Practice Start Test

Matrices and Determinants

JEE Mathematics · 271 questions · Page 15 of 28 · Click an option or "Show Solution" to reveal answer

Q141
Let for A=[123α31112],∣A∣=2A = \left[ \begin{array}{lll}1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2 \end{array} \right],|A| = 2. If ∣2 adj (2 adj (2A))∣=32n\mathrm{|2\,adj\,(2\,adj\,(2A))| = {32^n}}, then 3n+α3n + \alpha is equal to
A 11
B 9
C 12
D 10
Correct Answer
Option A
Solution
A=[123α31112]∣A∣=2\begin{aligned} & A=\left[\begin{array}{lll} 1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2 \end{array}\right] \\\\ & |A|=2 \end{aligned}
⇒1(6−1)−2(2α−1)+3(α−3)=2⇒5−4α+2+3α−9=2⇒−α−4=0⇒α=−4\begin{aligned} \Rightarrow&1(6-1)-2(2 \alpha-1)+3(\alpha-3)=2\\\\ \Rightarrow&5-4 \alpha+2+3 \alpha-9=2\\\\ \Rightarrow&-\alpha-4=0\\\\ \Rightarrow&\alpha=-4 \end{aligned}

Now,

∣2 adj (2 adj (2A))∣=32n\mathrm{|2\,adj\,(2\,adj\,(2A))| = {32^n}}

⇒\Rightarrow

23∣adj⁡(2adj⁡(2A))∣2^3|\operatorname{adj}(2 \operatorname{adj}(2 A))|

= 32n{32^n}

⇒8∣Adj⁡(2Adj⁡(2 A))∣=32n⇒8∣Adj⁡(2×22Adj⁡(A))∣=32n⇒8∣Adj⁡(23AdjA⁡)∣=32n⇒8∣26Adj⁡(AdjA⁡)∣=32n⇒23(26)3∣Adj⁡(Adj⁡)∣=32n⇒23⋅218∣ A∣4=32n⇒221⋅24=32n⇒225=32n⇒(25)5=32n⇒(32)5=32n\begin{aligned} \Rightarrow & 8|\operatorname{Adj}(2 \operatorname{Adj}(2 \mathrm{~A}))| = {32^n}\\\\ \Rightarrow & 8\left|\operatorname{Adj}\left(2 \times 2^2 \operatorname{Adj}(\mathrm{A})\right)\right| = {32^n} \\\\ \Rightarrow & 8\left|\operatorname{Adj}\left(2^3 \operatorname{AdjA}\right)\right| = {32^n} \\\\ \Rightarrow & 8\left|2^6 \operatorname{Adj}(\operatorname{AdjA})\right| = {32^n} \\\\ \Rightarrow & 2^3\left(2^6\right)^3|\operatorname{Adj}(\operatorname{Adj})| = {32^n} \\\\ \Rightarrow & 2^3 \cdot 2^{18}|\mathrm{~A}|^4 = {32^n} \\\\ \Rightarrow& 2^{21} \cdot 2^4 = {32^n}\\\\ \Rightarrow& 2^{25} = {32^n}\\\\ \Rightarrow& \left(2^5\right)^5 = {32^n}\\\\ \Rightarrow& (32)^5 = {32^n} \end{aligned}
∴n=5⇒3n+α=15−4=11\begin{array}{ll} \therefore n=5 \\\\ \Rightarrow 3 n+\alpha=15-4=11 \end{array}
Q142
If the system of equations 2x+y−z=52 x+y-z=5 2x−5y+λz=μ2 x-5 y+\lambda z=\mu x+2y−5z=7x+2 y-5 z=7 has infinitely many solutions, then (λ+μ)2+(λ−μ)2(\lambda+\mu)^{2}+(\lambda-\mu)^{2} is equal to
A 916
B 912
C 920
D 904
Correct Answer
Option A
Solution
2x+y−z=52x−5y+λz=μx+2y−5z=7\begin{aligned} & 2 x+y-z=5 \\ & 2 x-5 y+\lambda z=\mu \\ & x+2 y-5 z=7 \end{aligned}

For infinite solution Δ=0=Δ1=Δ2=Δ3\Delta=0=\Delta_1=\Delta_2=\Delta_3

Δ=∣21−12−5λ12−5∣=0\Delta=\left|\begin{array}{ccc} 2 & 1 & -1 \\ 2 & -5 & \lambda \\ 1 & 2 & -5 \end{array}\right|=0
⇒2(25−2λ)−(−10−λ)−(4+5)=0⇒50−4λ+10+λ−9=0⇒51=3λ⇒λ=17\begin{aligned} \Rightarrow& 2(25-2 \lambda)-(-10-\lambda)-(4+5)=0 \\\\ \Rightarrow& 50-4 \lambda+10+\lambda-9=0 \\\\ \Rightarrow& 51=3 \lambda \Rightarrow \lambda=17 \end{aligned}
Δ3=∣521μ2−5712∣=0\Delta_3=\left|\begin{array}{ccc} 5 & 2 & 1 \\ \mu & 2 & -5 \\ 7 & 1 & 2 \end{array}\right|=0
⇒2(−35−2μ)−(14−μ)+5(4+5)=0⇒−70−4μ−14+μ+45=0⇒−3μ=39⇒−μ=13\begin{aligned} \Rightarrow & 2(-35-2 \mu)-(14-\mu)+5(4+5)=0 \\\\ \Rightarrow & -70-4 \mu-14+\mu+45=0 \\\\ \Rightarrow & -3 \mu=39 \\\\ \Rightarrow & -\mu=13 \end{aligned}

Now (λ+μ)2+(λ−μ)2(\lambda+\mu)^2+(\lambda-\mu)^2

(17+13)2+(17−13)2900+16=916\begin{aligned} & (17+13)^2+(17-13)^2 \\\\ & 900+16 \\\\ & =916 \end{aligned}
Q143
For the system of linear equations 2x+4y+2az=b2 x+4 y+2 a z=b x+2y+3z=4x+2 y+3 z=4 2x−5y+2z=82 x-5 y+2 z=8 which of the following is NOT correct?
A It has infinitely many solutions if a=3,b=8a=3, b=8
B It has infinitely many solutions if a=3,b=6a=3, b=6
C It has unique solution if a=b=8a=b=8
D It has unique solution if a=b=6a=b=6
Correct Answer
Option B
Solution

The given system of equations is : 1.

2x+4y+2az=b2x + 4y + 2az = b

2.

x+2y+3z=4x + 2y + 3z = 4

3.

2x−5y+2z=82x - 5y + 2z = 8

We can write this in matrix form :

[242a1232−52][xyz]=[b48]\begin{bmatrix} 2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & -5 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} b \\ 4 \\ 8 \end{bmatrix}

First, we need to find the determinant of the coefficient matrix, which we'll call Δ. The coefficient matrix is :

[242a1232−52]\begin{bmatrix} 2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & -5 & 2 \end{bmatrix}

We find its determinant for 3 ×\times 3 matrices :

Δ=2(2)(2)+4(3)(2)+2a(1)(−5)−2a(2)(2)−4(1)(2)−2(3)(−5)\Delta = 2(2)(2) + 4(3)(2) + 2a(1)(-5) - 2a(2)(2) - 4(1)(2) - 2(3)(-5)
Δ=8+24−10a−8a−8+30\Delta = 8 + 24 - 10a - 8a - 8 + 30
Δ=−18a+54\Delta = -18a + 54
Δ=−18(a−3)\Delta = -18(a - 3)

Next, we substitute the third column of our matrix with the column of constants (b, 4, 8) and calculate the determinant Δ₃ :

[24b1242−58]\begin{bmatrix} 2 & 4 & b \\ 1 & 2 & 4 \\ 2 & -5 & 8 \end{bmatrix}

We find its determinant :

Δ3=2(2)(8)+4(4)(2)+b(1)(−5)−b(2)(2)−4(1)(8)−2(4)(−5)\Delta_3 = 2(2)(8) + 4(4)(2) + b(1)(-5) - b(2)(2) - 4(1)(8) - 2(4)(-5)
Δ3=32+32−5b−4b−32+40\Delta_3 = 32 + 32 - 5b - 4b - 32 + 40
Δ3=−9b+72\Delta_3 = -9b + 72
Δ3=9(8−b)\Delta_3 = 9(8 - b)

Now, let's analyze the options : 1.

For a=3 and b=8, we have Δ = 0 and Δ₃ = 0, which indicates an infinite number of solutions.

2.

For a=3 and b=6, we have Δ = 0 and Δ₃ ≠ 0, which indicates no solution.

3.

For a=8 and b=8, we have Δ ≠ 0, which indicates a unique solution.

4.

For a=6 and b=6, we have Δ ≠ 0, which indicates a unique solution.

Therefore, the statement that is NOT correct is Option B: "It has infinitely many solutions if a=3, b=6", because in this case the system actually has no solution.

Q144
Let B=[13α123αα4],α>2B=\left[\begin{array}{lll}1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4\end{array}\right], \alpha > 2 be the adjoint of a matrix AA and ∣A∣=2|A|=2. Then [α−2αα]B[α−2αα]\left[\begin{array}{ccc}\alpha & -2 \alpha & \alpha\end{array}\right] B\left[\begin{array}{c}\alpha \\ -2 \alpha \\ \alpha\end{array}\right] is equal to :
A 32
B −-16
C 0
D 16
Correct Answer
Option B
Solution
B=[13α123αα4],α>2B=\left[\begin{array}{lll} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{array}\right], \alpha>2

And adj⁡(A)=B,∣A∣=2\operatorname{adj}(A)=B,|A|=2

⇒∣adj⁡(A)∣=∣B∣⇒22=(8−3α)−3(4−3α)+α(−α)⇒α2−6α+8=0\begin{aligned} & \Rightarrow|\operatorname{adj}(A)|=|B| \\\\ & \Rightarrow 2^2=(8-3 \alpha)-3(4-3 \alpha)+\alpha(-\alpha) \\\\ & \Rightarrow \alpha^2-6 \alpha+8=0 \end{aligned}
⇒(α−4)(α−2)=0α=4,2 but α>2 so α=4\begin{aligned} \Rightarrow & (\alpha-4)(\alpha-2)=0 \\\\ & \alpha=4,2 \text{ but } \alpha>2 \text{ so } \alpha=4 \end{aligned}

Now

[α−2αα]B[α−2αα]=[4−84][134123444][4−84]=[12128][4−84]=48−96+32=−16\begin{aligned} & {\left[\begin{array}{ccc}\alpha & -2 \alpha & \alpha\end{array}\right] B\left[\begin{array}{c} \alpha \\ -2 \alpha \\ \alpha \end{array}\right]=\left[\begin{array}{lll} 4-8 & 4 \end{array}\right]\left[\begin{array}{lll} 1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4 \end{array}\right]\left[\begin{array}{c} 4 \\ -8 \\ 4 \end{array}\right]} \\\\ & =\left[\begin{array}{lll} 12 & 12 & 8 \end{array}\right]\left[\begin{array}{c} 4 \\ -8 \\ 4 \end{array}\right] \\\\ & = {48-96+32=-16} \end{aligned}
Q145
The number of symmetric matrices of order 3, with all the entries from the set {0,1,2,3,4,5,6,7,8,9}\{0,1,2,3,4,5,6,7,8,9\} is :
A 10910^{9}
B 9109^{10}
C 10610^{6}
D 6106^{10}
Correct Answer
Option C
Solution

Sure!

A symmetric matrix is a square matrix that is equal to its transpose.

For a matrix to be symmetric, the element at row i and column j must be equal to the element at row j and column i.

In other words,

Aij=AjiA_{ij} = A_{ji}

. For a 3 ×\times 3 symmetric matrix, it looks like this:

(abcbdecef)\begin{pmatrix} a & b & c \\ b & d & e \\ c & e & f \\ \end{pmatrix}

Notice that there are only 6 unique elements we need to fill because of the symmetry:

aa

in the (1,1) position

bb

in the (1,2) and (2,1) positions

cc

in the (1,3) and (3,1) positions

dd

in the (2,2) position

ee

in the (2,3) and (3,2) positions

ff

in the (3,3) position Each of these unique elements can take a value from the set

0,1,2,3,4,5,6,7,8,9{0,1,2,3,4,5,6,7,8,9}

, which has 10 elements.

We have 10 choices for each of the 6 unique elements, so the total number of symmetric matrices can be calculated as:

10×10×10×10×10×10=10610 \times 10 \times 10 \times 10 \times 10 \times 10 = 10^{6}

Thus, the total number of symmetric matrices of order 3 with entries from this set is

10610^{6}

.

Q146
Let A=[115101]A=\left[\begin{array}{cc}1 & \frac{1}{51} \\ 0 & 1\end{array}\right]. If B=[12−1−1]A[−1−211]\mathrm{B}=\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right] A\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right], then the sum of all the elements of the matrix ∑n=150Bn\sum\limits_{n=1}^{50} B^{n} is equal to
A 50
B 75
C 100
D 125
Correct Answer
Option C
Solution
 Let C=[12−1−1],D=[−1−211]DC=[12−1−1][−1−211]=[1001]=I\begin{aligned} & \text{ Let } C=\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right], \mathrm{D}=\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right] \\\\ & \mathrm{DC}=\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right]\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\mathrm{I} \end{aligned}
B=CADBn=(CAD)(CAD)(CAD)…(CAD)⏟n-times \begin{aligned} & \mathrm{B}=\mathrm{CAD} \\\\ & \mathrm{B}^{\mathrm{n}}=\underbrace{(\mathrm{CAD})(\mathrm{CAD})(\mathrm{CAD}) \ldots(\mathrm{CAD})}_{\text{n-times }} \end{aligned}
⇒Bn=CAnD\Rightarrow \mathrm{B}^{\mathrm{n}}=\mathrm{CA}^{\mathrm{n}} \mathrm{D}
A2=[115101][115101]=[125101]\mathrm{A}^2=\left[\begin{array}{cc} 1 & \frac{1}{51} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & \frac{1}{51} \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & \frac{2}{51} \\ 0 & 1 \end{array}\right]
A3=[115101][125101]=[135101]\mathrm{A}^3=\left[\begin{array}{cc} 1 & \frac{1}{51} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & \frac{2}{51} \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & \frac{3}{51} \\ 0 & 1 \end{array}\right]
 Similarly An=[1n5101]\text{ Similarly } A^{\mathrm{n}}=\left[\begin{array}{cc} 1 & \frac{\mathrm{n}}{51} \\ 0 & 1 \end{array}\right]
Bn=[12−1−1][1n5101][−1−211]=[1n51+2−1−n51−1][−1−211]=[n51+1n51−n511−n51]\begin{aligned} \mathrm{B}^{\mathrm{n}} & =\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right]\left[\begin{array}{cc} 1 & \frac{\mathrm{n}}{51} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right] \\\\ & =\left[\begin{array}{cc} 1 & \frac{\mathrm{n}}{51}+2 \\ -1 & -\frac{\mathrm{n}}{51}-1 \end{array}\right]\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right] \\\\ & =\left[\begin{array}{cc} \frac{\mathrm{n}}{51}+1 & \frac{\mathrm{n}}{51} \\ -\frac{\mathrm{n}}{51} & 1-\frac{\mathrm{n}}{51} \end{array}\right] \end{aligned}
∑n=150Bn=[50+50⋅512⋅5150⋅512⋅51−50⋅512⋅5150−50⋅512⋅51]=[7525−2525]\sum\limits_{n=1}^{50} B^n=\left[\begin{array}{cc} 50+\frac{50 \cdot 51}{2 \cdot 51} & \frac{50 \cdot 51}{2 \cdot 51} \\\\ \frac{-50 \cdot 51}{2 \cdot 51} & 50-\frac{50 \cdot 51}{2 \cdot 51} \end{array}\right]=\left[\begin{array}{cc} 75 & 25 \\ -25 & 25 \end{array}\right]

∴\therefore Sum of the elements = 100

Q147
If the system of linear equations 7x+11y+αz=135x+4y+7z=β175x+194y+57z=361 \begin{aligned} & 7 x+11 y+\alpha z=13 \\\\ & 5 x+4 y+7 z=\beta \\\\ & 175 x+194 y+57 z=361 \end{aligned} has infinitely many solutions, then α+β+2\alpha+\beta+2 is equal to :
A 6
B 4
C 5
D 3
Correct Answer
Option B
Solution

Given,

7x+11y+αz=135x+4y+7z=β175x+194y+57z=361\begin{aligned} & 7 x+11 y+\alpha z=13 \\\\ & 5 x+4 y+7 z=\beta \\\\ & 175 x+194 y+57 z=361 \end{aligned}
 For infinite solution, ∣711α54717519457∣=0\text{ For infinite solution, }\left|\begin{array}{ccc} 7 & 11 & \alpha \\ 5 & 4 & 7 \\ 175 & 194 & 57 \end{array}\right|=0
⇒∣711α5470−8157−25α∣=0\Rightarrow\left|\begin{array}{ccc} 7 & 11 & \alpha \\ 5 & 4 & 7 \\ 0 & -81 & 57-25 \alpha \end{array}\right|=0
⇒81(49−5α)+(57−25α)(−27)=0⇒270α=−2430⇒α=−9\begin{aligned} & \Rightarrow 81(49-5 \alpha)+(57-25 \alpha)(-27)=0 \\\\ & \Rightarrow 270 \alpha=-2430 \Rightarrow \alpha=-9 \end{aligned}

And Δ1=0\Delta_1=0

∣1311−9β4736119457∣=0⇒β=11\begin{aligned} & \left|\begin{array}{ccc} 13 & 11 & -9 \\ \beta & 4 & 7 \\ 361 & 194 & 57 \end{array}\right|=0 \\\\ & \Rightarrow \beta=11 \end{aligned}
∴α+β+2=4\therefore \alpha+\beta+2=4
Q148
∣x+1xxxx+λxxxx+λ2∣=98(103x+81)\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^{2}\end{array}\right|=\frac{9}{8}(103 x+81), then λ,λ3\lambda, \dfrac{\lambda}{3} are the roots of the equation :
A 4x2+24x−27=04 x^{2}+24 x-27=0
B 4x2−24x+27=04 x^{2}-24 x+27=0
C 4x2−24x−27=04 x^{2}-24 x-27=0
D 4x2+24x+27=04 x^{2}+24 x+27=0
Correct Answer
Option B
Solution
∣x+1xxxx+λxxxx+λ2∣=98(103x+81)\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^{2}\end{array}\right|=\frac{9}{8}(103 x+81)

Put x=0x=0

∣1000λ000λ2∣=98×81λ3=3623\begin{aligned} & \left|\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda^2 \end{array}\right|=\frac{9}{8} \times 81 \\\\ & \lambda^3=\frac{3^6}{2^3} \end{aligned}
⇒λ=92⇒λ3=32\begin{aligned} & \Rightarrow \lambda=\frac{9}{2} \\\\ & \Rightarrow \frac{\lambda}{3}=\frac{3}{2} \end{aligned}
 Equation: x2−(92+32)x+92×32=0⇒4x2−24x+27=0\begin{aligned} & \text{ Equation: } x^2-\left(\frac{9}{2}+\frac{3}{2}\right) x+\frac{9}{2} \times \frac{3}{2}=0 \\\\ & \Rightarrow 4 x^2-24 x+27=0 \end{aligned}
Q149
Let A\mathrm{A} be a 2×22 \times 2 matrix with real entries such that A′=αA+I\mathrm{A}'=\alpha \mathrm{A}+\mathrm{I}, where α∈R−{−1,1}\alpha \in \mathbb{R}-\{-1,1\}. If det⁡(A2−A)=4\operatorname{det}\left(A^{2}-A\right)=4, then the sum of all possible values of α\alpha is equal to :
A 2
B 32\dfrac{3}{2}
C 0
D 52\dfrac{5}{2}
Correct Answer
Option D
Solution

We have, AT=αA+IA^T=\alpha A+I, where AA is 2×22 \times 2 matrix and α∈R−{−1,1}\alpha \in R-\{-1,1\}

(AT)T=αAT+IA=αAT+IA=α(αA+I)+I[∵AT=αA+I]A=α2A+(α+1)IA(1−α2)=(α+1)IA=(α+1)(1−α)(1+α)I\begin{aligned} \left(A^T\right)^T & =\alpha A^T+I \\\\ A & =\alpha A^T+I \\\\ A & =\alpha(\alpha A+I)+I \left[\because A^T=\alpha A+I\right]\\\\ A & =\alpha^2 A+(\alpha+1) I \\\\ A & \left(1-\alpha^2\right)=(\alpha+1) I \\\\ A & =\frac{(\alpha+1)}{(1-\alpha)(1+\alpha)} I \end{aligned}
A=11−αI..........(i)∣A∣=1(1−α)2...........(ii)\begin{aligned} A & =\frac{1}{1-\alpha} I ..........(i)\\\\ |A| & =\frac{1}{(1-\alpha)^2} ...........(ii) \end{aligned}

Also,

∣A2−A∣=4∣A∣∣A−I∣=41(1−α)2∣(11−α−1)I∣=41(1−α)2∣(α1−α)I∣=41(1−α)2×α2(1−α)2=4α2(1−α)4=4α(1−α)2=±22(1−α)2=±α\begin{aligned} \left|A^2-A\right| & =4 \\\\ |A||A-I| & =4 \\\\ \frac{1}{(1-\alpha)^2}\left|\left(\frac{1}{1-\alpha}-1\right) I\right| & =4 \\\\ \frac{1}{(1-\alpha)^2}\left|\left(\frac{\alpha}{1-\alpha}\right) I\right| & =4 \\\\ \frac{1}{(1-\alpha)^2} \times \frac{\alpha^2}{(1-\alpha)^2} & =4 \\\\ \frac{\alpha^2}{(1-\alpha)^4} & =4 \\\\ \frac{\alpha}{(1-\alpha)^2} & = \pm 2 \\\\ 2(1-\alpha)^2 & = \pm \alpha \end{aligned}

If 2(1−α)2=α2(1-\alpha)^2=\alpha, then 2α2−5α+2=02 \alpha^2-5 \alpha+2=0 Sum of value of α=52[∵\alpha=\dfrac{5}{2} \quad\left[\because\right. Sum of zero =-Coefficient of x Coefficient of x2]\left.=\dfrac{\text{-Coefficient of } x}{\text{ Coefficient of } x^2}\right] If 2(1−α)2=−α2(1-\alpha)^2=-\alpha, then 2α2−3α+2=02 \alpha^2-3 \alpha+2=0 ∴\therefore No real value of α\alpha Hence, sum of all values of α=52\alpha=\dfrac{5}{2}

Q150
If A=15!6!7![5!6!7!6!7!8!7!8!9!]\mathrm{A}=\frac{1}{5 ! 6 ! 7 !}\left[\begin{array}{ccc}5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 !\end{array}\right], then ∣adj⁡(adj⁡(2 A))∣|\operatorname{adj}(\operatorname{adj}(2 \mathrm{~A}))| is equal to :
A 2122^{12}
B 2202^{20}
C 282^{8}
D 2162^{16}
Correct Answer
Option D
Solution

Given that

A=15!6!7![5!6!7!6!7!8!7!8!9!]⇒∣A∣=15!6!7!∣5!6!7!6!7!8!7!8!9!∣⇒∣A∣=15!6!7!×5!6!7!∣167×6178×7189×8∣\begin{aligned} & A=\frac{1}{5 ! 6 ! 7 !}\left[\begin{array}{lll} 5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 ! \end{array}\right] \\\\ & \Rightarrow|A|=\frac{1}{5 ! 6 ! 7 !}\left|\begin{array}{lll} 5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 ! \end{array}\right| \\\\ & \Rightarrow|A|=\frac{1}{5 ! 6 ! 7 !} \times 5 ! 6 ! 7 !\left|\begin{array}{lll} 1 & 6 & 7 \times 6 \\ 1 & 7 & 8 \times 7 \\ 1 & 8 & 9 \times 8 \end{array}\right| \end{aligned}
R3→R3−R2 and R2→R2−R1∣A∣=∣164201140116∣=2∣adj⁡(adj⁡(2A))∣=∣2A∣(n−1)2=∣2A∣4=(23∣A∣)4=212∣A∣4=216\begin{aligned} & R_3 \rightarrow R_3-R_2 \text{ and } R_2 \rightarrow R_2-R_1 \\\\ & |A|=\left|\begin{array}{lll} 1 & 6 & 42 \\ 0 & 1 & 14 \\ 0 & 1 & 16 \end{array}\right|=2 \\\\ & |\operatorname{adj}(\operatorname{adj}(2 A))|=|2 A|^{(n-1)^2} \\\\ & =|2 A|^4=\left(2^3|A|\right)^4=2^{12}|A|^4=2^{16} \end{aligned}
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