{array}\right]$

[Note: Answer given in the textbook is $A=\left[\begin{array}{cc}0 & \frac{1}{4} \\ \frac{1}{2} & 0 \\ 2 & \frac{1}{2}\end{array}\right]$

However, as per our calculation it is $\left[\begin{array}{cc}0 & \frac{1}{4} \\ \frac{1}{3} & 0 \\ 2 & \frac{1}{2}\end{array}\right]$.

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

$\therefore \mathrm{A}(\Delta \mathrm{L} \mathrm{MN})=\frac{1}{2}\left|\begin{array}{ccc}0 & \frac{1}{2} & 1 \\ 2 & -1 & 1 \\ -4 & \frac{7}{2} & 1\end{array}\right|$

$\begin{aligned} & \left.=\frac{1}{2}[0-] \frac{1}{2}(2+4)+1(7-4)\right] \\ & =\frac{1}{2}\left[-\frac{1}{2}(6)+1(3)\right] \\ & =\frac{1}{2}(-3+3)=0\end{aligned}$

∴ The points L, M, N are collinear.

∴ If A(∆PQR) = 0, then the points P,Q, R are collinear

$\therefore A(\triangle P Q R)=\frac{1}{2}\left|\begin{array}{ccc}3 & -5 & 1 \\ 6 & 1 & 1 \\ 4 & 2 & 1\end{array}\right|$

$\begin{aligned} & =\frac{1}{2}[3(1-2)-(-5)(6-4)+1(12-4)] \\ & =\frac{1}{2}[3(-1)+5(2)+1(8)] \\ & =\frac{1}{2}(-3+10+8)=\frac{15}{2} \neq 0\end{aligned}$∴ The points P, Q, R are non-collinear.

$\therefore A(\Delta \mathrm{ABC})=\frac{1}{2}\left|\begin{array}{ccc}3 & -1 & 1 \\ 0 & -3 & 1 \\ 12 & 5 & 1\end{array}\right|$

$\begin{aligned} & =\frac{1}{2}[3(-3-5)-(-1)(0-12)+1(0+36)] \\ & =\frac{1}{2}[3(-8)+1(-12)+1(36)] \\ & =\frac{1}{2}(-24-12+36) \\ & =0\end{aligned}$

∴ The points A, B, C are collinear.

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

∴ k(0 – 1) – 3(0 – 1) + 1(1 – 2) = 0 ∴ k(-1) – 3(-1) + 1(-1) = 0 ∴ -k + 3 – 1 = 0 ∴ k = 2.

$\left|\begin{array}{ccc}2 & 3 & -2 \\ 2 & 4 & -k \\ 1 & -2 & 3 k\end{array}\right|=0$

∴ 2(12k – 2k) – 3(6k + k) – 2(- 4 – 4) = 0 ∴ 2(10k) – 3(7k) – 2(- 8) = 0 ∴ 20k – 21k + 16 = 0 ∴ k = 16

$\left|\begin{array}{ccc}1 & 2 & -3 \\ 7 & 4 & -11 \\ 2 & 4 & -6\end{array}\right|$

= 1(-24 + 44) – 2(-42 + 22) – 3(28 – 8) = 1(20) – 2(-20) – 3(20) = 20 + 40 – 60 = 0 ∴ The given equations are consistent.

$\left|\begin{array}{ccc}2 & 3 & -4 \\ 1 & 2 & -3 \\ 3 & 4 & 5\end{array}\right|$

= 2(10 + 12) – 3(5 + 9) – 4(4 – 6) = 2 (22) – 3(14) – 4(-2) = 44 – 42 + 8 = 10 ≠ 0 ∴ The given equations are not consistent.

$D=\left|\begin{array}{ccc}2 & -1 & 3 \\ 3 & 1 & -2 \\ 11 & 2 & -3\end{array}\right|$

= 2(-3 + 4) – (-l)(-9 + 22) + 3(6-11) = 2(1)+1(13)+ 3(-5) = 2 + 13-15 = 0 ∴ The given equations are consistent.

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

∴ (x – 1)(x – 2)(x – 3) – 0] – x(0 – 0) + (x – 2)(0 – 0) = ∴ (x – 1)(x – 2)(x – 3) = 0 ∴ x — 1 = 0 or x-2 = 0 or x-3 = 0 ∴ x = 1 or x = 2 or x = 3

$\left|\begin{array}{ccc}x+2 & x+6 & x-1 \\ 4 & -7 & 3 \\ -3 & -4 & 7\end{array}\right|=0$

∴ (x + 2)(- 49 + 12) – (x + 6)(28 + 9) + (x- 1)(- 16 – 21) = 0 ∴ (x + 2) (-37) – (x + 6) (37) + (x – 1) (-37) = 0 ∴ -37(x + 2+ x + 6 + x – 1) = 0 ∴ 3x + 7 = 0

$\therefore x=\frac{-7}{3}$

$\cdots\left[\because \log _e b=\frac{\log _e b}{\log _e c}\right]$

Taking $\frac{1}{\log _e x}, \frac{1}{\log _e y}, \frac{1}{\log _e \mathrm{z}}$ common from $\mathrm{R}_1$,

$R_2, R_3$ respectively, we get

L.H.S.

$=\frac{1}{\log _e x \cdot \log _e y \cdot \log _e z}\left|\begin{array}{lll}\log _e x & \log _e y & \log _e z \\ \log _e x & \log _e y & \log _e z \\ \log _e x & \log _e y & \log _e z\end{array}\right|$

$=\frac{1}{\log _{\mathrm{e}} x \cdot \log _{\mathrm{e}} y_{\cdot} \log _{\mathrm{e}} z}(0)$

$\ldots\left[\because \mathbf{R}_1, \mathbf{R}_2, \mathbf{R}_3\right.$ are identical $]$

$=0=$ R.H.S.

Applying $C_1 \rightarrow C_1-\left(C_2+C_3\right)$, we get

L.H.S $=\left|\begin{array}{ccc}0 & a & b \\ -2 c & a+c & c \\ -2 c & c & b+c\end{array}\right|$

Taking $(-2)$ common from $C_1$, we get

L.H.S. $=-2\left|\begin{array}{ccc}0 & a & b \\ c & a+c & c \\ c & c & b+c\end{array}\right|$

Applying $C_2 \rightarrow C_2-C_1$ and $C_3 \rightarrow C_3-C_1$, we get

L.H.S. $=-2\left|\begin{array}{lll}0 & a & b \\ c & a & 0 \\ c & 0 & b\end{array}\right|$

= -2[0(ab – 0) – a(bc – 0) + b(0 – ac)] = -2(0 – abc – abc) = -2(-2abc) = 4abc = R.H.S.

Applying $\mathrm{Cx}_3-\mathrm{C}_3-9 \mathrm{C}_2$, we get

$D=\left|\begin{array}{lll}2 & 7 & 2 \\ 3 & 8 & 3 \\ 5 & 9 & 5\end{array}\right|$

$=0 \ldots\left[\because C_1\right.$ and $C_3$ are identical $]$

Taking $(3 x)$ common from $R_3$, we get

$D=3 x\left|\begin{array}{ccc}2 & 3 & 4 \\ 5 & 6 & 8 \\ 2 & 3 & 4\end{array}\right|$

= (3x)(0) = 0

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

= 0

Applying $\mathrm{C}_3 \rightarrow \mathrm{C}_3+\mathrm{C}_2$, we get.

$\mathrm{D}=\left|\begin{array}{ccc}1 & a & a+b+c \\ 1 & b & a+b+c \\ 1 & c & a+b+c\end{array}\right|$

Taking $(a+b+c)$ common from $C_3$, we get

$D=(a+b+c)\left|\begin{array}{lll}1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1\end{array}\right|$

= (a + b + c)(0)

$\ldots\left[\because C_1\right.$ and $C_3$ are identical $]$

= 0