Practice Start Test

Matrices and Determinants

JEE Mathematics · 271 questions · Page 2 of 28 · Click an option or "Show Solution" to reveal answer

Q11
Let AA be a 2×2a\,2 \times 2 matrix with real entries. Let II be the 2×22 \times 2 identity matrix. Denote by tr(A)(A), the sum of diagonal entries of aa. Assume that a2=I.{a^2} = I. Statement-1 : If A≠IA \ne I and A≠−IA \ne - I, then det(A)=−1(A)=-1 Statement- 2 : If A≠IA \ne I and A≠−IA \ne - I, then tr (A)(A) ≠0 \ne 0.
A statement - 1 is false, statement -2 is true
B statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1.
C statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1.
D statement - 1 is true, statement - 2 is false.
Correct Answer
Option D
Solution

Let

A=[abcd]A = \left[ \begin{array}{ll}a & b \\ c & d \end{array} \right]
   \,\,\,

then

A2=1{A^2} = 1
⇒a2+bc=1    ab+bd=0\Rightarrow {a^2} + bc = 1\,\,\,\,ab + bd = 0
ac+cd=0    bc+d2=1ac + cd = 0\,\,\,\,bc + {d^2} = 1

From these four relations,

a2+bc=bc+d2⇒a2=d2{a^2} + bc = bc + {d^2} \Rightarrow {a^2} = {d^2}

and

  b(a+d)=0=c(a+d)\,\,b\left( {a + d} \right) = 0 = c\left( {a + d} \right)
              ⇒a=−d\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a = - d

We can take

a=1,b=0,c=0,d=−1a = 1,b = 0,c = 0,d = - 1

as one possible set of values, then

A=[100−1]A = \left[ \begin{array}{ll}1 & 0 \\ 0 & { - 1} \end{array} \right]

Clearly

A≠I   A \ne I\,\,\,

and

    A≠−I  \,\,\,\,A \ne - I\,\,

and

   A=−1\,\,\,A = - 1

∴\therefore

     \,\,\,\,\,

Statement

11

is true. Also if

A≠I     tr(A)=0A \ne I\,\,\,\,\,tr\left( A \right) = 0

∴\therefore

     \,\,\,\,\,

Statement

22

is false.

Q12
Let AA be a  2×2\,2 \times 2 matrix with non-zero entries and let A2=I,{A^2} = I, where II is 2×22 \times 2 identity matrix. Define TrTr(A)=(A)= sum of diagonal elements of AA and ∣A∣=\left| A \right| = determinant of matrix AA. Statement- 1: TrTr(A)=0(A)=0. Statement- 2: ∣A∣=1\left| A \right| = 1 .
A statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1.
B statement - 1 is true, statement - 2 is false.
C statement - 1 is false, statement -2 is true
D statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1.
Correct Answer
Option B
Solution

Let

A=(abcd)A = \left( \begin{array}{ll}a & b \\ c & d \end{array} \right)

where

a,b,c,da,b,c,d
≠0\ne 0
A2=(abcd)(abcd){A^2} = \left( \begin{array}{ll}a & b \\ c & d \end{array} \right)\left( \begin{array}{ll}a & b \\ c & d \end{array} \right)
⇒A2=(a2+bcab+bdac+cdbc+d2)\Rightarrow {A^2} = \left( \begin{array}{ll}{{a^2} + bc} & {ab + bd} \\ {ac + cd} & {bc + {d^2}} \end{array} \right)
⇒a2+bc=1, bc+d2=1\Rightarrow {a^2} + bc = 1,\,bc + {d^2} = 1
ab+bd=ac+cd=0ab + bd = ac + cd = 0
c≠0     b≠0c \ne 0\,\,\,\,\,b \ne 0
⇒a+d=0⇒Tr(A)=0\Rightarrow a + d = 0 \Rightarrow Tr\left( A \right) = 0
∣A∣=ad−bc=−a2−bc=−1\left| A \right| = ad - bc = - {a^2} - bc = - 1
Q13
Consider the system of linear equations; x1+2x2+x3=32x1+3x2+x3=33x1+5x2+2x3=1\begin{array}{ll}{{x_1} + 2{x_2} + {x_3} = 3} \\ {2{x_1} + 3{x_2} + {x_3} = 3} \\ {3{x_1} + 5{x_2} + 2{x_3} = 1} \end{array} $ The system has :
A exactly 33 solutions
B a unique solution
C no solution
D infinitenumber of solutions
Correct Answer
Option C
Solution
D=∣121231352∣=0D = \left| \begin{array}{lll}1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & 5 & 2 \end{array} \right| = 0
D1∣321331152∣≠0{D_1}\left| \begin{array}{lll}3 & 2 & 1 \\ 3 & 3 & 1 \\ 1 & 5 & 2 \end{array} \right| \ne 0

⇒\Rightarrow Given system, does not have any solution. ⇒\Rightarrow No solution

Q14
The number of 3×33 \times 3 non-singular matrices, with four entries as 11 and all other entries as 00, is :
A 55
B 66
C at least 77
D less than 44
Correct Answer
Option C
Solution
[1.........1.........1]  \left[ \begin{array}{lll}1 & {...} & {...} \\ {...} & 1 & {...} \\ {...} & {...} & 1 \end{array} \right]\,\,

are

66

non-singular matrices because

66

blanks will be filled by

55

zeros and

11

one. Similarly,

[......1...1...1......]  \left[ \begin{array}{lll}{...} & {...} & 1 \\ {...} & 1 & {...} \\ 1 & {...} & {...} \end{array} \right]\,\,

are

66

non-singular matrices. So, required cases are more than

7,7,

non-singular

3×33 \times 3

matrices.

Q15
Let AA and BB be two symmetric matrices of order 33. Statement - 1 : A(BA)A(BA) and (AB)(AB)AA are symmetric matrices. Statement - 2 : ABAB is symmetric matrix if matrix multiplication of AA with BB is commutative.
A statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1.
B statement - 1 is true, statement - 2 is false.
C statement - 1 is false, statement -2 is true
D statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1.
Correct Answer
Option A
Solution

∴\therefore

A′=A,B′=BA' = A,B' = B

Now

   (A(BA))′=(BA)′A′\,\,\,\left( {A\left( {BA} \right)} \right)' = \left( {BA} \right)'A'
=(A′B′)A′=(AB)A=A(BA)= \left( {A'B'} \right)A' = \left( {AB} \right)A = A\left( {BA} \right)

Similarly

((AB)A)′=(AB)A\left( {\left( {AB} \right)A} \right)' = \left( {AB} \right)A

So,

A(BA)    A\left( {BA} \right)\,\,\,\,

and

A(BA)    A\left( {BA} \right)\,\,\,\,

are symmetric matrices. Again

(AB)′=B′A′=BA\left( {AB} \right)' = B'A' = BA

Now if

BA=ABBA=AB

, then

ABAB

is symmetric matrix.

Q16
Let A=(100210321)A = \left( \begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array} \right). If u1{u_1} and u2{u_2} are column matrices such that Au1=(100)A{u_1} = \left( \begin{array}{ll}1 \\ 0 \\ 0 \end{array} \right) and Au2=(010),A{u_2} = \left( \begin{array}{ll}0 \\ 1 \\ 0 \end{array} \right), then u1+u2{u_1} + {u_2} is equal to :
A (−110)\left( \begin{array}{ll}-1 \\ 1 \\ 0 \end{array} \right)
B (−11−1)\left( \begin{array}{ll}-1 \\ 1 \\ -1 \end{array} \right)
C (−1−10)\left( \begin{array}{ll}-1 \\ -1 \\ 0 \end{array} \right)
D (1−1−1)\left( \begin{array}{ll}1 \\ -1 \\ -1 \end{array} \right)
Correct Answer
Option D
Solution

Let

Au1=(100)      Au2=(010)A{u_1} = \left( \begin{array}{ll}1 \\ 0 \\ 0 \end{array} \right)\,\,\,\,\,\,A{u_2} = \left( \begin{array}{ll}0 \\ 1 \\ 0 \end{array} \right)

Then,

Au1+Au2=(100)+(010)A{u_1} + A{u_2} = \left( \begin{array}{ll}1 \\ 0 \\ 0 \end{array} \right) + \left( \begin{array}{ll}0 \\ 1 \\ 0 \end{array} \right)
⇒A(u1+u2)=(110)                ...(1)\Rightarrow A\left( {{u_1} + {u_2}} \right) = \left( \begin{array}{ll}1 \\ 1 \\ 0 \end{array} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

Also,

A=(100210321)A = \left( \begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array} \right)
⇒∣A∣=1(1)−0(2)+0(4−3)=1\Rightarrow \left| A \right| = 1\left( 1 \right) - 0\left( 2 \right) + 0\left( {4 - 3} \right) = 1

We know,

A−1=1∣A∣ adjA⇒A−1=adj(A){A^{ - 1}} = {1 \over {\left| A \right|}}\,adjA \Rightarrow {A^{ - 1}} = adj\left( A \right)

( as

∣A∣=1\left| A \right| = 1

) Now, from equation

(1)(1)

, we have

u1+u2=A−1(110){u_1} + {u_2} = {A^{ - 1}}\left( \begin{array}{ll}1 \\ 1 \\ 0 \end{array} \right)
=[100−2101−21](110)= \left[ \begin{array}{lll}1 & 0 & 0 \\ { - 2} & 1 & 0 \\ 1 & { - 2} & 1 \end{array} \right]\left( \begin{array}{ll}1 \\ 1 \\ 0 \end{array} \right)
=[1−1−1]= \left[ \begin{array}{ll}1 \\ { - 1} \\ { - 1} \end{array} \right]
Q17
If P=[1α3133244]P = \left[ \begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{array} \right] is the adjoint of a 3×33 \times 3 matrix AA and ∣A∣=4,\left| A \right| = 4, then α\alpha is equal to :
A 44
B 1111
C 55
D 00
Correct Answer
Option B
Solution
∣P∣=1(12−12)−α(4−6)+\left| P \right| = 1\left( {12 - 12} \right) - \alpha \left( {4 - 6} \right) +
           3(4−6)=2α−6\,\,\,\,\,\,\,\,\,\,\,3\left( {4 - 6} \right) = 2\alpha - 6

Now,

adj  A=P adj\,\,A = P\,
        ⇒∣adj A∣=∣P∣\,\,\,\,\,\,\,\, \Rightarrow \left| {adj\,A} \right| = \left| P \right|
⇒∣A∣2=∣P∣\Rightarrow {\left| A \right|^2} = \left| P \right|
⇒∣P∣=16\Rightarrow \left| P \right| = 16
⇒2α−6=16\Rightarrow 2\alpha - 6 = 16
⇒α=11\Rightarrow \alpha = 11
Q18
If AA is a 3×33 \times 3 non-singular matrix such that AA′=A′AAA'=A'A and B=A−1A′,B = {A^{ - 1}}A', then BB′BB' equals:
A B−1{B^{ - 1}}
B (B−1)′\left( {{B^{ - 1}}} \right)'
C I+BI+B
D II
Correct Answer
Option D
Solution
BB′=B(A−1A′)′=B(A′)′(A−1)′=BA(A−1)′BB' = B\left( {{A^{ - 1}}A'} \right)' = B\left( {A'} \right)'\left( {{A^{ - 1}}} \right)' = BA\left( {{A^{ - 1}}} \right)'
=(A−1A′)(A(A−1)′)= \left( {{A^{ - 1}}A'} \right)\left( {A\left( {{A^{ - 1}}} \right)'} \right)
=A−1A.A′.(A−1)′      = {A^{ - 1}}A.A'.\left( {{A^{ - 1}}} \right)'\,\,\,\,\,\,
{\left\{ {} \right.

as

      \,\,\,\,\,\,
AA′=A′AAA' = A'A
 }\left. \, \right\}
=I(A−1A)′= I\left( {{A^{ - 1}}A} \right)'
=I.I=I2=I= I.I = {I^2} = I
Q19
If B=[52α1021α3−1]B = \left[ \begin{array}{lll}5 & {2\alpha } & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & { - 1} \end{array} \right] is the inverse of a 3 × 3 matrix A, then the sum of all values of α\alpha for which det(A) + 1 = 0, is :
A 2
B - 1
C 0
D 1
Correct Answer
Option D
Solution

Given |A| + 1 = 0 ⇒\Rightarrow |A| = -1

∣B∣=∣A−1∣=1∣A∣=−1\left| B \right| = \left| {{A^{ - 1}}} \right| = {1 \over {\left| A \right|}} = - 1
∣52α1021α3−1∣\left| \begin{array}{lll}5 & {2\alpha } & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & { - 1} \end{array} \right|

= -1 ⇒\Rightarrow

5(−2−3)+2α(α)+1(−2α)=−15( - 2 - 3) + 2\alpha (\alpha ) + 1( - 2\alpha ) = - 1

⇒\Rightarrow

2α2−2α−24=02{\alpha ^2} - 2\alpha - 24 = 0

∴\therefore sum of value of α\alpha =

−(−2)2=1{{ - ( - 2)} \over 2} = 1
Q20
If A=[12221−2a2b]A = \left[ \begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & { - 2} \\ a & 2 & b \end{array} \right] is a matrix satisfying the equation AAT=9I,A{A^T} = 9\text{I}, where II is 3×33 \times 3 identity matrix, then the ordered pair (a,b)(a, b) is equal to :
A (2,1)(2, 1)
B (−2,−1)(-2, -1)
C (2,−1)(2, -1)
D (−2,1)(-2, 1)
Correct Answer
Option B
Solution
[12221−2a2b][12a2122−2b]=[900090009]\left[ \begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & { - 2} \\ a & 2 & b \end{array} \right]\left[ \begin{array}{lll}1 & 2 & a \\ 2 & 1 & 2 \\ 2 & { - 2} & b \end{array} \right] = \left[ \begin{array}{lll}9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array} \right]
⇒[1+4+42+2−4a+4+2b2+2−44+1+42a+2−2ba+4+2b2a+2−2ba2+4+b2]\Rightarrow \left[ \begin{array}{lll}{1 + 4 + 4} & {2 + 2 - 4} & {a + 4 + 2b} \\ {2 + 2 - 4} & {4 + 1 + 4} & {2a + 2 - 2b} \\ {a + 4 + 2b} & {2a + 2 - 2b} & {{a^2} + 4 + {b^2}} \end{array} \right]
                  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
                  =[900090009]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ \begin{array}{lll}9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array} \right]
⇒a+4+2b=0\Rightarrow a + 4 + 2b = 0
⇒a+2b=−4              ...(i)\Rightarrow a + 2b = - 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)
2a+2−2b=0⇒2a−2b=−22a + 2 - 2b = 0 \Rightarrow 2a - 2b = - 2
                 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
        ,\,\,\,\,\,\,\,\,\\,
                 ⇒a−b=−1                 ...(ii)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a - b = - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

On solving

(i)(i)

and

(ii)(ii)

we get

                 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
        \,\,\,\,\,\,\,\,
                 −1+b+2b=−4            ...(i)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1 + b + 2b = - 4\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)
                 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
                 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
b=−1b=-1

and

a=−2a=-2
(a,b)=(−2,−1)\left( {a,b} \right) = \left( { - 2, - 1} \right)
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