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Matrices and Determinants

JEE Mathematics · 271 questions · Page 17 of 28 · Click an option or "Show Solution" to reveal answer

Q161
Let the system of equations x+2y+3z=5,2x+3y+z=9,4x+3y+λz=μx+2 y+3 z=5,2 x+3 y+z=9,4 x+3 y+\lambda z=\mu have infinite number of solutions. Then λ+2μ\lambda+2 \mu is equal to :
A 22
B 17
C 15
D 28
Correct Answer
Option B
Solution
x+2y+3z=52x+3y+z=94x+3y+λz=μ\begin{aligned} & x+2 y+3 z=5 \\\\ & 2 x+3 y+z=9 \\\\ & 4 x+3 y+\lambda z=\mu \end{aligned}

For infinite following Δ=Δ1=Δ2=Δ3=0\Delta=\Delta_1=\Delta_2=\Delta_3=0 Δ=∣12323143λ∣=0⇒λ=−13Δ1=∣523931μ3−13∣=0⇒μ=15Δ2=∣153291415−13∣=0\begin{aligned} & \Delta=\left|\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1 \\ 4 & 3 & \lambda\end{array}\right|=0 \Rightarrow \lambda=-13 \\\\ & \Delta_1=\left|\begin{array}{llc}5 & 2 & 3 \\ 9 & 3 & 1 \\ \mu & 3 & -13\end{array}\right|=0 \Rightarrow \mu=15 \\\\ & \Delta_2=\left|\begin{array}{ccc}1 & 5 & 3 \\ 2 & 9 & 1 \\ 4 & 15 & -13\end{array}\right|=0\end{aligned}

Δ3=∣1252394315∣=0\Delta_3=\left|\begin{array}{ccc} 1 & 2 & 5 \\ 2 & 3 & 9 \\ 4 & 3 & 15 \end{array}\right|=0

For λ=−13,μ=15\lambda=-13, \mu=15 system of equation has infinite solution hence λ+2μ=17\lambda+2 \mu=17

Q162
If A=[21−12],B=[1011],C=ABAT\mathrm{A}=\left[\begin{array}{cc}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right], \mathrm{C}=\mathrm{ABA}^{\mathrm{T}} and X=ATC2 A\mathrm{X}=\mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{~A}, then det⁡X\operatorname{det} \mathrm{X} is equal to :
A 243
B 729
C 27
D 891
Correct Answer
Option B
Solution

The solution involves understanding matrix operations and properties such as multiplication, transpose, and determinant.

Given A\mathrm{A}, B\mathrm{B}, and that C=ABAT\mathrm{C} = \mathrm{ABA}^{\mathrm{T}}, and X=ATC2A\mathrm{X} = \mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{A}, we find det⁡X\operatorname{det} \mathrm{X} as follows: First, we express ∣C∣|\mathrm{C}|, the determinant of C\mathrm{C}, in terms of the determinants of A\mathrm{A} and B\mathrm{B}, using the property that ∣ABC∣=∣A∣⋅∣B∣⋅∣C∣|\mathrm{ABC}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{C}|: ∣C∣=∣ABAT∣=∣A∣⋅∣B∣⋅∣AT∣=∣A∣2⋅∣B∣since ∣AT∣=∣A∣.\begin{aligned} |\mathrm{C}| &= |\mathrm{ABA}^{\mathrm{T}}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{A}^{\mathrm{T}}| \\\\ &= |\mathrm{A}|^2 \cdot |\mathrm{B}| \quad \text{since } |\mathrm{A}^{\mathrm{T}}| = |\mathrm{A}|. \end{aligned} Next, we find ∣X∣|\mathrm{X}|, using the property ∣ABC∣=∣A∣⋅∣B∣⋅∣C∣|\mathrm{ABC}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{C}| and substituting ∣C∣|\mathrm{C}|: ∣X∣=∣ATC2A∣=∣AT∣⋅∣C∣2⋅∣A∣=∣A∣2⋅∣C∣2.\begin{aligned} |\mathrm{X}| &= |\mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{A}| = |\mathrm{A}^{\mathrm{T}}| \cdot |\mathrm{C}|^2 \cdot |\mathrm{A}| \\\\ &= |\mathrm{A}|^2 \cdot |\mathrm{C}|^2. \end{aligned} Substituting the expression for ∣C∣|\mathrm{C}| obtained earlier: ∣X∣=∣A∣2⋅(∣A∣2⋅∣B∣)2=(∣A∣3⋅∣B∣)2.\begin{aligned} |\mathrm{X}| &= |\mathrm{A}|^2 \cdot (|\mathrm{A}|^2 \cdot |\mathrm{B}|)^2 \\\\ &= (|\mathrm{A}|^3 \cdot |\mathrm{B}|)^2. \end{aligned} The determinants of A\mathrm{A} and B\mathrm{B} are calculated as:

∣A∣=∣21−12∣=2+1=3|A|=\left|\begin{array}{cc} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{array}\right|=2+1=3
∣B∣=∣1011∣=1|B|=\left|\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right|=1

Finally, substituting these values into our expression for ∣X∣|\mathrm{X}|: ∣X∣=(33⋅1)2=729.\begin{aligned} |\mathrm{X}| &= (3^3 \cdot 1)^2 \\\\ &= 729. \end{aligned} Therefore, det⁡X=729\operatorname{det} \mathrm{X} = 729.

Q163
If the system of equations 2x+3y−z=5x+αy+3z=−43x−y+βz=7 \begin{aligned} & 2 x+3 y-z=5 \\\\ & x+\alpha y+3 z=-4 \\\\ & 3 x-y+\beta z=7 \end{aligned} has infinitely many solutions, then 13αβ13 \alpha \beta is equal to :
A 1110
B 1120
C 1210
D 1220
Correct Answer
Option B
Solution

 Given 2x+3y−z=5x+αy+3z=−43x−y+βz=7Δ2=∣2−1513−43β7∣Δ2=2(21+4β)+1(7+12)+5(β−9)Δ2=42+8β+19+5β−45Δ2=13β+16Δ2=0\begin{aligned} & \text{ Given } 2 x+3 y-z=5 \\\\ & x+\alpha y+3 z=-4 \\\\ & 3 x-y+\beta z=7 \\\\ & \Delta_2=\left|\begin{array}{ccc}2 & -1 & 5 \\ 1 & 3 & -4 \\ 3 & \beta & 7\end{array}\right| \\\\ & \Delta_2=2(21+4 \beta)+1(7+12)+5(\beta-9) \\\\& \Delta_2=42+8 \beta+19+5 \beta-45 \\\\ & \Delta_2=13 \beta+16 \\\\ & \Delta_2=0\end{aligned} ∴β=−1613Δ3=∣2351α−43−17∣Δ3=2(7α−4)−3(7+12)+5(−1−3α)Δ3=14α−8−57−5−15αΔ3=−α−70\begin{aligned} & \therefore \beta=-\dfrac{16}{13} \\\\ & \Delta_3=\left|\begin{array}{lll}2 & 3 & 5 \\ 1 & \alpha & -4 \\ 3 & -1 & 7\end{array}\right| \\\\ & \Delta_3=2(7 \alpha-4)-3(7+12)+5(-1-3 \alpha) \\\\ & \Delta_3=14 \alpha-8-57-5-15 \alpha \\\\ & \Delta_3=-\alpha-70\end{aligned} Δ3=0α=−7013αβ=(13)(−70)(−1613)=+1120\begin{aligned} & \Delta_3=0 \\\\ & \alpha=-70 \\\\ & 13 \alpha \beta=(13)(-70)\left(-\dfrac{16}{13}\right) \\\\ & =+1120\end{aligned}

Q164
Consider the matrix f(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]f(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]. Given below are two statements : Statement I : f(−x) f(-x) is the inverse of the matrix f(x)f(x). Statement II : f(x)f(y)=f(x+y)f(x) f(y)=f(x+y). In the light of the above statements, choose the correct answer from the options given below :
A Statement I is false but Statement II is true
B Both Statement I and Statement II are false
C Both Statement I and Statement II are true
D Statement I is true but Statement II is false
Correct Answer
Option C
Solution
f(−x)=[cos⁡xsin⁡x0−sin⁡xcos⁡x0001]f(x)⋅f(−x)=[100010001]=I\begin{aligned} & f(-x)=\left[\begin{array}{ccc} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & f(x) \cdot f(-x)=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I \end{aligned}

Hence statement- I is correct Now, checking statement II

f(y)=[cos⁡y−sin⁡y0sin⁡ycos⁡y0001]f(x)⋅f(y)=[cos⁡(x+y)−sin⁡(x+y)0sin⁡(x+y)cos⁡(x+y)0001]⇒f(x)⋅f(y)=f(x+y)\begin{aligned} & f(y)=\left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & f(x) \cdot f(y)=\left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \Rightarrow f(x) \cdot f(y)=f(x+y) \end{aligned}

Hence statement-II is also correct.

Q165
The values of α\alpha, for which ∣132α+32113α+132α+33α+10∣=0\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0, lie in the interval
A (−2,1)(-2,1)
B (−32,32)\left(-\dfrac{3}{2}, \dfrac{3}{2}\right)
C (−3,0)(-3,0)
D (0,3)(0,3)
Correct Answer
Option C
Solution
∣132α+32113α+132α+33α+10∣=0\left|\begin{array}{ccc} 1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0 \end{array}\right|=0
⇒(2α+3){7α6}−(3α+1){−76}=0⇒(2α+3)⋅7α6+(3α+1)⋅76=0⇒2α2+3α+3α+1=0⇒2α2+6α+1=0⇒α=−3+72,−3−72\begin{aligned} & \Rightarrow(2 \alpha+3)\left\{\frac{7 \alpha}{6}\right\}-(3 \alpha+1)\left\{\frac{-7}{6}\right\}=0 \\ & \Rightarrow(2 \alpha+3) \cdot \frac{7 \alpha}{6}+(3 \alpha+1) \cdot \frac{7}{6}=0 \\ & \Rightarrow 2 \alpha^2+3 \alpha+3 \alpha+1=0 \\ & \Rightarrow 2 \alpha^2+6 \alpha+1=0 \\ & \Rightarrow \alpha=\frac{-3+\sqrt{7}}{2}, \frac{-3-\sqrt{7}}{2} \end{aligned}

Hence option (C) is correct.

Q166
Let AA be a 3×33 \times 3 real matrix such that A(101)=2(101),A(−101)=4(−101),A(010)=2(010). A\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right)=2\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right), A\left(\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right)=4\left(\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right), A\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right)=2\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right) \text{. } Then, the system (A−3I)(xyz)=(123)(A-3 I)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right) has :
A exactly two solutions
B infinitely many solutions
C unique solution
D no solution
Correct Answer
Option C
Solution
 Let A=[x1y1z1x2y2z2x3y3z3]\text{ Let } A=\left[\begin{array}{lll} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{array}\right]
 Given A[101]=[202] ..... (1)\text{ Given } A\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \\ 2 \end{array}\right] \quad \text{ ..... (1)}
∴[x1+z1x2+z2x3+z3]=[202]\therefore\left[\begin{array}{l} \mathrm{x}_1+\mathrm{z}_1 \\ \mathrm{x}_2+\mathrm{z}_2 \\ \mathrm{x}_3+\mathrm{z}_3 \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \\ 2 \end{array}\right]
∴x1+z1=2.... (2)x2+z2=0.... (3)x3+z3=0.... (4)\begin{aligned} \therefore \mathrm{x}_1+\mathrm{z}_1 & =2 \quad \text{.... (2)}\\ \mathrm{x}_2+\mathrm{z}_2 & =0 \quad \text{.... (3)}\\ \mathrm{x}_3+\mathrm{z}_3 & =0 \quad \text{.... (4)} \end{aligned}
 Given A[−101]=[−404]∴[−x1+z1−x2+z2−x3+z3]=[404]\begin{aligned} & \text{ Given } A\left[\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} -4 \\ 0 \\ 4 \end{array}\right] \\ & \therefore\left[\begin{array}{l} -\mathrm{x}_1+\mathrm{z}_1 \\ -\mathrm{x}_2+\mathrm{z}_2 \\ -\mathrm{x}_3+\mathrm{z}_3 \end{array}\right]=\left[\begin{array}{l} 4 \\ 0 \\ 4 \end{array}\right] \end{aligned}
⇒−x1+z1=−4... (5)−x2+x2=0... (6)−x3+z3=4\begin{array}{r} \Rightarrow-\mathrm{x}_1+\mathrm{z}_1=-4 \quad \text{... (5)}\\ -\mathrm{x}_2+\mathrm{x}_2=0 \quad \text{... (6)}\\ -\mathrm{x}_3+\mathrm{z}_3=4 \end{array}
 Given A[010]=[020]∴[y1y2y3]=[020]\begin{aligned} & \text{ Given } A\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]=\left[\begin{array}{l} 0 \\ 2 \\ 0 \end{array}\right] \\ & \therefore\left[\begin{array}{l} \mathrm{y}_1 \\ \mathrm{y}_2 \\ \mathrm{y}_3 \end{array}\right]=\left[\begin{array}{l} 0 \\ 2 \\ 0 \end{array}\right] \end{aligned}
∴y1=0,y2=2,y3=0∴ from (2),(3),(4),(5),(6) and (7)x1=3x,x2=0,x3=−1y1=0,y2=2,y3=0z1=−1,z2=0,z3=3\begin{aligned} & \therefore \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 \\ & \therefore \text{ from }(2),(3),(4),(5),(6) \text{ and }(7) \\ & \mathrm{x}_1=3 \mathrm{x}, \mathrm{x}_2=0, \mathrm{x}_3=-1 \\ & \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 \\ & \mathrm{z}_1=-1, \mathrm{z}_2=0, \mathrm{z}_3=3 \end{aligned}
∴A=[30−1020−103]∴ Now (A−31)[xyz]=[−123]∴[00−10−10−100][xyz]=[−123][−z−y−x]=[123][z=−1],[y=−2],[x=−3]\begin{gathered} \therefore A=\left[\begin{array}{ccc} 3 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 3 \end{array}\right] \\ \therefore \text{ Now }(A-31)\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} -1 \\ 2 \\ 3 \end{array}\right] \\ \therefore\left[\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} -1 \\ 2 \\ 3 \end{array}\right] \\ {\left[\begin{array}{c} -z \\ -y \\ -x \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]} \\ {[z=-1],[y=-2],[x=-3]} \end{gathered}
Q167
If the system of linear equations x−2y+z=−42x+αy+3z=53x−y+βz=3\begin{aligned} & x-2 y+z=-4 \\ & 2 x+\alpha y+3 z=5 \\ & 3 x-y+\beta z=3 \end{aligned} has infinitely many solutions, then 12α+13β12 \alpha+13 \beta is equal to
A 60
B 54
C 64
D 58
Correct Answer
Option D
Solution
D=∣1−212α33−1β∣=1(αβ+3)+2(2β−9)+1(−2−3α)=αβ+3+4β−18−2−3α\begin{aligned} & D=\left|\begin{array}{ccc} 1 & -2 & 1 \\ 2 & \alpha & 3 \\ 3 & -1 & \beta \end{array}\right| \\ & =1(\alpha \beta+3)+2(2 \beta-9)+1(-2-3 \alpha) \\ & =\alpha \beta+3+4 \beta-18-2-3 \alpha \end{aligned}

For infinite solutions

D=0,D1=0,D2=0\mathrm{D}=0, \mathrm{D}_1=0, \mathrm{D}_2=0

and

D3=0D=0\begin{aligned} & \mathrm{D}_3=0 \\ & \mathrm{D}=0 \end{aligned}
αβ−3α+4β=17\alpha \beta-3 \alpha+4 \beta=17

..... (1)

D1=∣−4−215α33−1β∣=0D2=∣1−4125333β∣=0⇒1(5β−9)+4(2β−9)+1(6−15)=013β−9−36−9=0\begin{aligned} & \mathrm{D}_1=\left|\begin{array}{ccc} -4 & -2 & 1 \\ 5 & \alpha & 3 \\ 3 & -1 & \beta \end{array}\right|=0 \\ & \mathrm{D}_2=\left|\begin{array}{ccc} 1 & -4 & 1 \\ 2 & 5 & 3 \\ 3 & 3 & \beta \end{array}\right|=0 \\ & \Rightarrow 1(5 \beta-9)+4(2 \beta-9)+1(6-15)=0 \\ & 13 \beta-9-36-9=0 \end{aligned}
13β=54,β=541313 \beta=54, \beta=\frac{54}{13}

put in (1)

5413α−3α+4(5413)=17\frac{54}{13} \alpha-3 \alpha+4\left(\frac{54}{13}\right)=17
54α−39α+216=22154 \alpha-39 \alpha+216=221
15α=5α=1315 \alpha=5 \quad \alpha=\frac{1}{3}

Now,

12α+13β=12⋅13+13⋅541312 \alpha+13 \beta=12 \cdot \frac{1}{3}+13 \cdot \frac{54}{13}
=4+54=58=4+54=58
Q168
 Let A=[1000αβ0βα] and ∣2 A∣3=221 where α,β∈Z, Then a value of α is \text{ Let } A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{array}\right] \text{ and }|2 \mathrm{~A}|^3=2^{21} \text{ where } \alpha, \beta \in Z \text{, Then a value of } \alpha \text{ is }
A 9
B 17
C 3
D 5
Correct Answer
Option D
Solution
∣A∣=α2−β2∣2 A∣3=221⇒∣A∣=24α2−β2=16(α+β)(α−β)=16⇒α=4 or 5\begin{aligned} & |\mathrm{A}|=\alpha^2-\beta^2 \\ & |2 \mathrm{~A}|^3=2^{21} \Rightarrow|\mathrm{A}|=2^4 \\ & \alpha^2-\beta^2=16 \\ & (\alpha+\beta)(\alpha-\beta)=16 \Rightarrow \alpha=4 \text{ or } 5 \end{aligned}
Q169
Let A\mathrm{A} be a square matrix such that AAT=I\mathrm{AA}^{\mathrm{T}}=\mathrm{I}. Then 12A[(A+AT)2+(A−AT)2]\dfrac{1}{2} A\left[\left(A+A^T\right)^2+\left(A-A^T\right)^2\right] is equal to
A A2+AT\mathrm{A}^2+\mathrm{A}^{\mathrm{T}}
B A3+I\mathrm{A}^3+\mathrm{I}
C A3+AT\mathrm{A}^3+\mathrm{A}^{\mathrm{T}}
D A2+I\mathrm{A}^2+\mathrm{I}
Correct Answer
Option C
Solution
AAT=I=ATA\mathrm{AA}^{\mathrm{T}}=\mathrm{I}=\mathrm{A}^{\mathrm{T}} \mathrm{A}

On solving given expression, we get

12 A[ A2+(AT)2+2 A AT+A2+(AT)2−2 A AT]=A[A2+(AT)2]=A3+AT\begin{aligned} & \frac{1}{2} \mathrm{~A}\left[\mathrm{~A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2+2 \mathrm{~A} \mathrm{~A}^{\mathrm{T}}+\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2-2 \mathrm{~A} \mathrm{~A}^{\mathrm{T}}\right] \\ & =\mathrm{A}\left[\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2\right]=\mathrm{A}^3+\mathrm{A}^{\mathrm{T}} \end{aligned}
Q170
Let A=[2126211332]A=\left[\begin{array}{ccc}2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2\end{array}\right] and P=[120502715]P=\left[\begin{array}{lll}1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5\end{array}\right]. The sum of the prime factors of ∣P−1AP−2I∣\left|P^{-1} A P-2 I\right| is equal to
A 66
B 27
C 23
D 26
Correct Answer
Option D
Solution
∣P−1AP−2I∣=∣P−1AP−2P−1P∣=∣P−1( A−2I)P∣=∣P−1∣∣A−2I∣∣P∣=∣A−2I∣=∣0126011330∣=69\begin{aligned} \left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{I}\right| & =\left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{P}^{-1} \mathrm{P}\right| \\ & =\left|\mathrm{P}^{-1}(\mathrm{~A}-2 \mathrm{I}) \mathrm{P}\right| \\ & =\left|\mathrm{P}^{-1}\right||\mathrm{A}-2 \mathrm{I}||\mathrm{P}| \\ & =|\mathrm{A}-2 \mathrm{I}| \\ & =\left|\begin{array}{ccc} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{array}\right|=69 \end{aligned}

So, Prime factor of 69 is 3 & 23 So, sum = 26

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