Mathematics - Determinants and Matrices
Exam Duration: 45 Mins Total Questions : 30
If the equations x=ay+z, y=z+ax, az=x+y are consistent having a non-zero solution, the value of a2+b2+c2 is
- (a)
1
- (b)
2
- (c)
3
- (d)
none of these
The value of determinant
\(\left| \begin{matrix} 0 & a-b & a-c \\ b-a & 0 & b-c \\ c-a & c-b & 0 \end{matrix} \right| \) is
- (a)
abc
- (b)
a+b+c
- (c)
ab+bc+ca
- (d)
0
Let the three digit numbers A28,3B9,62C, where A,B,C are integers between 0 and 9, be divisible by a fixed integer k; then the determinant
\(\left| \begin{matrix} A & 3 & 6 \\ 8 & 9 & C \\ 2 & B & 2 \end{matrix} \right| \) is divisible by
- (a)
k+1
- (b)
k-1
- (c)
k
- (d)
k2
If the system of linear equations
x+2ay+az=0
x+3by+bz=0
x+4cy+cz=0
has a non -zero solution, then a,b,c
- (a)
are in A.P.
- (b)
are in G.P.
- (c)
are in H.P.
- (d)
satisfy a+2b+3c=0
If a,b,c are in A.P. then value of
\(\triangle =\left| \begin{matrix} 4 & 5 & 6\quad a \\ 5 & 6 & 7\quad b \\ 6 & 7 & 8\quad c \\ x & y & z\quad 0 \end{matrix} \right| \) is
- (a)
0
- (b)
1
- (c)
2
- (d)
5
A square matrix can always be expressed as
- (a)
sum of diagonal matrix and a symmetric matrix
- (b)
difference of symmetric matrix and a skew-symmetric matrix.
- (c)
sum of symmetric matrix and a skew-symmetric matrix.
- (d)
none of these
If A is a square matrix such then AAt=I,then A is a
- (a)
symmetric matrix
- (b)
skew-symmetric matrix
- (c)
diagonal matrix
- (d)
none of these
If B is a symmetric matrix, then AB At is a
- (a)
symmetric matrix
- (b)
skew-symmetric matrix
- (c)
diagonal matrix
- (d)
scalar matrix
If \(F(\alpha )=\begin{bmatrix} cos{ \alpha } & -sin{ \alpha } & 0 \\ sin{ \alpha } & cos{ \alpha } & 0 \\ 0 & 0 & 1 \end{bmatrix}\), then the value of {F(\(\alpha\))}-1 is equal to
- (a)
F(\(-\alpha\))
- (b)
\(F(\alpha^{-1})\)
- (c)
\(F(2\alpha)\)
- (d)
None of these
If A=\(\begin{bmatrix} ab & b^{ 2 } \\ -a^{ 2 } & -ab \end{bmatrix}\) and An=0, then the minimum values of n are
- (a)
2
- (b)
3
- (c)
4
- (d)
5
If a and B are square matrices of size nXn such that A2-B2=(A-B)(A+B), then which of the following will be always true?
- (a)
AB=BA
- (b)
Either of A or B is a zero matrix
- (c)
Either of A or B is an identity matrix
- (d)
A=B
If A2-A+I=0 then the inverse of A is
- (a)
I-A
- (b)
A-I
- (c)
A
- (d)
A+I
If f(y)=\(\begin{vmatrix} a & -1 & 0 \\ ay & a & -1 \\ ay^{ 2 } & ay & a \end{vmatrix}\) then f(2y)-f(y) is divisible by
- (a)
y3
- (b)
0
- (c)
2a+3y
- (d)
y2
Let A=\(\begin{bmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{bmatrix}\) If \(\begin{vmatrix} A^{ 2 } \end{vmatrix}=25\) then \(\begin{vmatrix} \alpha \end{vmatrix}\) equals
- (a)
52
- (b)
1
- (c)
\(1\over 5\)
- (d)
5
What is wrong in the following computation?
\({ \left[ \begin{matrix} 1 & 0.01 \\ 1 & 1 \end{matrix} \right] }^{ n }={ \left\{ \left[ \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right] +{ 10 }^{ -2 }\left[ \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right] \right\} }^{ n }\)
\(={ \left[ \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right] }^{ n }+n\times { 10 }^{ -2 }{ \left[ \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right] }^{ n-1 }{ \left[ \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right] }\)
\(Since,\quad { \left[ \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right] }^{ k }={ \left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] ,fork\ge }2\quad \)
- (a)
\(Since{ \left[ \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right] }^{ k }={ \left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] },k\ge 2,isnottrue\)
- (b)
Computation of second term on RHS is not valid
- (c)
First term should be calculated completely
- (d)
None of the above
The equations \(\lambda x-y=2,2x-3y=-\lambda ,3x-2y=-1\) are consistent for
- (a)
\(\lambda =-4\)
- (b)
\(\lambda =-1,4\)
- (c)
\(\lambda =-1\)
- (d)
\(\lambda =1,-4\)
\({ f }_{ i }=\sum _{ i=0 }^{ 2 }{ { a }_{ ij }{ x }^{ i },j=1,2,3 } and\quad { f }_{ i }^{ \prime }and\quad { f }_{ i }^{ \prime \prime }are\quad denoted\quad by\quad \frac { { df }_{ i } }{ dx } and\quad \frac { { d }^{ 2 }{ f }_{ j } }{ { dx }^{ 2 } } respectively\quad then\quad g\left( x \right) =\left| \begin{matrix} { f }_{ 1 } & { f }_{ 2 } & { f }_{ 3 } \\ { f }_{ 1 }^{ \prime } & { f }_{ 2 }^{ \prime } & { f }_{ 3 }^{ \prime } \\ { f }_{ 1 }^{ \prime \prime } & { f }_{ 2 }^{ \prime \prime } & { f }_{ 3 }^{ \prime \prime } \end{matrix} \right| is\)
- (a)
a constant
- (b)
a linear in x
- (c)
a quadratic in x
- (d)
a cubic in x
With 1,\(\omega,\omega^2\) as cube roots of unity,inverse of following matrices exists?
- (a)
\(\left[ \begin{matrix} 1 & \omega \\ \omega & { \omega }^{ 2 } \end{matrix} \right] \)
- (b)
\(\left[ \begin{matrix} { \omega }^{ 2 } & 1 \\ 1 & \omega \end{matrix} \right] \)
- (c)
\(\left[ \begin{matrix} \omega & { \omega }^{ 2 } \\ { \omega }^{ 2 } & 1 \end{matrix} \right] \)
- (d)
none of the above
If A = \(\left[ \begin{matrix} a & 0 & 0 \\ 0 & b & c \\ 0 & 0 & C \end{matrix} \right] \), Then A-1 is equal to
- (a)
\(\left[ \begin{matrix} a & 0 & 0 \\ 0 & b & c \\ 0 & 0 & C \end{matrix} \right] \)
- (b)
\(\left[ \begin{matrix} a^2 & 0 & 0 \\ 0 & ab & 0 \\ 0 & 0 & ac \end{matrix} \right] \)
- (c)
\(\left[ \begin{matrix} 1/a & 0 & 0 \\ 0 & 1/b & 0 \\ 0 & 0 & 1/C \end{matrix} \right] \)
- (d)
\(\left[ \begin{matrix} -a & 0 & 0 \\ 0 & -b & 0 \\ 0 & 0 & -C \end{matrix} \right] \)
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)
TIlt' values of ex for which the system of equations x + y + z = 1, x + 2y + 4z = \(\alpha\), x + 4y + 10z = \(\alpha\) 2 consistent, are given by
- (a)
1, 2
- (b)
-1, 2
- (c)
1, -2
- (d)
-1, -2
The coefficient of x in the determinant \(\left| \begin{matrix} { \left( 1+x \right) }^{ { a }_{ 1 }{ b }_{ 1 } } & { \left( 1+x \right) }^{ { a }_{ 1 }{ b }_{ 2 } } & { \left( 1+x \right) }^{ { a }_{ 1 }{ b }_{ 3 } } \\ { \left( 1+x \right) }^{ { a }_{ 2 }{ b }_{ 1 } } & { \left( 1+x \right) }^{ { a }_{ 2 }{ b }_{ 2 } } & { \left( 1+x \right) }^{ { a }_{ 2 }{ b }_{ 3 } } \\ { \left( 1+x \right) }^{ { a }_{ 3 }{ b }_{ 1 } } & { \left( 1+x \right) }^{ { a }_{ 3 }{ b }_{ 2 } } & { \left( 1+x \right) }^{ { a }_{ 3 }{ b }_{ 3 } } \end{matrix} \right| \) is
- (a)
4
- (b)
2
- (c)
0
- (d)
-2
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
The determinant \(\Delta =\left| \begin{matrix} b & c & b\lambda +c \\ c & d & c\lambda +d \\ b\lambda +c & c\lambda +d & a{ \lambda }^{ 3 }+3c\lambda \end{matrix} \right| \) is equal to zero, if
- (a)
b, c, d are in AP
- (b)
b, c, d are in GP
- (c)
b, c, d are in HP
- (d)
\(\lambda\) is a root of ax3- bx2 + cx - d = 0
If matrix A =\(\left[ \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right] \)where a,b,c are real positive numbers, abc =1 and ATA =I, then the value of a3+b3+c3 is
- (a)
1
- (b)
2
- (c)
3
- (d)
4
If A =\(\left[ \begin{matrix} 2 & 0 & 0 \\ 2 & 2 & 0 \\ 2 & 2 & 2 \end{matrix} \right] \), then adj (adj A) is equal to
- (a)
\(8\left[ \begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{matrix} \right] \)
- (b)
\(8\left[ \begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{matrix} \right] \)
- (c)
\(16\left[ \begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{matrix} \right] \)
- (d)
none of these
\(\left| \begin{matrix} z & x & y \\ { z }^{ 2 } & { x }^{ 2 } & { y }^{ 2 } \\ { z }^{ 4 } & { x }^{ 4 } & { y }^{ 4 } \end{matrix} \right| =\)
- (a)
\(\left| \begin{matrix} x & y & z \\ { x }^{ 2 } & { y }^{ 2 } & { z }^{ 2 } \\ { x }^{ 4 } & { y }^{ 4 } & { z }^{ 4 } \end{matrix} \right| \)
- (b)
\(\left| \begin{matrix} { x }^{ 2 } & { y }^{ 2 } & { z }^{ 2 } \\ { x }^{ 4 } & { y }^{ 4 } & { z }^{ 4 } \\ x & y & z \end{matrix} \right| \)
- (c)
xyz(x-y)(y-z)(z-x)(x+y+z)
- (d)
All of these
If the points (a1, b1), (a2, b2) and (a1+a2, b1+b2) are collinear, then
- (a)
a1b2 = a2b1
- (b)
a1+a2 = b1 + b2
- (c)
a2b2 = a1b1
- (d)
a1+b1 = a2 + b2
The value of determinant\(\left| \begin{matrix} a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c \end{matrix} \right| \) is
- (a)
a3 + b3 + c3
- (b)
3bc
- (c)
a3 + b3 + c3 -3abc
- (d)
none of these
An amount of Rs.5000 is put into three investments of 6%, 7% and 85 per annum respectively. The total annual income from these investments is Rs.358. If the total annual income from first two investments is Rs.70 more than the income from the third, find the amount of each investment by the matrix method.
- (a)
Rs.500,Rs.2200, Rs.1800
- (b)
Rs.1000,Rs.2200, Rs.1800
- (c)
Rs.1200,Rs.1400,Rs.2000
- (d)
none of these