Practice Start Test

Matrices and Determinants

JEE Mathematics · 271 questions · Page 20 of 28 · Click an option or "Show Solution" to reveal answer

Q191
Let α,β (α≠β) \alpha, \beta \ (\alpha \neq \beta) be the values of m m , for which the equations x+y+z=1 x+y+z=1 , x+2y+4z=m x+2y+4z=m and x+4y+10z=m2 x+4y+10z=m^2 have infinitely many solutions. Then the value of ∑n=110(nα+nβ) \sum\limits_{n=1}^{10} (n^{\alpha}+n^{\beta}) is equal to :
A 3410
B 560
C 3080
D 440
Correct Answer
Option D
Solution
Δ=∣1111241410∣=1(20−16)−1(10−4)+1(4−2)=4−6+2=0 For infinite solutions Δx=Δy=Δz=0 m2−3x+2=0 m=1,2α=1,β=2∴∑n=110(nα+nβ)=∑n=110n1+∑n=110n2=10(11)2+10(11)(21)6=55+385=440\begin{aligned} &\begin{aligned} \Delta & =\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{array}\right|=1(20-16)-1(10-4)+1(4-2) \\ & =4-6+2=0 \end{aligned}\\ &\text{ For infinite solutions }\\ &\begin{aligned} & \Delta_{\mathrm{x}}=\Delta_{\mathrm{y}}=\Delta_{\mathrm{z}}=0 \\ & \mathrm{~m}^2-3 \mathrm{x}+2=0 \\ & \mathrm{~m}=1,2 \\ & \alpha=1, \beta=2 \\ & \therefore \sum_{\mathrm{n}=1}^{10}\left(\mathrm{n}^\alpha+\mathrm{n}^\beta\right)=\sum_{\mathrm{n}=1}^{10} \mathrm{n}^1+\sum_{\mathrm{n}=1}^{10} \mathrm{n}^2 \\ &= \frac{10(11)}{2}+\frac{10(11)(21)}{6} \\ &= 55+385 \\ &= 440 \end{aligned} \end{aligned}
Q192
Let A=[aij]\mathrm{A}=\left[a_{i j}\right] be a matrix of order 3×33 \times 3, with aij=(2)i+ja_{i j}=(\sqrt{2})^{i+j}. If the sum of all the elements in the third row of A2A^2 is α+β2,α,β∈Z\alpha+\beta \sqrt{2}, \alpha, \beta \in \mathbf{Z}, then α+β\alpha+\beta is equal to :
A 210
B 280
C 224
D 168
Correct Answer
Option C
Solution
A=[(2)2(2)3(2)4(2)3(2)4(2)5(2)4(2)5(2)6]A=[2224224424428]A2=22[12222222224][12222222224]\begin{aligned} & A=\left[\begin{array}{lll} (\sqrt{2})^2 & (\sqrt{2})^3 & (\sqrt{2})^4 \\ (\sqrt{2})^3 & (\sqrt{2})^4 & (\sqrt{2})^5 \\ (\sqrt{2})^4 & (\sqrt{2})^5 & (\sqrt{2})^6 \end{array}\right] \\ & A=\left[\begin{array}{ccc} 2 & 2 \sqrt{2} & 4 \\ 2 \sqrt{2} & 4 & 4 \sqrt{2} \\ 4 & 4 \sqrt{2} & 8 \end{array}\right] \\ & A^2=2^2\left[\begin{array}{ccc} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4 \end{array}\right]\left[\begin{array}{ccc} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4 \end{array}\right] \end{aligned}
=4[−−−−−−(2+4+8)(22+42+82)(4+8+16)] Sum of elements of 3rd  row =4(14+142+28)=4(42+142)=168+562α+β2∴α+β=168+56=224\begin{aligned} &=4\left[\begin{array}{ccc} - & - & - \\ - & - & - \\ (2+4+8) & (2 \sqrt{2}+4 \sqrt{2}+8 \sqrt{2}) & (4+8+16) \end{array}\right]\\ &\text{ Sum of elements of } 3^{\text{rd }} \text{ row }=4(14+14 \sqrt{2}+28)\\ &\begin{aligned} & =4(42+14 \sqrt{2}) \\ & =168+56 \sqrt{2} \\ & \alpha+\beta \sqrt{2} \\ \therefore \quad & \alpha+\beta=168+56=224 \end{aligned} \end{aligned}
Q193
Let M and m respectively be the maximum and the minimum values of f(x)=∣1+sin⁡2xcos⁡2x4sin⁡4xsin⁡2x1+cos⁡2x4sin⁡4xsin⁡2xcos⁡2x1+4sin⁡4x∣,x∈Rf(x)=\left|\begin{array}{ccc}1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & \cos ^2 x & 1+4 \sin 4 x\end{array}\right|, x \in R Then M4−m4 M^4 - m^4 is equal to :
A 1280
B 1040
C 1215
D 1295
Correct Answer
Option A
Solution
∣1+sin⁡2xcos⁡2x4sin⁡4xsin⁡2x1+cos⁡2x4sin⁡4xsin⁡2xcos⁡2x1+4sin⁡4x∣,x∈RR2→R2−R1&R3→R3−R1f(x)∣1+sin⁡2xcos⁡2x4sin⁡4x−110−101∣\begin{aligned} & \left|\begin{array}{ccc} 1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & \cos ^2 x & 1+4 \sin 4 x \end{array}\right|, x \in R \\ & R_2 \rightarrow R_2-R_1 \& R_3 \rightarrow R_3-R_1 \\ & f(x)\left|\begin{array}{ccc} 1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right| \end{aligned}

Expand about R1\mathrm{R}_1, we get

f(x)=2+4sin⁡4xf(x)=2+4 \sin 4 x

∴M=max⁡\therefore M=\max value of f(x)=6f(x)=6 m=min\mathrm{m}=\mathrm{min} value of f(x)=−2\mathrm{f}(\mathrm{x})=-2

∴M4−m4=1280\therefore \mathrm{M}^4-\mathrm{m}^4=1280
Q194
If the system of linear equations : x+y+2z=62x+3y+az=a+1−x−3y+bz=2 b\begin{aligned} & x+y+2 z=6 \\ & 2 x+3 y+\mathrm{az}=\mathrm{a}+1 \\ & -x-3 y+\mathrm{b} z=2 \mathrm{~b} \end{aligned} where a,b∈Ra, b \in \mathbf{R}, has infinitely many solutions, then 7a+3b7 a+3 b is equal to :
A 12
B 9
C 22
D 16
Correct Answer
Option D
Solution

We begin with the system:

x+y+2z=6,2x+3y+az=a+1,−x−3y+bz=2b.\begin{aligned} x+y+2z &= 6, \\ 2x+3y+az &= a+1, \\ -x-3y+bz &= 2b. \end{aligned}

Step 1. Solve the first equation for

xx

:

x=6−y−2z.x = 6 - y - 2z.

Step 2. Substitute

x=6−y−2zx = 6-y-2z

into the second equation:

2(6−y−2z)+3y+az=a+1.2(6-y-2z) + 3y + az = a+1.

Expanding and simplifying:

12−2y−4z+3y+az=a+1⟹y+(a−4)z=a−11.12 - 2y - 4z + 3y + az = a+1 \quad \Longrightarrow \quad y + (a-4)z = a - 11.

Call this Equation (I). Step 3. Substitute

x=6−y−2zx = 6-y-2z

into the third equation:

−(6−y−2z)−3y+bz=2b.-(6-y-2z) - 3y + bz = 2b.

Expanding and simplifying:

−6+y+2z−3y+bz=2b⟹−2y+(b+2)z=2b+6.-6 + y + 2z - 3y + bz = 2b \quad \Longrightarrow \quad -2y + (b+2)z = 2b + 6.

Call this Equation (II). Step 4. For the system to have infinitely many solutions, the two equations in

yy

and

zz

must be dependent—that is, one must be a constant multiple of the other. Assume there exists a constant

kk

such that

−2=k⋅1⟹k=−2.-2 = k \cdot 1 \quad \Longrightarrow \quad k = -2.

Apply this to the coefficient of

zz

and the constant term. For the

zz

-coefficient in Equations (I) and (II):

b+2=k(a−4)=−2(a−4)=−2a+8.b+2 = k(a-4) = -2(a-4) = -2a+8.

Thus,

b=−2a+6.b = -2a+6.

For the constant term:

2b+6=k(a−11)=−2(a−11)=−2a+22.2b+6 = k(a-11) = -2(a-11) = -2a + 22.

Substitute

b=−2a+6b = -2a+6

into this equation:

2(−2a+6)+6=−2a+22⟹−4a+12+6=−2a+22.2(-2a+6) + 6 = -2a + 22 \quad \Longrightarrow \quad -4a + 12 + 6 = -2a + 22.

Simplify:

−4a+18=−2a+22.-4a + 18 = -2a + 22.

Solve for

aa

:

−4a+18+4a=−2a+22+4a⟹18=2a+22,-4a + 18 + 4a = -2a + 22 + 4a \quad \Longrightarrow \quad 18 = 2a + 22,
2a=18−22=−4⟹a=−2.2a = 18 - 22 = -4 \quad \Longrightarrow \quad a = -2.

Substitute

a=−2a = -2

into

b=−2a+6b = -2a+6

:

b=−2(−2)+6=4+6=10.b = -2(-2) + 6 = 4 + 6 = 10.

Step 5. We now compute

7a+3b=7(−2)+3(10)=−14+30=16.7a + 3b = 7(-2) + 3(10) = -14 + 30 = 16.

Thus, the value of

7a+3b7a+3b

is

16.\boxed{16}.
Q195
For a 3×33 \times 3 matrix MM, let trace (M)(M) denote the sum of all the diagonal elements of MM. Let AA be a 3×33 \times 3 matrix such that ∣A∣=12|A|=\dfrac{1}{2} and trace (A)=3(A)=3. If B=adj⁡(adj⁡(2A))B=\operatorname{adj}(\operatorname{adj}(2 A)), then the value of ∣B∣+|B|+ trace (B)(B) equals :
A 56
B 132
C 174
D 280
Correct Answer
Option D
Solution
B=adj⁡(adj⁡(2A))=det⁡(2A)⋅(2A)B = \operatorname{adj}(\operatorname{adj}(2A)) = \det(2A) \cdot (2A)

Since A A is a 3×3 3 \times 3 matrix with

det⁡(A)=12,\det(A) = \frac{1}{2},

the determinant of 2A 2A is computed as

det⁡(2A)=23det⁡(A)=8⋅12=4.\det(2A) = 2^3 \det(A) = 8 \cdot \frac{1}{2} = 4.

Thus,

B=4⋅(2A)=8A.B = 4 \cdot (2A) = 8A.

Now, compute the determinant and the trace of B B : Determinant of B B :

det⁡(B)=det⁡(8A)=83det⁡(A)=512⋅12=256.\det(B) = \det(8A) = 8^3 \det(A) = 512 \cdot \frac{1}{2} = 256.

Trace of B B :

trace⁡(B)=trace⁡(8A)=8⋅trace⁡(A)=8⋅3=24.\operatorname{trace}(B) = \operatorname{trace}(8A) = 8 \cdot \operatorname{trace}(A) = 8 \cdot 3 = 24.

Finally, adding these results:

det⁡(B)+trace⁡(B)=256+24=280.\det(B) + \operatorname{trace}(B) = 256 + 24 = 280.
Q196
If the system of equations x+2y−3z=22x+λy+5z=514x+3y+μz=33 \begin{aligned} & x+2 y-3 z=2 \\ & 2 x+\lambda y+5 z=5 \\ & 14 x+3 y+\mu z=33 \end{aligned} has infinitely many solutions, then λ+μ\lambda+\mu is equal to :
A 13
B 10
C 12
D 11
Correct Answer
Option C
Solution
\begin{aligned} &\begin{aligned} & \mathrm{D}=\left|\begin{array}{rrr} 1 & 2 & -3 \\ 2 & \lambda & 5 \\ 14 & 3 & \mu \end{array}\right|=0, \lambda \mu+42 \lambda-4 \mu+107=0 \\ & \mathrm{D}_1=2 \lambda \mu+99 \lambda-10 \mu+255 \\ & \mathrm{D}_2=13-\mu \\ & D_3=5 \lambda+5 \\ & D_2=0 \Rightarrow \mu=13 \& D_3=0 \Rightarrow \lambda=-1 \end{aligned}\\ &\text{ check & verify for these values } \mathrm{D} \& \mathrm{D}_2=0 \end{aligned}
Q197
Let A=[12−201]\mathrm{A}=\left[\begin{array}{cc}\dfrac{1}{\sqrt{2}} & -2 \\ 0 & 1\end{array}\right] and P=[cos⁡θ−sin⁡θsin⁡θcos⁡θ],θ>0\mathrm{P}=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right], \theta>0. If B=PAP⊤,C=P⊤B10P\mathrm{B}=\mathrm{PAP}{ }^{\top}, \mathrm{C}=\mathrm{P}^{\top} \mathrm{B}^{10} \mathrm{P} and the sum of the diagonal elements of CC is mn\dfrac{m}{n}, where gcd⁡(m,n)=1\operatorname{gcd}(m, n)=1, then m+nm+n is :
A 127
B 2049
C 258
D 65
Correct Answer
Option D
Solution
P=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]∵PTP=I B=PAPT\begin{aligned} & \mathrm{P}=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ & \because \mathrm{P}^{\mathrm{T}} \mathrm{P}=\mathrm{I} \\ & \mathrm{~B}=\mathrm{PAPT} \end{aligned}

Pre multiply by PT\mathrm{P}^{\mathrm{T}} ( Given)

PT B=PTPAPT=APT\mathrm{P}^{\mathrm{T}} \mathrm{~B}=\mathrm{P}^{\mathrm{T}} \mathrm{PA} \mathrm{P}^{\mathrm{T}}=\mathrm{AP}^{\mathrm{T}}

Now post multiply by P

PTBP=APTP=A\mathrm{P}^{\mathrm{T}} \mathrm{BP}=\mathrm{AP}^{\mathrm{T}} \mathrm{P}=\mathrm{A}
A2=PT B2P\mathrm{A}^2=\mathrm{P}^{\mathrm{T}} \mathrm{~B}^2 \mathrm{P}

Similarly A10=PTB10P=CA^{10}=P^T B^{10} P=C

A=[12−201] (Given) ⇒A2=[12−2−201]\begin{aligned} & A=\left[\begin{array}{cc} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{array}\right] \text{ (Given) } \\ & \Rightarrow A^2=\left[\begin{array}{cc} \frac{1}{2} & -\sqrt{2}-2 \\ 0 & 1 \end{array}\right] \end{aligned}

Similarly check A3\mathrm{A}^3 and so on since C=A10\mathrm{C}=\mathrm{A}^{10} ⇒\Rightarrow Sum of diagonal elements of C is (12)10+1\left(\dfrac{1}{\sqrt{2}}\right)^{10}+1

=132+1=3332=mn g cd( m,n)=1 (Given) ⇒m+n=65\begin{aligned} & =\frac{1}{32}+1=\frac{33}{32}=\frac{\mathrm{m}}{\mathrm{n}} \\ & \mathrm{~g} \mathrm{~cd}(\mathrm{~m}, \mathrm{n})=1 \text{ (Given) } \\ & \Rightarrow \mathrm{m}+\mathrm{n}=65 \end{aligned}
Q198
If the system of equations (λ−1)x+(λ−4)y+λz=5λx+(λ−1)y+(λ−4)z=7(λ+1)x+(λ+2)y−(λ+2)z=9 \begin{aligned} & (\lambda-1) x+(\lambda-4) y+\lambda z=5 \\ & \lambda x+(\lambda-1) y+(\lambda-4) z=7 \\ & (\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9 \end{aligned} has infinitely many solutions, then λ2+λ\lambda^2+\lambda is equal to
A 20
B 10
C 6
D 12
Correct Answer
Option D
Solution
(λ−1)x+(λ−4)y+λz=5λx+(λ−1)y+(λ−4)z=7(λ+1)x+(λ+2)y−(λ+2)z=9 For infinitely many solutions D=∣λ−1λ−4λλλ−1λ−4λ+1λ+2−(λ+2)∣=0(λ−3)(2λ+1)=0Dx=∣5λ−4λ7λ−1λ−49λ+2−(λ+2)∣=02(3−λ)(23−2λ)=0λ=3∴λ2+λ=9+3=12\begin{aligned} &\begin{aligned} & (\lambda-1) x+(\lambda-4) y+\lambda z=5 \\ & \lambda x+(\lambda-1) y+(\lambda-4) z=7 \\ & (\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9 \end{aligned}\\ &\text{ For infinitely many solutions }\\ &\begin{aligned} & \mathrm{D}=\left|\begin{array}{ccc} \lambda-1 & \lambda-4 & \lambda \\ \lambda & \lambda-1 & \lambda-4 \\ \lambda+1 & \lambda+2 & -(\lambda+2) \end{array}\right|=0 \\ & (\lambda-3)(2 \lambda+1)=0 \\ & \mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ccc} 5 & \lambda-4 & \lambda \\ 7 & \lambda-1 & \lambda-4 \\ 9 & \lambda+2 & -(\lambda+2) \end{array}\right|=0 \\ & 2(3-\lambda)(23-2 \lambda)=0 \\ & \lambda=3 \\ & \therefore \lambda^2+\lambda=9+3=12 \end{aligned} \end{aligned}
Q199
If A,B,and⁡(adj⁡(A−1)+adj⁡(B−1))\mathrm{A}, \mathrm{B}, \operatorname{and}\left(\operatorname{adj}\left(\mathrm{A}^{-1}\right)+\operatorname{adj}\left(\mathrm{B}^{-1}\right)\right) are non-singular matrices of same order, then the inverse of A(adj⁡(A−1)+adj⁡(B−1))−1BA\left(\operatorname{adj}\left(A^{-1}\right)+\operatorname{adj}\left(B^{-1}\right)\right)^{-1} B, is equal to
A AB−1∣A∣+BA−1∣B∣\dfrac{A B^{-1}}{|A|}+\dfrac{B A^{-1}}{|B|}
B adj⁡(B−1)+adj⁡(A−1)\operatorname{adj}\left(\mathrm{B}^{-1}\right)+\operatorname{adj}\left(\mathrm{A}^{-1}\right)
C AB−1+A−1 B\mathrm{AB}^{-1}+\mathrm{A}^{-1} \mathrm{~B}
D 1∣AB∣(adj⁡(B)+adj⁡(A))\dfrac{1}{|A B|}(\operatorname{adj}(B)+\operatorname{adj}(A))
Correct Answer
Option D
Solution
[A(adj⁡( A−1)+adj⁡(B−1))−1⋅ B]−1 B−1⋅(adj⁡( A−1)+adj⁡(B−1))⋅A−1 B−1adj⁡( A−1)A−1+B−1(adj⁡( B−1))⋅A−1 B−1∣ A−1∣I+∣B−1∣IA−1 B−1∣ A∣+A−1∣ B∣⇒adjB⁡∣B∣∣A∣+adj⁡∣A∣∣B∣=1∣ A∣∣B∣(adjB⁡+adjA⁡)\begin{aligned} & {\left[\mathrm{A}\left(\operatorname{adj}\left(\mathrm{~A}^{-1}\right)+\operatorname{adj}\left(\mathrm{B}^{-1}\right)\right)^{-1} \cdot \mathrm{~B}\right]^{-1}} \\ & \mathrm{~B}^{-1} \cdot\left(\operatorname{adj}\left(\mathrm{~A}^{-1}\right)+\operatorname{adj}\left(\mathrm{B}^{-1}\right)\right) \cdot \mathrm{A}^{-1} \\ & \mathrm{~B}^{-1} \operatorname{adj}\left(\mathrm{~A}^{-1}\right) \mathrm{A}^{-1}+\mathrm{B}^{-1}\left(\operatorname{adj}\left(\mathrm{~B}^{-1}\right)\right) \cdot \mathrm{A}^{-1} \\ & \mathrm{~B}^{-1}\left|\mathrm{~A}^{-1}\right| \mathrm{I}+\left|\mathrm{B}^{-1}\right| \mathrm{IA}^{-1} \\ & \frac{\mathrm{~B}^{-1}}{|\mathrm{~A}|}+\frac{\mathrm{A}^{-1}}{|\mathrm{~B}|} \\ & \Rightarrow \frac{\operatorname{adjB}}{|\mathrm{B}||\mathrm{A}|}+\frac{\operatorname{adj}}{|\mathrm{A}||\mathrm{B}|} \\ & =\frac{1}{|\mathrm{~A}||\mathrm{B}|}(\operatorname{adjB}+\operatorname{adjA}) \end{aligned}
Q200
The system of equations x+y+z=6,x+2y+5z=9,x+5y+λz=μ,\begin{aligned} & x+y+z=6, \\ & x+2 y+5 z=9, \\ & x+5 y+\lambda z=\mu, \end{aligned} has no solution if
A λ=17,μ=18\lambda=17, \mu=18
B λ=17,μ≠18\lambda=17, \mu \neq 18
C λ=15,μ≠17\lambda=15, \mu \neq 17
D λ≠17,μ≠18\lambda \neq 17, \mu \neq 18
Correct Answer
Option B
Solution
D=∣11112515λ∣=0λ=17Dz=∣11612915μ∣≠0μ≠18\begin{aligned} & \mathrm{D}=\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 1 & 5 & \lambda \end{array}\right|=0 \\ & \lambda=17 \\ & D_z=\left|\begin{array}{lll} 1 & 1 & 6 \\ 1 & 2 & 9 \\ 1 & 5 & \mu \end{array}\right| \neq 0 \\ & \mu \neq 18 \end{aligned}
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