JEE Main Mathematics - Determinants and Matrices
Exam Duration: 60 Mins Total Questions : 30
If \(\triangle =\left| \begin{matrix} ({ a }^{ x }+{ a }^{ -x })^{ 2 } & ({ a }^{ x }-{ a }^{ -x })^{ 2 } & 1 \\ ({ b }^{ x }{ +b }^{ -x })^{ 2 } & ({ b }^{ x }-{ b }^{ -x })^{ 2 } & 1 \\ ({ c }^{ x }+c^{ -x })^{ 2 } & ({ c }^{ x }-c^{ -x })^{ 2 } & 1 \end{matrix} \right| \) then
- (a)
\(\triangle =\)xyz
- (b)
\(\triangle =\)0
- (c)
\(\triangle =\)axbycz
- (d)
\(\triangle =\)(ax+by+cz)2
Applying C1⟶C1-C2 and using (ax+a-x)2-(ax-a-x)2=4, we get
Δ=\(\left| \begin{matrix} 4 & ({ a }^{ x }-{ a }^{ -x }) & 1 \\ 4 & ({ b }^{ y }-{ b }^{ -y })^{ 2 } & 1 \\ 4 & ({ c }^{ z }-{ c }^{ -z })^{ 2 } & 1 \end{matrix} \right| \)
=4\(\left| \begin{matrix} 1 & ({ a }^{ x }-{ a }^{ -x })^{ 2 } & 1 \\ 1 & ({ b }^{ y }-{ b }^{ -y })^{ 2 } & 1 \\ 1 & ({ (c }^{ z }-{ c }^{ -z })^{ 2 } & 1 \end{matrix} \right| \)=0
Hence, the correct alternative is (b).
If D=diagonal [d1,d2,.....dn], when d1,d2,d3,......dn \(\neq \)0 then D-1 is
- (a)
D
- (b)
diagonal[d-1,d2-2,d3-1,........dn-1]
- (c)
In
- (d)
none of these
Adj.D=\(\left[ \begin{matrix} { D }_{ 1 } \\ 0 \\ \begin{matrix} 0 \\ ... \\ \begin{matrix} ... \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} D_{ 2 } \\ 0 \\ ... \end{matrix} \\ \begin{matrix} ... \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 0 \\ 0 \\ \begin{matrix} { D }_{ 3 } \\ ... \\ \begin{matrix} ... \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \\ ... \end{matrix} \\ \begin{matrix} ... \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} ... \\ ... \\ \begin{matrix} ... \\ ... \\ \begin{matrix} ... \\ \begin{matrix} { D }_{ n-1 } \\ 0 \end{matrix} \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 0 \\ 0 \\ \begin{matrix} 0 \\ ... \\ \begin{matrix} ... \\ \begin{matrix} 0 \\ { D }_{ n } \end{matrix} \end{matrix} \end{matrix} \end{matrix} \right] \)
=\(\left[ \begin{matrix} { D }_{ 1 } \\ 0 \\ \begin{matrix} 0 \\ ... \\ \begin{matrix} ... \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} { D }_{ 2 } \\ 0 \\ ... \end{matrix} \\ \begin{matrix} ... \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 0 \\ 0 \\ \begin{matrix} { D }_{ 3 } \\ ... \\ \begin{matrix} ... \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \\ ... \end{matrix} \\ \begin{matrix} ... \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} ... \\ ... \\ \begin{matrix} ... \\ ... \\ \begin{matrix} ... \\ \begin{matrix} { D }_{ n-1 } \\ 0 \end{matrix} \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 0 \\ 0 \\ \begin{matrix} 0 \\ - \\ \begin{matrix} - \\ \begin{matrix} 0 \\ { D }_{ n } \end{matrix} \end{matrix} \end{matrix} \end{matrix} \right] \)
=A-1=\(\frac { 1 }{ { d }_{ 1 }{ d }_{ 2 }{ d }_{ 3 }...{ d }_{ n } } \)
\(\left[ \begin{matrix} { d }_{ 1 }^{ -1 } \\ 0 \\ \begin{matrix} 0 \\ - \\ \begin{matrix} - \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} { d }_{ 2 }^{ -1 } \\ 0 \\ - \end{matrix} \\ \begin{matrix} - \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 0 \\ 0 \\ \begin{matrix} { d }_{ 3 }^{ -1 } \\ - \\ \begin{matrix} - \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \\ - \end{matrix} \\ \begin{matrix} - \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} ... \\ ... \\ \begin{matrix} ... \\ ... \\ \begin{matrix} ... \\ \begin{matrix} { d }_{ n-1 }^{ -1 } \\ ... \end{matrix} \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 0 \\ 0 \\ \begin{matrix} 0 \\ - \\ \begin{matrix} - \\ \begin{matrix} 0 \\ { d }_{ n }^{ -1 } \end{matrix} \end{matrix} \end{matrix} \end{matrix} \right] \)
A-1=Diag.[\({ d }_{ 1 }^{ -1 }\quad { d }_{ 2 }^{ -1 }{ d }_{ 3 }^{ -1 }....{ d }_{ n }^{ -1 }\)]
If A=\(\begin{bmatrix} 1 & \tan { \frac { \theta }{ 2 } } \\ -\tan { \frac { \theta }{ 2 } } & 1 \end{bmatrix}\) and AB=I then B is equal to
- (a)
\(cos^{ 2 }{ \frac { \theta }{ 2 } }A^{ 2 }\)
- (b)
\(cos^{ 2 }{ \frac { \theta }{ 2 } }.A^{ T }\)
- (c)
\(cos^{ 2 }{ \frac { \theta }{ 2 } }.I+A\)
- (d)
None of these
We have, A=\(\left[ \begin{matrix} 1 & tan\frac { \theta }{ 2 } \\ -tan\frac { \theta }{ 2 } & 1 \end{matrix} \right] \)
∵ AB=I ⇒ B=A-1
Now, consider A=IA
⇒ A=IA
⇒ \(\left[ \begin{matrix} 1 & tan\frac { \theta }{ 2 } \\ -tan\frac { \theta }{ 2 } & 1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
On applying R2⟶R2+tanፀ/2R1, we get
\(\left[ \begin{matrix} 1 & tan\frac { \theta }{ 2 } \\ 0 & 1+tan^{ 2 }\frac { \theta }{ 2 } \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ tan\frac { \theta }{ 2 } & 1 \end{matrix} \right] \)
\(\left[ \begin{matrix} 1 & tan\theta /2 \\ 0 & { sec }^{ 2 }\theta /2 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ tan\theta /2 & 1 \end{matrix} \right] \)
\(\left[ \begin{matrix} 1 & tan\theta /2 \\ 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ tan\theta /2{ cos }^{ 2 }\theta /2 & cos^{ 2 }\theta /2 \end{matrix} \right] \)
⇒ \(\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 1-{ tan }^{ 2 }\theta /2{ cos }^{ 2 }\theta /2 & -tan\theta /2{ cos }^{ 2 }\theta /2 \\ tan\theta /2{ cos }^{ 2 }\theta /2 & cos^{ 2 }\theta /2 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} cos^{ 2 }\theta /2 & -tan\theta /2cos^{ 2 }\theta /2 \\ tan\theta /2cos^{ 2 }\theta /2 & cos^{ 2 }\theta /2 \end{matrix} \right] \)
=cos2ፀ/2\(\left[ \begin{matrix} 1 & -tan\theta /2 \\ tan\theta /2 & 1 \end{matrix} \right] \)A
⇒ I=A-1A
∴ B=A-1=cos2ፀ/2AT
[∵ A-1=\(\frac { 1 }{ 1+{ sec }^{ 2 }\theta /2 } \left[ \begin{matrix} 1 & -tan\theta /2 \\ tan\theta /2 & 1 \end{matrix} \right] =cos^{ 2 }\theta /2\)AT]
If \(\omega\) is an imaginary cube root of unity then the value of \(\begin{vmatrix} a & b\omega ^{ 2 } & a\omega \\ b\omega & c & b\omega ^{ 2 } \\ c\omega ^{ 2 } & a\omega & c \end{vmatrix}\) is
- (a)
\(a^3+b^3+c^2-2abc\)
- (b)
a2b+b2c
- (c)
0
- (d)
a2+b2+c2
\(\Delta =\begin{vmatrix} a(1+\omega ) & b\omega ^{ 2 } & a\omega \\ b(\omega +\omega ^{ 2 }) & c & b\omega ^{ 2 } \\ c(\omega ^{ 2 }+1) & a\omega & c \end{vmatrix}\)
=\(\begin{vmatrix} -a\omega ^{ 2 } & b\omega ^{ 2 } & a\omega \\ -b & c & b\omega ^{ 2 } \\ -c\omega & a\omega & c \end{vmatrix}\)=\(\omega ^{ 2 }.\omega \begin{vmatrix} -a & b & a\omega ^{ 2 } \\ -b & c & b\omega ^{ 2 } \\ -c & a & c\omega ^{ 2 } \end{vmatrix}\)
=\(-\omega 2\begin{vmatrix} a & b & a \\ b & c & b \\ c & a & c \end{vmatrix}\)=0
Let \(\left| \begin{matrix} x & 2 & x \\ { x }^{ 2 } & x & 6 \\ x & x & 6 \end{matrix} \right| ={ Ax }^{ 4 }+{ Bx }^{ 3 }+{ Cx }^{ 2 }+{ Dx+E }\) then the value of 5A + 4B + 3C + 2D + E is equal to
- (a)
-17
- (b)
-14
- (c)
-11
- (d)
-8
Let
\(\triangle \left( x \right) =\left| \begin{matrix} x & 2 & x \\ { x }^{ 2 } & x & 6 \\ x & x & 6 \end{matrix} \right| \Rightarrow \triangle \left( 1 \right) =0\\ \therefore \quad { \triangle }^{ \prime }\left( x \right) =\left| \begin{matrix} 1 & 0 & 1 \\ { x }^{ 2 } & x & 6 \\ x & x & 6 \end{matrix} \right| +\left| \begin{matrix} { x }^{ 2 } & 2 & x \\ 2x & 1 & 0 \\ x & x & 6 \end{matrix} \right| +\left| \begin{matrix} x & 2 & x \\ { x }^{ 2 } & x & 6 \\ 1 & 1 & 0 \end{matrix} \right| \\ { \triangle }^{ \prime }\left( 1 \right) =0+\left| \begin{matrix} 1 & 2 & 1 \\ 2 & 1 & 0 \\ 1 & 1 & 6 \end{matrix} \right| +\left| \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & 6 \\ 1 & 1 & 0 \end{matrix} \right| \\ =-17+6=-11\\ \)
\(Now,\quad \triangle \left( 1 \right) +{ \triangle }^{ \prime }\left( 1 \right) =5A+4B+3C+4D+E=-11\)
If \(f\left( x \right) =\left| \begin{matrix} { \left( 1+x \right) }^{ a } & { \left( 1+2x \right) }^{ b } & 1 \\ 1 & { \left( 1+x \right) }^{ a } & { \left( 1+2x \right) }^{ b } \\ { \left( 1+2x \right) }^{ b } & 1 & { \left( 1+x \right) }^{ a } \end{matrix} \right| ;\) a,b being positive integers, then
- (a)
constant term of f(x) is 4
- (b)
coefficient of x in f(x) is 0
- (c)
constant term in f(x) is (a - b)
- (d)
constant term in f(x) is (a + b)
Let \(\left| \begin{matrix} { \left( 1+x \right) }^{ a } & { \left( 1+2x \right) }^{ b } & 1 \\ 1 & { \left( 1+x \right) }^{ a } & { \left( 1+2x \right) }^{ b } \\ { \left( 1+2x \right) }^{ b } & 1 & { \left( 1+x \right) }^{ a } \end{matrix} \right| =A+Bx+C{ x }^{ 2 }+...\)
Put x = 0, then A=0
Differentiating both sides w.r.t.x and then put x = 0
\(\left| \begin{matrix} a & 2b & 0 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right| +\left| \begin{matrix} 1 & 1 & 1 \\ 0 & a & 2b \\ 1 & 1 & 1 \end{matrix} \right| +\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 2b & 0 & a \end{matrix} \right| =B\)
\(\therefore \quad \quad 0+0+0=B\\ \Rightarrow \quad B=0\)
Hence, constant term of f(x) = 0 and coefficient of x in f(x)=0
\({ f }_{ i }=\sum _{ i=0 }^{ 2 }{ { a }_{ ij }{ x }^{ i },j=1,2,3 } and\quad { f }_{ i }^{ \prime }and\quad { f }_{ i }^{ \prime \prime }are\quad denoted\quad by\quad \frac { { df }_{ i } }{ dx } and\quad \frac { { d }^{ 2 }{ f }_{ j } }{ { dx }^{ 2 } } respectively\quad then\quad g\left( x \right) =\left| \begin{matrix} { f }_{ 1 } & { f }_{ 2 } & { f }_{ 3 } \\ { f }_{ 1 }^{ \prime } & { f }_{ 2 }^{ \prime } & { f }_{ 3 }^{ \prime } \\ { f }_{ 1 }^{ \prime \prime } & { f }_{ 2 }^{ \prime \prime } & { f }_{ 3 }^{ \prime \prime } \end{matrix} \right| is\)
- (a)
a constant
- (b)
a linear in x
- (c)
a quadratic in x
- (d)
a cubic in x
\(\because \quad { g }_{ (x) }^{ \prime }\quad =\quad \left| \begin{matrix} { f }_{ 1 }^{ \prime } & { f }_{ 2 }^{ \prime } & { f }_{ 3 }^{ \prime } \\ { f }_{ 1 }^{ \prime } & { f }_{ 2 }^{ \prime } & { f }_{ 3 }^{ \prime } \\ { f }_{ 1 }^{ \prime \prime } & { f }_{ 2 }^{ \prime \prime } & { f }_{ 3 }^{ \prime \prime } \end{matrix} \right| +\left| \begin{matrix} { f }_{ 1 } & { f }_{ 2 } & { f }_{ 3 } \\ { f }_{ 1 }^{ \prime \prime } & { f }_{ 2 }^{ \prime \prime } & { f }_{ 3 }^{ \prime \prime } \\ { f }_{ 1 }^{ \prime \prime } & { f }_{ 2 }^{ \prime \prime } & { f }_{ 3 }^{ \prime \prime } \end{matrix} \right| +\left| \begin{matrix} { f }_{ 1 } & { f }_{ 2 } & { f }_{ 3 } \\ { f }_{ 1 }^{ \prime } & { f }_{ 2 }^{ \prime } & { f }_{ 3 }^{ \prime } \\ { f }_{ 1 }^{ \prime \prime } & { f }_{ 2 }^{ \prime \prime \prime } & { f }_{ 3 }^{ \prime \prime \prime } \end{matrix} \right| \)
= 0 + 0 + 0
\(=0\left( \because \quad { f }_{ 1 }\quad is\quad a\quad quadratic\quad function\quad \therefore \quad { f }_{ 1 }^{ \prime \prime \prime }=0 \right) \)
\(\therefore g\left( x \right) =c=Constant\)
Consider the system of equations a1x + b1y + c1z = 0 , a2x + b2 Y + c2z = 0, a3x + b3Y + c3z = 0. If \(\begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{vmatrix}=0\) then the system has
- (a)
more than two solutions
- (b)
one trivial and one non-trivial solutions
- (c)
no solution
- (d)
only trivial solution (0, 0, 0)
\(\because \begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{vmatrix}=0\)
ie, non-trivial solution.
The system of equations has more than two solutions.
The value of the determinant \(\left| \begin{matrix} 1 & { e }^{ { i\pi }/{ 3 } } & { e }^{ { i\pi }/{ 4 } } \\ { e }^{ -{ i\pi }/{ 3 } } & 1 & { e }^{ { 2i\pi }/{ 3 } } \\ { e }^{ -{ i\pi }/{ 4 } } & { e }^{ { -2i\pi }/{ 3 } } & 1 \end{matrix} \right| \) where \(i=\sqrt { -1 }\) is
- (a)
\(2+\sqrt { 2 }\)
- (b)
\(-\left( 2+\sqrt { 2 } \right)\)
- (c)
\(-2+\sqrt { 3 }\)
- (d)
\(2-\sqrt { 3 }\)
Let\(\quad \Delta =\left| \begin{matrix} 1 & { e }^{ { i\pi }/{ 3 } } & { e }^{ { i\pi }/{ 4 } } \\ { e }^{ -{ i\pi }/{ 3 } } & 1 & { e }^{ { 2i\pi }/{ 3 } } \\ { e }^{ -{ i\pi }/{ 4 } } & { e }^{ { -2i\pi }/{ 3 } } & 1 \end{matrix} \right|\)
\( \\ Taking\quad { e }^{ { -i\pi }/{ 3 } }\quad common\quad from\quad { R }_{ 2 }\\ \therefore \quad \Delta ={ e }^{ { -i\pi }/{ 3 } }\left| \begin{matrix} 1 & { e }^{ { i\pi }/{ 3 } } & { e }^{ { i\pi }/{ 4 } } \\ 1 & { e }^{ { i\pi }/{ 3 } } & { e }^{ i\pi } \\ { e }^{ -{ i\pi }/{ 4 } } & { e }^{ { -2i\pi }/{ 3 } } & 1 \end{matrix} \right|\)
\( \\ Multipling\quad { e }^{ { -i\pi }/{ 3 } }\quad in\quad { C }_{ 2 },\quad Then\)
\(\\ \Delta =\left| \begin{matrix} 1 & 1 & { e }^{ { i\pi }/{ 4 } } \\ 1 & 1 & { e }^{ i\pi } \\ { e }^{ -{ i\pi }/{ 4 } } & { e }^{ -i\pi } & 1 \end{matrix} \right| \\\)
\( =\left| \begin{matrix} 1 & 1 & \frac { 1+i }{ \sqrt { 2 } } \\ 1 & 1 & -1 \\ \frac { 1-i }{ \sqrt { 2 } } & -1 & 1 \end{matrix} \right|\)
\( \\ =1(1-1)-1(1+\frac { 1-i }{ \sqrt { 2 } } )+(\frac { 1+i }{ \sqrt { 2 } } )(-1-\frac { 1-i }{ \sqrt { 2 } } )\)
\(\\ =0-1-(\frac { 1-i }{ \sqrt { 2 } } )-(\frac { 1+i }{ \sqrt { 2 } } )-\frac { 2 }{ 2 }\)
\( \\ =-2-\sqrt { 2 } =-(2+\sqrt { 2 }\))
If the system of equations 2x - y + z = 0, x - 2y + z = 0, tx - y + 2z = 0 has infinitely many solutions and f(x) be a continuous function such that f(5 + x) + f(x) = 2, then \(\int _{ 0 }^{ -2t }{ f\left( x \right) dx } \) is equal to
- (a)
0
- (b)
-2t
- (c)
5
- (d)
t
\(For\quad infinitely\quad many\quad solutions\\ \Delta ={ \Delta }_{ 1 }={ \Delta }_{ 2 }={ \Delta }_{ 3 }=0\\ \Delta =0\Rightarrow \left| \begin{matrix} 2 & -1 & 1 \\ 1 & -2 & 1 \\ t & -1 & 2 \end{matrix} \right| =0\Rightarrow t=5\\ For\quad t=5,{ \Delta }_{ 1 }={ \Delta }_{ 2 }={ \Delta }_{ 3 }=0\\ Now,\int _{ 0 }^{ -2t }{ f\left( x \right) dx\quad =\int _{ 0 }^{ -10 }{ f\left( x \right) dx\quad = } } \int _{ 0 }^{ -5 }{ f\left( x \right) dx } +\int _{ -5 }^{ -10 }{ f\left( x \right) dx } \\ =\int _{ -5 }^{ -10 }{ f\left( x+5 \right) dx } +\int _{ -5 }^{ -10 }{ f\left( x \right) dx } \\ =\int _{ -5 }^{ -10 }{ \left\{ f\left( x+5 \right) \quad +\quad f\left( x \right) \right\} dx } =\int _{ -5 }^{ -10 }{ 2dx } =-10\\ =-2t\)
Which of the following is correct?
- (a)
If A is a square matrix, (A + A' ) is a symmetric matrix
- (b)
If A is a square matrix, (A - A') is a skew symmetric matrix
- (c)
Every square matrix can be expressed as the sum of a symmetric and skew symmetric matrix
- (d)
Some elements of the skew symmetric matrix must be zero
\(\because\) (A+A')' = A' + (A')' = A' + A
(A - A')' = A' - (A')' = A' - A = - (A - A')
and A = \(1\over2\) (A+A')+\(1\over2\) (A - A')
in skew-symmetric matrix \(\left[ \begin{matrix} 0 & -c & b \\ c & 0 & a \\ -b & -a & 0 \end{matrix} \right] \) diagonal elements
Let A, Band C be 2 x 2 matrices with entries from the set of real numbers. Define 0 as follows AoB = \(1\over2\) (AB + BA), then
- (a)
AoB=BoA
- (b)
AoI = A, I is an identity matrix of order 2
- (c)
AoA = A2
- (d)
Ao(B + C) = AoB + AoC
\(\therefore\) AoB = \(1\over2\) (AB + BA) = \(1\over2\) (BA + AB) = BoA
AoI = \(1\over2\)(AI + I A) = \(1\over2\) (A + A) = A
AoA = \(1\over2\)(A A + A A) = \(1\over2\)(A2 + A2) = A2
and Ao(B + C) = \(1\over2\) {A (B + C) + (B + C)A}
= \(1\over2\) {AB + AC + BA + CA}
= \(1\over2\) (AB + BA) + \(1\over2\) (AC + CA)
= AoB + AoC
if \(\sin { 2x=1 }\) then \({ \left| \begin{matrix} 0 & \cos { x } & -\sin { x } \\ \sin { x } & 0 & \cos { x } \\ \cos { x } & \sin { x } & 0 \end{matrix} \right| }^{ 2 }\) equals
- (a)
3
- (b)
0
- (c)
1
- (d)
none of these
\(\therefore \quad \sin { 2x=1 } \\ x={ \pi }/{ 4 }\\ Then,\quad { \left| \begin{matrix} 0 & \cos { x } & -\sin { x } \\ \sin { x } & 0 & \cos { x } \\ \cos { x } & \sin { x } & 0 \end{matrix} \right| }^{ 2 }={ \left| \begin{matrix} 0 & \frac { 1 }{ \sqrt { 2 } } & -\frac { 1 }{ \sqrt { 2 } } \\ \frac { 1 }{ \sqrt { 2 } } & 0 & \frac { 1 }{ \sqrt { 2 } } \\ \frac { 1 }{ \sqrt { 2 } } & \frac { 1 }{ \sqrt { 2 } } & 0 \end{matrix} \right| }^{ 2 }\\ ={ \left( \frac { 1 }{ \sqrt { 2 } } \times \frac { 1 }{ \sqrt { 2 } } \times \frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }{ \left| \begin{matrix} 0 & 1 & -1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right| }^{ 2 }\\ =\frac { 1 }{ 8 } { \left\{ 0-1\left( 0-1 \right) -1\left( 1 \right) \right\} }^{ 2 }\\ =0\)
The roots of the equation \(\left| \begin{matrix} { 3x }^{ 2 } & { x }^{ 2 }+x\cos { \theta } +\cos ^{ 2 }{ \theta } & { x }^{ 2 }+x\sin { \theta +\sin ^{ 2 }{ \theta } } \\ { x }^{ 2 }+x\cos { \theta } +\cos ^{ 2 }{ \theta } & 3\cos ^{ 2 }{ \theta } & 1+\frac { \sin { 2\theta } }{ 2 } \\ { x }^{ 2 }+x\sin { \theta +\sin ^{ 2 }{ \theta } } & 1+\frac { \sin { 2\theta } }{ 2 } & 3\sin ^{ 2 }{ \theta } \end{matrix} \right| =0\) are
- (a)
\(\sin { \theta } ,\cos { \theta }\)
- (b)
\(\sin ^{ 2 }{ \theta } ,\cos ^{ 2 }{ \theta }\)
- (c)
\(\sin { \theta } ,\cos ^{ 2 }{ \theta }\)
- (d)
\(\sin ^{ 2 }{ \theta } ,\cos { \theta }\)
\(\Rightarrow \left| \begin{matrix} { x }^{ 2 } & x & 1 \\ \cos ^{ 2 }{ \theta } & \cos { \theta } & 1 \\ \sin ^{ 2 }{ \theta } & \sin { \theta } & 1 \end{matrix} \right| \times \left| \begin{matrix} 1 & x & { x }^{ 2 } \\ 1 & \cos { \theta } & \cos ^{ 2 }{ \theta } \\ 1 & \sin { \theta } & \sin ^{ 2 }{ \theta } \end{matrix} \right| =0\quad \quad (row\quad by\quad row)\\ or\quad \left| \begin{matrix} 1 & x & { x }^{ 2 } \\ 1 & \cos { \theta } & \cos ^{ 2 }{ \theta } \\ 1 & \sin { \theta } & \sin ^{ 2 }{ \theta } \end{matrix} \right| =0\\ Applying\quad { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\quad and\quad { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }\quad then\\ \Rightarrow \left| \begin{matrix} 1 & x & { x }^{ 2 } \\ 0 & \cos { \theta -x } & \cos ^{ 2 }{ \theta } -{ x }^{ 2 } \\ 0 & \sin { \theta } -x & \sin ^{ 2 }{ \theta } -{ x }^{ 2 } \end{matrix} \right| \\ \Rightarrow \left( \cos { \theta -x } \right) \left( \sin { \theta } -x \right) \left\{ \sin { \theta } +x-\left( \cos { \theta +x } \right) \right\} =0\\ \therefore \quad x=\sin { \theta ,\cos { \theta } } \)
The determinant \(\left| \begin{matrix} a & b & a\alpha +b \\ b & c & b\alpha +c \\ a\alpha +b & b\alpha +c & 0 \end{matrix} \right| \) is equal to zero, if
- (a)
a, b, c are in AP
- (b)
a, b, c are in GP
- (c)
a, b, c are in HP
- (d)
\(\alpha\) is a root of ax2 + bx + c = 0
\(Applying\quad { C }_{ 3 }\rightarrow { C }_{ 3 }-\alpha { C }_{ 1 }-{ C }_{ 2 }\\ Then\quad \left| \begin{matrix} a & b & \begin{matrix} 0 \\ \vdots \end{matrix} \\ b & c & \begin{matrix} 0 \\ \vdots \end{matrix} \\ \begin{matrix} a\alpha +b & \cdots \end{matrix} & \begin{matrix} b\alpha +c & \dots \end{matrix} & \left( a{ \alpha }^{ 2 }+2b\alpha +c \right) \end{matrix} \right| =0\\ \Rightarrow \left( a{ \alpha }^{ 2 }+2b\alpha +c \right) \left( { b }^{ 2 }-ac \right) =0\\ a,b,c\quad are\quad in\quad Gp\quad and\quad a{ \alpha }^{ 2 }+2b\alpha +c=0\)
If f(x) and g(x) are functions such that f(x + y) = f(x) g(y) + g(x) f(y), then \(\left| \begin{matrix} f\left( \alpha \right) & g\left( \alpha \right) & f\left( \alpha +\theta \right) \\ f\left( \beta \right) & g\left( \beta \right) & f\left( \beta +\theta \right) \\ f\left( \gamma \right) & g\left( \gamma \right) & f\left( \gamma +\theta \right) \end{matrix} \right| \) is independent of
- (a)
\(\alpha\)
- (b)
\(\beta\)
- (c)
\(\gamma\)
- (d)
\(\theta\)
\(Now,\quad Applying\quad { C }_{ 3 }\rightarrow { C }_{ 3 }-g\left( \theta \right) { C }_{ 1 }-f\left( \theta \right) { C }_{ 2 }\\ \left| \begin{matrix} f\left( \alpha \right) & g\left( \alpha \right) & 0 \\ f\left( \beta \right) & g\left( \beta \right) & 0 \\ f\left( \gamma \right) & g\left( \gamma \right) & 0 \end{matrix} \right| =0\\ \left[ \because \quad f\left( x+y \right) =f\left( x \right) g\left( y \right) +g\left( x \right) f\left( y \right) \right] \)
If \(x\epsilon N\) and \(^{ x }{ { C }_{ i }, }^{ { x }^{ 2 } }{ { C }_{ i }, }^{ { x }^{ 3 } }{ { C }_{ i }, }(i=1,2,3)\) are binomial coefficients, then \(12\left| \begin{matrix} ^{ x }{ { C }_{ 1 } } & ^{ x }{ { C }_{ 2 } } & ^{ x }{ { C }_{ 3 } } \\ ^{ { x }^{ 2 } }{ { C }_{ 1 } } & ^{ { x }^{ 2 } }{ { C }_{ 2 } } & ^{ { x }^{ 2 } }{ { C }_{ 3 } } \\ ^{ { x }^{ 3 } }{ { C }_{ 1 } } & ^{ { x }^{ 3 } }{ { C }_{ 2 } } & ^{ { x }^{ 3 } }{ { C }_{ 3 } } \end{matrix} \right| \) is divisible by
- (a)
x3
- (b)
x6
- (c)
x9
- (d)
x12
\(12\left| \begin{matrix} x & \frac { x\left( x-1 \right) }{ 2 } & \frac { x\left( x-1 \right) \left( x-2 \right) }{ 6 } \\ { x }^{ 2 } & \frac { { x }^{ 2 }\left( { x }^{ 2 }-1 \right) }{ 2 } & \frac { { x }^{ 2 }\left( { x }^{ 2 }-1 \right) \left( { x }^{ 2 }-2 \right) }{ 6 } \\ { x }^{ 3 } & \frac { { x }^{ 3 }\left( { x }^{ 3 }-1 \right) }{ 2 } & \frac { { x }^{ 3 }\left( { x }^{ 3 }-1 \right) \left( { x }^{ 3 }-2 \right) }{ 6 } \end{matrix} \right| \\ =x.{ x }^{ 2 }.{ x }^{ 3 }\left| \begin{matrix} 1 & \left( x-1 \right) & \left( x-1 \right) \left( x-2 \right) \\ 1 & \left( { x }^{ 2 }-1 \right) & \left( { x }^{ 2 }-1 \right) \left( { x }^{ 2 }-2 \right) \\ 1 & \left( { x }^{ 3 }-1 \right) & \left( { x }^{ 3 }-1 \right) \left( { x }^{ 3 }-2 \right) \end{matrix} \right| \\ ={ x }^{ 6 }{ \left( x-1 \right) }^{ 2 }\left| \begin{matrix} 1 & 1 & x-2 \\ 1 & \left( x+1 \right) & \left( x+1 \right) \left( { x }^{ 2 }-2 \right) \\ 1 & { x }^{ 2 }+x+1 & { (x }^{ 2 }+x+1)\left( { x }^{ 3 }-2 \right) \end{matrix} \right| \\ Now,\quad applying\quad { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\quad and\quad { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }\\ ={ x }^{ 6 }{ \left( x-1 \right) }\left| \begin{matrix} \begin{matrix} 1 & \cdots \end{matrix} & 1 & x-2 \\ \begin{matrix} \vdots \\ 0 \\ \vdots \end{matrix} & x & x\left( { x }^{ 2 }+x-3 \right) \\ 1 & { x }^{ 2 }+x & x\left( { x }^{ 4 }+{ x }^{ 3 }+{ x }^{ 2 }-2x-3 \right) \end{matrix} \right| \\ ={ x }^{ 8 }\left( x-1 \right) \left\{ { x }^{ 4 }+{ x }^{ 3 }+{ x }^{ 2 }-2x-3-{ x }^{ 3 }-{ x }^{ 2 }+3x-{ x }^{ 2 }-x+3 \right\} \\ ={ x }^{ 8 }\left( x-1 \right) \left( { x }^{ 4 }-{ x }^{ 2 } \right) \\ ={ x }^{ 10 }\left( x-1 \right) \left( { x }^{ 2 }-1 \right) ={ x }^{ 10 }{ \left( x-1 \right) }^{ 2 }\left( x+1 \right) \)
Let f1 (x) = x + a, f2(x) = X2 + bx + c and \(\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ { f }_{ 1 }\left( { x }_{ 1 } \right) & { f }_{ 1 }\left( { x }_{ 2 } \right) & { f }_{ 1 }\left( { x }_{ 3 } \right) \\ { f }_{ 2 }\left( { x }_{ 1 } \right) & { f }_{ 2 }\left( { x }_{ 2 } \right) & { f }_{ 2 }\left( { x }_{ 3 } \right) \end{matrix} \right| \) then
- (a)
\(\Delta\) is independent of a
- (b)
\(\Delta\) is independent of band c
- (c)
\(\Delta\) is independent of x1 , x2, x3
- (d)
none of the above
\(Applying\quad { C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 1 }\quad and\quad { C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 },\quad then\\ \Delta =\left| \begin{matrix} 1 & \begin{matrix} \cdots & 0 & \cdots \end{matrix} & 0 \\ \begin{matrix} \vdots \\ { f }_{ 1 }\left( { x }_{ 1 } \right) \\ \vdots \end{matrix} & { f }_{ 1 }\left( { x }_{ 2 } \right) -{ f }_{ 2 }\left( { x }_{ 1 } \right) & { f }_{ 2 }\left( { x }_{ 3 } \right) -{ f }_{ 2 }\left( { x }_{ 1 } \right) \end{matrix} \right| \\ =\left| \begin{matrix} { x }_{ 2 }-{ x }_{ 1 } & { x }_{ 3 }-{ x }_{ 1 } \\ { x }_{ 2 }-{ x }_{ 1 }\left( { x }_{ 2 }-{ x }_{ 1 } \right) \left( { x }_{ 1 }+{ x }_{ 2 }+b \right) & { x }_{ 3 }-{ x }_{ 1 }\left( { x }_{ 3 }-{ x }_{ 1 } \right) \left( { x }_{ 1 }+{ x }_{ 3 }+b \right) \end{matrix} \right| \\ =\left( { x }_{ 2 }-{ x }_{ 1 } \right) \left( { x }_{ 3 }-{ x }_{ 1 } \right) \left| \begin{matrix} 1 & 1 \\ { x }_{ 1 }+{ x }_{ 2 }+b & { x }_{ 1 }+{ x }_{ 3 }+b \end{matrix} \right| \\ =\left( { x }_{ 2 }-{ x }_{ 1 } \right) \left( { x }_{ 3 }-{ x }_{ 1 } \right) \left( { x }_{ 3 }-{ x }_{ 2 } \right) \)
Let \(\Delta \neq 0\) and \({ \Delta }^{ c }\) denotes the determinant of cofactors, then \({ \Delta }^{ c }={ \Delta }^{ n-1 }\) where n (>0) is the order of \(\Delta \)
On the basis of above information, answer the question :
Suppose \(a,b,c\epsilon R,\)a + b + c > 0, A = bc - a2, B = ca _ b2 and C = ab -c2 and \(\left| \begin{matrix} A & B & C \\ B & C & A \\ C & A & B \end{matrix} \right| =49\) then \(\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right| \) equals
- (a)
-7
- (b)
7
- (c)
-2401
- (d)
2401
\(\left| \begin{matrix} A & B & C \\ B & C & A \\ C & A & B \end{matrix} \right| \\ =\left| \begin{matrix} bc-{ a }^{ 2 } & ca-{ b }^{ 2 } & ab-{ c }^{ 2 } \\ ca-{ b }^{ 2 } & ab-{ c }^{ 2 } & bc-{ a }^{ 2 } \\ ab-{ c }^{ 2 } & bc-{ a }^{ 2 } & ca-{ b }^{ 2 } \end{matrix} \right| \\ ={ \left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right| }^{ 2 }=49\\ \therefore \left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right| =7\quad \left( \because a+b+c>0 \right) \)
If A =[aij]4\(\times\)3 where aij =\(\frac { i-j }{ i+j } \) then find A
- (a)
\(\left[ \begin{matrix} 0 & -1/3 & -1/2 \\ 1/2 & 0 & 1/5 \\ 1/3 & 1/5 & 0 \\ 3/5 & 1/3 & 1/7 \end{matrix} \right] \)
- (b)
\(\left[ \begin{matrix} 0 & -1/3 & -1/2 \\ 1/3 & 0 & 1/5 \\ 1/2 & 1/5 & 0 \\ 3/5 & 1/3 & 1/7 \end{matrix} \right] \)
- (c)
\(\left[ \begin{matrix} 0 & -1/3 & -1/2 \\ 2 & 0 & 5 \\ 3 & 5 & 0 \\ 3/5 & 3 & 7 \end{matrix} \right] \)
- (d)
\(\left[ \begin{matrix} 0 & 1/3 & 1/2 \\ -1/3 & 0 & 1/5 \\ -1/2 & -1/5 & 0 \\ -3/5 & -1/3 & -1/7 \end{matrix} \right] \)
Here, a ij =\(\frac { i-j }{ i+j } \)
∴ \({ a }_{ 11 }=\frac { 1-1 }{ 1+1 } =0,{ a }_{ 12 }=\frac { 1-2 }{ 1+2 } ,a13=\frac { 1-3 }{ 1+3 } =\frac { -1 }{ 2 } \)
\({ a }_{ 21 }=\frac { 2-1 }{ 2+1 } =\frac { 1 }{ 3 } ,{ a }_{ 22 }=\frac { 2-2 }{ 2+2 } =0,{ a }_{ 23 }=\frac { 2-3 }{ 2+3 } =\frac { -1 }{ 5 } ,{ a }_{ 31 }=\frac { 3-1 }{ 3+1 } =\frac { 1 }{ 2 } ,{ a }_{ 32 }=\frac { 3-2 }{ 3+2 } =\frac { 1 }{ 5 } ,{ a }_{ 33 }=\frac { 3-3 }{ 3+3 } =0\)
\({ a }_{ 41 }=\frac { 4-1 }{ 4+1 } =\frac { 3 }{ 5 } ,{ a }_{ 42 }=\frac { 4-2 }{ 4+2 } =\frac { 1 }{ 3 } ,{ a }_{ 43 }=\frac { 4-3 }{ 4+3 } =\frac { 1 }{ 7 } \)
So, required matrix is \(\left[ \begin{matrix} 0 & -1/3 & -1/2 \\ 1/3 & 0 & 1/5 \\ 1/2 & 1/5 & 0 \\ 3/5 & 1/3 & 1/7 \end{matrix} \right] \)
Two matrices of same order are said to be equal if the _____of the two matrices are equal.
- (a)
corresponding elements
- (b)
diagonal elements
- (c)
only non-diagonal elements
- (d)
none of these
For what values of x and y are the matrices\(A=\begin{bmatrix} 2x+1 & 3y \\ 0 & { y }^{ 2 }-5y \end{bmatrix},B=\begin{bmatrix} x+3 & { y }^{ 2 }+2 \\ 0 & -6 \end{bmatrix}\) equal?
- (a)
2,3
- (b)
3,4
- (c)
2,2
- (d)
3,3
Since the corresponding elements of two equal matrices are equal, therefore
A=B⇒2x+1 =x+3⇒x=2, 3y=y2+2 y2-3y+2 = 0 ⇒ y=1,2 and y2-5y=-6 ⇒ y2=5y+6 = 0 ⇒ y= 2,3
Since 3y = y2+2 and y2-5y=-6 must hold good simultaneously so, we take the common solution of these two equations. Thereofore y =2
Hence, A = B if x=2, y=2
Evaluate the following determinant \(\left| \begin{matrix} cos\ \theta & -sin\ \theta \\ sin\ \theta & cos\ \theta \end{matrix} \right| \)
- (a)
0
- (b)
1
- (c)
2
- (d)
3
We have, \(\left| \begin{matrix} cos\ \theta & -sin\ \theta \\ sin\ \theta & cos\ \theta \end{matrix} \right| \) =cos2 \(\theta\) sin2 \(\theta\) = 1
If \(\left| \begin{matrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{matrix} \right| =k\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right| ,\) then k =
- (a)
0
- (b)
1
- (c)
2
- (d)
3
\(\left| \begin{matrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{matrix} \right| =\left| \begin{matrix} b & c & a \\ c & a & b \\ a & b & c \end{matrix} \right| +\left| \begin{matrix} c & a & b \\ a & b & c \\ b & c & a \end{matrix} \right| \)
\(=\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right| +\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right| =2\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right| \)
\(\therefore\) k = 2
If A =\(\left[ \begin{matrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{matrix} \right] \) , then AT+A=I2, If
- (a)
θ= nπ, n∈ Z
- (b)
θ= (2n+1)\(\frac { \pi }{ 2 } \) ,n∈Z
- (c)
θ=2nπ+\(\frac { \pi }{ 3 } \),n∈Z
- (d)
None of these
We have
A = \(\left[ \begin{matrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{matrix} \right] \) ⇒ AT = \(\left[ \begin{matrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{matrix} \right] \)
Now, AT+A =I2 (given)
⇒ \(\left[ \begin{matrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{matrix} \right] +\left[ \begin{matrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
⇒ \(\left[ \begin{matrix} 2cos\theta & 0 \\ 0 & 2cos\theta \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
⇒ 2 cos =1 ⇒ cos = \(\frac { 1 }{ 2 } \)
⇒ θ=2nπ+\(\frac { \pi }{ 3 } \),n∈Z
If A is any square matrix, then which of the following is a skew-symmetric
- (a)
A+AT
- (b)
A-AT
- (c)
AAT
- (d)
ATA
(A-AT)T=AT-(AT)T=AT-A=-(A-AT) Hence, (A-AT) is a skew symmetric
If A is a symmetric and n∈N, then An is
- (a)
symmetric matrix
- (b)
a diagonal matrix
- (c)
skew-symmetric matrix
- (d)
None of these
Since, A is symmetric ⇒A'=A
Now, (An)' = (A')n =An
Therefore, An is a symmetric matrix
If the equations a(y+z)=x, b(z+x)=y, c(x+y)=z have non-trivial solutions, then the value of \(\frac { 1 }{ 1+a } +\frac { 1 }{ 1+b } +\frac { 1 }{ 1+c } \) is
- (a)
1
- (b)
2
- (c)
-1
- (d)
-2
\(\left| \begin{matrix} -1 & a & a \\ b & -1 & b \\ c & c & -1 \end{matrix} \right| =0\)
Applying C2 ⟶C2-C1 and C3 ⟶C3-C1, we get
\(\left| \begin{matrix} -1 & a+1 & a+1 \\ b & -(b+1) & b \\ c & c & -(1+c) \end{matrix} \right| =0\)
Taking (a+1), (b+1) and (C+1) common from R1, R2 and R3 respectively
\(\left| \begin{matrix} -\frac { 1 }{ a+1 } & 1 & 1 \\ \frac { b }{ b+1 } & -1 & 0 \\ \frac { c }{ c+1 } & 0 & -1 \end{matrix} \right| =0\)
Expanding along C1, we get
\(-\frac { 1 }{ a+1 } +\frac { b }{ b+1 } +\frac { c }{ c+1 } =0\)
\(\Rightarrow -\frac { 1 }{ a+1 } +1-\frac { b }{ b+1 } +1-\frac { c }{ c+1 } =0\)
\(\therefore \frac { 1 }{ a+1 } +\frac { b }{ b+1 } +\frac { c }{ c+1 } =2\)
For what value of k the following system of linear equations will have infinite solutions?
x-y+z=3
2x+y-z=2
and -3x - 2ky + 6z = 3
- (a)
k\(\neq \)2
- (b)
k = 0
- (c)
k = 3
- (d)
k\(\in \)[2,3]
The given system of equations has infinite solution.
\(\therefore \left| \begin{matrix} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -3 & -2k & 6 \end{matrix} \right| =0\)
\(\Rightarrow\) 6k - 18 = 0 \(\Rightarrow\) k = 3
The area of a triangle with vertices (-3,0), (3,0) and (0, k) is 9 sq. units. The value of k will be
- (a)
9
- (b)
3
- (c)
-9
- (d)
6
Area of triangle =\(\frac { 1 }{ 2 } \left| \begin{matrix} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{matrix} \right| =\pm 9\)
\(\Rightarrow -k(-3-30=\pm 18\Rightarrow 6k=\pm 18\Rightarrow k=\pm 3\)