Matrices – JEE Main Previous Year Question with Solutions

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Q. Let A be the set of all 3  3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0.

(A) The number of matrices in A is –

(A) 12            (B) 6          (C) 9                  (D) 3

(B) The number of matrices A in A for which the system of linear equations

$A\left[\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right]=\left[\begin{array}{l}{1} \\ {0} \\ {0}\end{array}\right]$ has a unique solution, is –

(A) less than 4

(B) at least 4 but less than 7

(C) at least 7 but less than 10

(D) at least 10

(C) The number of matrices A in A for which the system of linear equations $A\left[\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right]=\left[\begin{array}{l}{1} \\ {0} \\ {0}\end{array}\right]$ is inconsistent, is –

(A) 0 (B) more than 2 (C) 2 (D) 1

[JEE 2009, 4+4+4]

Sol. ( (A) A ,(b) B, (c) B )

Q. (A) The number of 3  3 matrices A whose entries are either 0 or 1 and for which the system $A\left[\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right]=\left[\begin{array}{l}{1} \\ {0} \\ {0}\end{array}\right]$ has exactly two distinct solutions, is

(A) 0 (B) $2^{9}-1$ (C) 168 (D) 2

(B) Let $\mathrm{k}$ be a positive real number and let $\mathrm{A}=\left[\begin{array}{ccc}{2 \mathrm{k}-1} & {2 \sqrt{\mathrm{k}}} & {2 \sqrt{\mathrm{k}}} \\ {2 \sqrt{\mathrm{k}}} & {1} & {-2 \mathrm{k}} \\ {-2 \sqrt{\mathrm{k}}} & {2 \mathrm{k}} & {-1}\end{array}\right]$ and

$\mathrm{B}=\left[\begin{array}{ccc}{0} & {2 \mathrm{k}-1} & {\sqrt{\mathrm{k}}} \\ {1-2 \mathrm{k}} & {0} & {2 \sqrt{\mathrm{k}}} \\ {-\sqrt{\mathrm{k}}} & {-2 \sqrt{\mathrm{k}}} & {0}\end{array}\right]$

If $\operatorname{det}(\operatorname{adj} \mathrm{A})+\operatorname{det}(\operatorname{adj} \mathrm{B})=10^{6},$ then $[\mathrm{k}]$ is equal to

[Note : adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k].

(C) Let p be an odd prime number and Tp be the following set of 2  2 matrices:

$\mathrm{T}_{\mathrm{p}}=\left\{\mathrm{A}=\left[\begin{array}{ll}{\mathrm{a}} & {\mathrm{b}} \\ {\mathrm{c}} & {\mathrm{a}}\end{array}\right]: \mathrm{a}, \mathrm{b}, \mathrm{c} \in\{0,1,2, \ldots \ldots, \mathrm{p}-1\}\right.$

(i) The number of $A$ in $T_{p}$ such that $A$ is either symmetric or skew symmetric or both, and det(A) divisible by p is –

(A) $(\mathrm{p}-1)^{2}$

(B) 2 (p – 1)

(C) $(p-1)^{2}+1$

(D) 2p –1

(ii) The number of $A$ in $T_{p}$ such that the trace of $A$ is not divisible by $p$

but det (A) is divisible by p is –

[Note : The trace of a matrix is the sum of its diagonal entries.]

(A) $(p-1)\left(p^{2}-p+1\right)$

(B) $\mathrm{p}^{3}-(\mathrm{p}-1)^{2}$

(C) $(p-1)^{2}$

(D) $(p-1)\left(p^{2}-2\right)$

(iii) The number of $A$ in $T_{p}$ such that $\operatorname{det}(A)$ is not divisible by $p$ is –

(A) $2 p^{2}$

(B) $\mathrm{p}^{3}-5 \mathrm{p}$

(C) $\mathrm{p}^{3}-3 \mathrm{p}$

(D) $\mathrm{p}^{3}-\mathrm{p}^{2}$

[JEE 2010, 3+3+3+3+3]

Sol. ($(a) A,(b) 4 ;(c)(i) D,(i i) C,(\text { iii }) D$)

For the matrix $(6),(9),(10),(11)$ the system of linear equation is incosistent.

(A) The given matrix system is a linear system in $\mathrm{x}, \mathrm{y}, \mathrm{z},$ hence it can have either a unique solution or no-solution or infinitely many solutions. It can never have exactly two distinct solutions.

Q. Let $\mathrm{M}$ and $\mathrm{N}$ be two $3 \times 3$ non-singular skew-symmetric matrices such that $\mathrm{MN}=\mathrm{NM}$. If $\mathrm{P}^{\mathrm{T}}$ denotes the transpose of $\mathrm{P},$ then $\mathrm{M}^{2} \mathrm{N}^{2}\left(\mathrm{M}^{\mathrm{T}} \mathrm{N}\right)^{-1}\left(\mathrm{MN}^{-1}\right)^{\mathrm{T}}$ is equal to –

(A) $\mathbf{M}^{2}$

(B) $-N^{2}$

(C) $-M^{2}$

(D) MN

[JEE 2011, 4]

Sol. (Bonus)

(Comment : Although 3 3 skew symmetric matrices can never be non-singular. Therefore the information given in question is wrong. Now if we consider only non singular skew symmetric matrices M & N, then the solution is-)

Q. Let $\omega \neq 1$ be a cube root of unity and $S$ be the set of all non-singular matrices of the form $\left[\begin{array}{lll}{1} & {a} & {b} \\ {\omega} & {1} & {c} \\ {\omega^{2}} & {\omega} & {1}\end{array}\right],$ where each of a,b and $c$ is either $\omega$ or $\omega^{2} .$ Then the number of distinct matrices in the set $\mathrm{S}$ is-

(A) 2              (B) 6              (C) 4              (D) 8

[JEE 2011, 3, (–1)]

Sol. (A)

Q. Let M be 3  3 matrix satisfying $\mathbf{M}\left[\begin{array}{l}{0} \\ {1} \\ {0}\end{array}\right]=\left[\begin{array}{r}{-1} \\ {2} \\ {3}\end{array}\right], \mathbf{M}\left[\begin{array}{c}{1} \\ {-1} \\ {0}\end{array}\right]=\left[\begin{array}{r}{1} \\ {1} \\ {-1}\end{array}\right]$ and $\mathbf{M}\left[\begin{array}{l}{1} \\ {1} \\ {1}\end{array}\right]=\left[\begin{array}{c}{0} \\ {0} \\ {12}\end{array}\right]$ Then the sum of the diagonal entries of M is

[JEE 2011, 4]

Sol. 9

Q. Let $\mathrm{P}=\left[\mathrm{a}_{\mathrm{ij}}\right]$ be a $3 \times 3$ matrix and let $\mathrm{Q}=\left[\mathrm{b}_{\mathrm{ij}}\right],$ where $\mathrm{b}_{\mathrm{ij}}=2^{\mathrm{i}+\mathrm{j}} \mathrm{a}_{\mathrm{ij}}$ for $1 \leq \mathrm{i}, \mathrm{j} \leq 3$ If the determinant of $\mathrm{P}$ is $2,$ then the determinant of the matrix $\mathrm{Q}$ is –

(A) $2^{10}$

(B) $2^{11}$

(C) $2^{12}$

$(D) 2^{13}$

[JEE 2011, 4]

Sol. (D)

Q. If $\mathrm{P}$ is a $3 \times 3$ matrix such that $\mathrm{P}^{\mathrm{T}}=2 \mathrm{P}+\mathrm{I},$ where $\mathrm{P}^{\mathrm{T}}$ is the transpose of $\mathrm{P}$ and $\mathrm{I}$ is the $3 \times 3 \times 3$ identity matrix, then there exists a column matrix

$\mathrm{X}=\left[\begin{array}{l}{\mathrm{x}} \\ {\mathrm{y}} \\ {\mathrm{z}}\end{array}\right] \neq\left[\begin{array}{l}{0} \\ {0} \\ {0}\end{array}\right]$ such that

(A) $\mathrm{PX}=\left[\begin{array}{l}{0} \\ {0} \\ {0}\end{array}\right]$ (B) PX = X (C) PX = 2X (D) PX = –X

[JEE 2012, 3M, –1M]

Sol. (D)

$\mathrm{P}^{\mathrm{T}}=2 \mathrm{P}+\mathrm{I}$

$\Rightarrow \mathrm{P}=2 \mathrm{P}^{\mathrm{T}}+\mathrm{I}$

$\Rightarrow \mathrm{P}=2(2 \mathrm{P}+\mathrm{I})+\mathrm{I}$

$\Rightarrow P=4 P+3 I$

$\Rightarrow P=-I$

$\Rightarrow \mathrm{PX}=-\mathrm{X}$

Q. If the adjoint of a $3 \times 3$ matrix $\mathrm{P}$ is $\left[\begin{array}{ccc}{1} & {4} & {4} \\ {2} & {1} & {7} \\ {1} & {1} & {3}\end{array}\right],$ then the possible value(s) of the determinant of $\mathrm{P}$ is (are)

(A) –2               (B) –1               (C) 1                     (D) 2

[JEE 2012, 4M]

Sol. (A,D)

Q. For 3  3 matrices M and N, which of the following statement(s) is (are) NOT correct ?

(A) $\mathbf{N}^{\mathrm{T}} \mathbf{M}$ N is symmetric or skew symmetric, according as M is symmetric or

skew symmetric

(B) MN – NM is skew symmetric for all symmetric matrices M and N

(C) MN is symmetric for all symmetric matrices M and N

Sol. (C,D)

(A) $\mathrm{B}=\mathrm{N}^{\mathrm{T}} \mathrm{MN}$

$\mathrm{B}^{\mathrm{T}}=\left(\mathrm{N}^{\mathrm{T}} \mathrm{MN}\right)^{\mathrm{T}}=\mathrm{N}^{\mathrm{T}} \mathrm{M}^{\mathrm{T}} \mathrm{N}$

Now it depands on $\mathrm{M}$ if $\mathrm{M}=\mathrm{M}^{\mathrm{T}}$

So A is true

$(\mathrm{B}) \mathrm{B}=(\mathrm{MN}-\mathrm{NM})$

$\mathrm{B}=(\mathrm{MN}-\mathrm{NM})^{\mathrm{T}}$

$=\mathbf{N}^{\mathrm{T}} \mathbf{M}^{\mathrm{T}}-\mathbf{M}^{\mathrm{N}} \mathbf{N}^{\mathrm{T}}$

$=\mathrm{NM}-\mathrm{MN}=-(\mathrm{B})$ Skew symmetric

B is ture

$(\mathrm{C}) \mathrm{B}=\mathrm{MN}$

$\mathrm{B}^{\mathrm{T}}=(\mathrm{MN})^{\mathrm{T}}$

$\mathrm{B}^{\mathrm{T}}=\mathrm{N}^{\mathrm{T}} \mathrm{M}^{\mathrm{T}}$

$\mathrm{B}^{\mathrm{T}}=\mathrm{NM} \neq \mathrm{B}$ so wrong statement

(D) Obviousely wrong

because adj $(\mathrm{BA})=\operatorname{adj}(\mathrm{A}) \cdot \operatorname{adJ}(\mathrm{B})$

Q. Let M be a 2  2 symmetric matrix with integer entries. Then M is invertible if

(A) the first column of M is the transpose of the second row of M

(B) the second row of M is the transpose of the first column of M

(C) M is a diagonal matrix with nonzero entries in the main diagonal

(D) the product of entries in the main diagonal of M is not the square of an integer

Sol. (C,D)

Q. Let $\mathrm{M}$ and $\mathrm{N}$ be two $3 \times 3$ matrices such that $\mathrm{MN}=\mathrm{NM}$. Further, if $\mathrm{M} \neq \mathrm{N}^{2}$ and $\mathrm{M}^{2}=$ $\mathrm{N}^{4},$ then

(A) determinant of $\left(\mathrm{M}^{2}+\mathrm{MN}^{2}\right)$ is 0

(B) there is a $3 \times 3$ non-zero matrix $\mathrm{U}$ such that $\left(\mathrm{M}^{2}+\mathrm{MN}^{2}\right) \mathrm{U}$ is zero Matrix

(C) determinant of $\left(\mathrm{M}^{2}+\mathrm{MN}^{2}\right) \geq 1$

(D) for a $3 \times 3$ matrix $\mathrm{U},$ if $\left(\mathrm{M}^{2}+\mathrm{MN}^{2}\right)$ U equals the zero matrix then $\mathrm{U}$ is the zero matrix

Sol. (A,B)

$\mathrm{M}^{2}=\mathrm{N}^{4}$

$\mathrm{M}^{2}=\mathrm{N}^{4}=0(\therefore \mathrm{MN}=\mathrm{NM})$

$\left(\mathrm{M}+\mathrm{N}^{2}\right)\left(\mathrm{M}-\mathrm{N}^{2}\right)=0$

So $\left(\mathrm{M}+\mathrm{N}^{2}\right)=0$

$\mathrm{Now} \mathrm{M} \cdot\left(\mathrm{M}+\mathrm{N}^{2}\right)=0$

$\mathrm{M}^{2}+\mathrm{MN}^{2}=0$

$\left|\mathrm{M}^{2}+\mathrm{MN}^{2}\right|=0$

Option A is right So we know $A \cdot B=0$

when $\mathrm{B} \neq 0$

$\Rightarrow(\mathrm{A})=0$

So in $\mathrm{U}\left(\mathrm{M}^{2}+\mathrm{MN}^{2}\right) \mathrm{U}=0$

Because $\mathrm{U} \neq 0$ But $\left|\mathrm{M}^{2}+\mathrm{MN}^{2}\right|=0$

So option $\mathrm{B}$ is also right

Q. Let X and Y be two arbitrary, 3  3, non-zero, skew-symmetric matrices and Z be an arbitrary 3 × 3, non-zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric ?

(A) $\mathrm{Y}^{3} \mathrm{Z}^{4}-\mathrm{Z}^{4} \mathrm{Y}^{3}$

(B) $\mathrm{X}^{44}+\mathrm{Y}^{44}$

(C) $\mathrm{X}^{4} \mathrm{Z}^{3}-\mathrm{Z}^{3} \mathrm{X}^{4}$

(D) $\mathrm{X}^{23}+\mathrm{Y}^{23}$

Sol. (C,D)

$\mathrm{x}^{\mathrm{T}}=-\mathrm{x}, \mathrm{y}^{\mathrm{T}}=-\mathrm{y}, \mathrm{z}^{\mathrm{T}}=\mathrm{z}$

(A) Let $P=y^{3} z^{4}-z^{4} y^{3}$

$\mathrm{P}^{\mathrm{T}}=\left(\mathrm{y}^{3} \mathrm{z}^{4}\right)^{\mathrm{T}}-\left(\mathrm{z}^{4} \mathrm{y}^{3}\right)^{\mathrm{T}}$

$=-z^{4} y^{3}+y^{3} z^{4}=P \Rightarrow$ symmetric

(B) $\quad$ Let $\mathrm{P}=\mathrm{x}^{44}+\mathrm{y}^{44}$

$\mathrm{P}^{\mathrm{T}}=\left(\mathrm{X}^{44}\right)^{\mathrm{T}}+\left(\mathrm{y}^{44}\right)^{\mathrm{T}}=\mathrm{P} \Rightarrow$ symmetric

(C) Let $P=x^{4} z^{3}-z^{3} x^{4}$

$\mathrm{P}^{\mathrm{T}}=\left(\mathrm{z}^{3}\right)^{\mathrm{T}}\left(\mathrm{x}^{4}\right)^{\mathrm{T}}-\left(\mathrm{x}^{4}\right)^{\mathrm{T}}\left(\mathrm{z}^{3}\right)^{\mathrm{T}}$

$=\mathrm{z}^{3} \mathrm{x}^{4}-\mathrm{x}^{4} \mathrm{z}^{3}=-\mathrm{P} \Rightarrow$ skew symmetric

(D) Let $P=x^{23}+y^{23}$ $\mathrm{P}^{\mathrm{T}}=-\mathrm{x}^{23}-\mathrm{y}^{23}=-\mathrm{P} \Rightarrow$ skew symmetric

Q. Let $P=\left[\begin{array}{ccc}{3} & {-1} & {-2} \\ {2} & {0} & {\alpha} \\ {3} & {-5} & {0}\end{array}\right],$ where $\alpha \in R,$ Suppose $Q=\left[q_{i j}\right]$ is a matrix such that $P Q=k I$ where $\mathrm{k} \in \mathrm{R}, \mathrm{k} \neq 0$ and $\mathrm{I}$ is the identity matrix of order $3 .$ If $\mathrm{q}_{23}=-\frac{\mathrm{k}}{8}$ and $\operatorname{det}(\mathrm{Q})=\frac{\mathrm{k}^{2}}{2}$ then-

(A)  = 0, k = 8

(B) $4 \alpha-k+8=0$

(C) $\operatorname{det}(\operatorname{Padj}(\mathrm{Q}))=2^{9}$

(D) det(Qadj(P)) = $2^{13}$

Sol. (B,C)

$\mathrm{PQ}=\mathrm{kI}$

$|\mathrm{P}| \cdot|\mathrm{Q}|=\mathrm{k}^{3} \Rightarrow|\mathrm{P}|=2 \mathrm{k} \neq 0 \Rightarrow \mathrm{P}$ is an invertible matrix

$\because \mathrm{PQ}=\mathrm{kI}$

$\therefore \mathrm{Q}=\mathrm{k} \mathrm{P}^{-1} \mathrm{I}$

$\therefore \mathrm{Q}=\frac{\mathrm{adj.P}}{2}$

$\because \mathrm{q}_{23}=-\frac{\mathrm{k}}{8}$

$\therefore \frac{-(3 \alpha+4)}{2}=-\frac{k}{8} \Rightarrow k=4$

$\therefore|\mathrm{P}|=2 \mathrm{k} \Rightarrow \mathrm{k}=10+6 \alpha \ldots(\mathrm{i})$

Put value of $k$ in (i).. we get $\alpha=-1$

$\therefore 4 \alpha-k+8=0$

$\& \operatorname{det}(\mathrm{P}(\mathrm{adj} . \mathrm{Q}))=|\mathrm{P}||\operatorname{adj} . \mathrm{Q}|=2 \mathrm{k} \cdot\left(\frac{\mathrm{k}^{2}}{2}\right)^{2}=\frac{\mathrm{k}^{5}}{2}=2^{9}$

Q. Let $P=\left[\begin{array}{lll}{1} & {0} & {0} \\ {4} & {1} & {0} \\ {16} & {4} & {1}\end{array}\right]$ and $I$ be the identity matrix of order $3 .$ If $Q=\left[q_{i j}\right]$ is a matrix such that $\mathrm{P}^{50}-\mathrm{Q}=\mathrm{I},$ then $\frac{\mathrm{q}_{31}+\mathrm{q}_{32}}{\mathrm{q}_{21}}$ equals

(A) 52             (B) 103               (C) 201                  (D) 205

Sol. (A)

Q. Which of the following is(are) NOT the square of a 3  3 matrix with real entries ?

Sol. (A,B)

Q. How many $3 \times 3$ matrices $\mathrm{M}$ with entries from $\{0,1,2\}$ are there, for which the sum of the diagonal entries of $\mathrm{M}^{\mathrm{T}} \mathrm{M}$ is $5 ?$

(A) 198              (B) 126                (C) 135                 (D) 16

Sol. (A)

Q. Let S be the of all column matrices $\left[\begin{array}{l}{b_{1}} \\ {b_{2}} \\ {b_{3}}\end{array}\right]$ such that $b_{1}, b_{2}, b_{3} \in \square$ and the system of equations (in real variables)

$-x+2 y+5 z=b_{1}$

$2 x-4 y+3 z=b_{2}$

$x-2 y+2 z=b_{3}$

has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution of each $\left[\begin{array}{l}{b_{1}} \\ {b_{2}} \\ {b_{3}}\end{array}\right] \in S ?$

(A) $\mathrm{x}+2 \mathrm{y}+3 \mathrm{z}=\mathrm{b}_{1}, 4 \mathrm{y}+5 \mathrm{z}=\mathrm{b}_{2}$ and $\mathrm{x}+2 \mathrm{y}+6 \mathrm{z}=\mathrm{b}_{3}$

(B) $x+y+3 z=b_{1}, 5 x+2 y+6 z=b_{2}$ and $-2 x-y-3 z=b_{3}$

(C) $-\mathrm{x}+2 \mathrm{y}-5 \mathrm{z}=\mathrm{b}_{1}, 2 \mathrm{x}-4 \mathrm{y}+10 \mathrm{z}=\mathrm{b}_{2}$ and $\mathrm{x}-2 \mathrm{y}+5 \mathrm{z}=\mathrm{b}_{3}$

(D) $x+2 y+5 z=b_{1}, 2 x+3 z=b_{2}$ and $x+4 y-5 z=b_{3}$