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JEE • MATHEMATICS • MATRICES

Matrices practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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301 Easy
301 of 995
Application +4 | -1
If A=[n000n000n]A=\left[\begin{array}{lll}n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n\end{array}\right] and B=[a1a2a3b1b2b3c1c2c3]B=\left[\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right], then ABA B equal to
View answer and explanation
Correct option: B
nBn B
Option B (nBnB) is correct because multiplying a matrix B by a scalar matrix A (nI) results in each element of B being multiplied by n.
302 Medium
302 of 995
Understanding +4 | -1
From the matrix equation AB=ACAB = AC we can conclude B=CB = C provided that
View answer and explanation
Correct option: B
AA is non-singular
Option B (AA is non-singular) is correct because if AA is non-singular, it has an inverse, allowing us to multiply both sides of the equation AB=ACAB = AC by A−1A^{-1} to get B=CB = C.
303 Easy
303 of 995
Application +4 | -1
Let X=[x1x2x3],  A=[1−12201321]X = \left[ {\begin{array}{cccccccccccccccccccc} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right],\;A = \left[ {\begin{array}{cccccccccccccccccccc} 1&{ - 1}&2\\ 2&0&1\\ 3&2&1 \end{array}} \right] and B=[314]B = \left[ {\begin{array}{cccccccccccccccccccc} 3\\ 1\\ 4 \end{array}} \right]. If AX=BAX = B, then XX is equal to
View answer and explanation
Correct option: D
[−123]\left[ {\begin{array}{cccccccccccccccccccc} { - 1}\\ 2\\ 3 \end{array}} \right]
Option D ([-1], [2], [3]) is correct because it is the solution to the matrix equation AX = B, obtained by finding the inverse of A and multiplying it by B.
304 Easy
304 of 995
Understanding +4 | -1
If AA and BB are 2×22 \times 2 matrices, then which of the following is true?
View answer and explanation
Correct option: C
(A−B)(A+B)=A2+AB−BA−B2(\mathrm{A}-\mathrm{B})(\mathrm{A}+\mathrm{B})=\mathrm{A}^2+\mathrm{AB}-\mathrm{BA}-\mathrm{B}^2
Option C, (A−B)(A+B)=A2+AB−BA−B2(\mathrm{A}-\mathrm{B})(\mathrm{A}+\mathrm{B})=\mathrm{A}^2+\mathrm{AB}-\mathrm{BA}-\mathrm{B}^2, is correct because matrix multiplication is not commutative, so AB is not necessarily equal to BA.
305 Medium
305 of 995
Analyzing +4 | -1
If A=[21−1352161]A=\left[\begin{array}{rrr}2 & 1 & -1 \\ 3 & 5 & 2 \\ 1 & 6 & 1\end{array}\right], then tr⁡(Aadj⁡(adj⁡A)\operatorname{tr}(\operatorname{Aadj}(\operatorname{adj} \mathrm{A})) is equal to (where tr⁡(P)\operatorname{tr}(\mathrm{P}) denotes the trace of the matrix P i.e the sum of all the diagonal elements of the matrix P and adj (P) denotes the adjoint of matrix P)
View answer and explanation
Correct option: D
−1624-1624
Option D (-1624) is correct because it is the trace of the matrix Aadj(adj A) which can be simplified using properties of adjoint and determinant.
306 Easy
306 of 995
Understanding +4 | -1
The inverse of a skew symmetric matrix of odd order is
View answer and explanation
Correct option: D
does not exist
Option D (does not exist) is correct because a skew-symmetric matrix of odd order has a determinant of 0, and therefore is not invertible.
307 Easy
307 of 995
Understanding +4 | -1
Consider the following statements

I. For multiplication of two matrices AA and BB, the number of columns in AA should be less than the number of rows in B{B}.

II. For getting the elements of the product matrix, we take rows of AA and column of BB, multiply them elementwise and take the sum.

Choose the correct option.
View answer and explanation
Correct option: B
Only II is true
Option B (Only II is true) is correct because for matrix multiplication, the number of columns in the first matrix must equal the number of rows in the second.
308 Easy
308 of 995
Analyzing +4 | -1
If AA and BB are square matrices such that A2006=0A^{2006}=0 and AB=A+B,A B=A+B, then det⁡(B)\operatorname{det}(B) equals to
View answer and explanation
Correct option: B
0
Option B (0) is correct because given A2006 = 0, det(A) = 0, which leads to det(B) = 0.
309 Easy
309 of 995
Understanding +4 | -1
Choose the correct answer in the following questions : If A,BA, B are symmetric matrices of same order then AB−BA\mathrm{AB}-\mathrm{BA} is a
View answer and explanation
Correct option: A
Skew-symmetric matrix
Option A (Skew-symmetric matrix) is correct because if A and B are symmetric, then (AB - BA)' = -(AB - BA), which is the definition of a skew-symmetric matrix.
310 Easy
310 of 995
Understanding +4 | -1
If I=(1001),J=(01−10)I=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right), J=\left(\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right), and B=(cos⁡θsin⁡θ−sin⁡θcos⁡θ)B=\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right), then BB equals
View answer and explanation
Correct option: A
(cos⁡θ)I+(sin⁡θ)J(\cos \theta) I+(\sin \theta) J
Option A ((cos⁡θ\cos \theta)I + (sinθsin \theta)J) is correct because it accurately represents the given matrix B as a linear combination of the identity matrix I and the matrix J.
311 Easy
311 of 995
Application +4 | -1
The matrix A=[1212−12−12]A = \left[ {\begin{array}{cccccccccccccccccccc} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ { - \frac{1}{{\sqrt 2 }}}&{ - \frac{1}{{\sqrt 2 }}} \end{array}} \right] is
View answer and explanation
Correct option: C
Nilpotent
Option C (Nilpotent) is correct because A squared equals the zero matrix, which is the definition of a nilpotent matrix.
312 Easy
312 of 995
Application +4 | -1
If [2174]A[−325−3]=[1001]\left[\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right] A\left[\begin{array}{rr}-3 & 2 \\ 5 & -3\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right], then matrix AA equals
View answer and explanation
Correct option: A
[75−11−8]\left[\begin{array}{cc}7 & 5 \\ -11 & -8\end{array}\right]
Option A ([75−11−8]\left[\begin{array}{cc}7 & 5 \\ -11 & -8\end{array}\right]) is correct because it is the result of calculating the inverse matrices and performing the matrix multiplication.
313 Easy
313 of 995
Application +4 | -1
Let the numbers 2, b, c be in an AP and
A=[1112bc4b2c2].Ifdet⁡(A)∈[2,16],then cA = \left[ {\begin{array}{cccccccccccccccccccc} 1&1&1\\ 2&b&c\\ 4&{{b^2}}&{{c^2}} \end{array}} \right].\mathrm{ If}\det (A) \in [2,16],\mathrm{then}\,\mathrm{ }c lies in the interval
View answer and explanation
Correct option: C
[4,6]\left[ {4,6} \right]
Option C ([4,6]) is correct because the value of c lies in this interval after solving for the common difference of the AP and using the determinant condition.
314 Easy
314 of 995
Application +4 | -1
A root of the equation ∣0x−ax−bx+a0x−cx+bx+c0∣=0,\left| {\begin{array}{cccccccccccccccccccc} 0&{x - a}&{x - b}\\ {x + a}&0&{x - c}\\ {x + b}&{x + c}&0 \end{array}} \right| = 0, is
View answer and explanation
Correct option: C
0
Option C (0) is correct because expanding the determinant leads to an equation where x=0 is a root.
315 Easy
315 of 995
Application +4 | -1
If A=[3201],  A = \left[ {\begin{array}{cccccccccccccccccccc} 3&2\\ 0&1 \end{array}} \right],\; then (A−1)3{\left( {{A^{ - 1}}} \right)^3} is equal to
View answer and explanation
Correct option: A
127[1−26027]\frac{1}{{27}}\left[ {\begin{array}{cccccccccccccccccccc} 1&{ - 26}\\ 0&{27} \end{array}} \right]
Option A (127[1−26027]\frac{1}{{27}}\left[ {\begin{array}{cccccccccccccccccccc} 1&{ - 26}\\ 0&{27} \end{array}} \right]) is correct because it is the result of cubing the inverse of the given matrix.
316 Easy
316 of 995
Analyzing +4 | -1
Let α\alpha be a root of the equation x2+x+1=0{x^2} + x + 1 = 0 and the matrix A=13[1111αα21α2α4]A = \frac{1}{{\sqrt 3 }}\left[ {\begin{array}{cccccccccccccccccccc} 1&1&1\\ 1&\alpha &{{\alpha ^2}}\\ 1&{{\alpha ^2}}&{{\alpha ^4}} \end{array}} \right], then the matrix A31{A^{31}} is equal to
View answer and explanation
Correct option: A
A3{A^3}
Option A (A3{A^3}) is correct because after calculating A2A^2 and using the property α3=1{\alpha^3} = 1, we find that A4=I{A^4} = I, so A31=A28⋅A3=A3{A^{31}} = {A^{28}} \cdot {A^3} = {A^3}.
317 Easy
317 of 995
Application +4 | -1
If A=[1234]A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right], then A2−5A−IA^2-5 A-I equals
View answer and explanation
Correct option: B
II
Option B (II) is correct because after calculating A2−5AA^2 - 5A, the result is 2I2I. Subtracting II from 2I2I yields II.
318 Medium
318 of 995
Understanding +4 | -1
Consider the system of equations a1x+b1y+c1z=0a_1 x+b_1 y+c_1 z=0, a2x+b2y+c2z=0,a3x+b3y+c3z=0a_2 x+b_2 y+c_2 z=0, a_3 x+b_3 y+c_3 z=0. If ∣a1b1c1a2b2c2a3c3c3∣=0\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & c_3 & c_3\end{array}\right|=0, then the system has
View answer and explanation
Correct option: A
More than 2 solutions
Option A (More than 2 solutions) is correct because when the determinant of the coefficient matrix is zero, the system has non-trivial solutions, implying more than one solution.
319 Easy
319 of 995
Analyzing +4 | -1
If a,ba, b and cc are non-zero real numbers and if the equations (a−1)x=y+z,(b−1)y=z+x(a-1) x=y+z,(b-1) y=z+x, (c−1)z=x+y(c-1) z=x+y has a non-trivial solution, then ab+bc+caa b+b c+c a equals to
View answer and explanation
Correct option: B
abc{abc}
Option B (abc{abc}) is correct because, for a non-trivial solution, the determinant of the coefficient matrix must be zero, leading to the equation ab+bc+ca=abcab + bc + ca = abc.
320 Easy
320 of 995
Application +4 | -1
Let A=[12−3−2033−31]A = \left[ {\begin{array}{cccccccccccccccccccc} 1&2&{ - 3}\\ { - 2}&0&3\\ 3&{ - 3}&1 \end{array}} \right] be a matrix, then (determinant of A)×\times (adjoint of inverse A) is equal to
View answer and explanation
Correct option: B
[12−3−2033−31]\left[ {\begin{array}{cccccccccccccccccccc} 1&2&{ - 3}\\ { - 2}&0&3\\ 3&{ - 3}&1 \end{array}} \right]
Option B is correct. The expression simplifies to the original matrix A, as the determinant of A times the adjoint of the inverse of A equals A.
321 Easy
321 of 995
Application +4 | -1
If ω\omega be a complex cube root of unity, then ∣1ω−ω2/21111−10∣\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{ - {\omega ^2}/2}\\ 1&1&1\\ 1&{ - 1}&0 \end{array}} \right| is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because simplifying the determinant using the property 1+ω+ω2=01 + \omega + \omega^2 = 0 results in a zero determinant.
322 Easy
322 of 995
Application +4 | -1
Let [x]\left[ x \right] represent the greatest integer less than or equal to xx, then the value of the determinant ∣[e][π][π2−6][π][π2−6][e][π2−6][e][π]∣\left| {\begin{array}{cccccccccccccccccccc} {\left[ e \right]}&{\left[ \pi \right]}&{\left[ {{\pi ^2} - 6} \right]}\\ {\left[ \pi \right]}&{\left[ {{\pi ^2} - 6} \right]}&{\left[ e \right]}\\ {\left[ {{\pi ^2} - 6} \right]}&{\left[ e \right]}&{\left[ \pi \right]} \end{array}} \right| is
View answer and explanation
Correct option: A
−8- 8
Option A (-8) is correct because substituting the integer values of e, pi, and pi2 - 6 into the determinant and evaluating it yields -8.
323 Medium
323 of 995
Analyzing +4 | -1
Suppose p,q,r≠0p, q, r \neq 0 and system of equations
(p+a)x+by+cz=0ax+(q+b)y+cz=0ax+by+(r+c)z=0\begin{array}{l} (p+a) x+b y+c z=0 \\ a x+(q+b) y+c z=0 \\ a x+b y+(r+c) z=0 \end{array}

has a non-trivial solution, then value of ap+bq+cr\frac{a}{p}+\frac{b}{q}+\frac{c}{r} is
View answer and explanation
Correct option: A
-1
Option A (-1) is correct because the determinant of the coefficient matrix must be zero for a non-trivial solution, leading to the equation a/p + b/q + c/r = -1.
324 Easy
324 of 995
Application +4 | -1
If AA is a matrix such that A2=A+1,  {A^2} = A + 1,\; where II is the unit matrix, then A5  {A^5}\; is equal to
View answer and explanation
Correct option: C
5A+3I5A + 3I
Option C (5A+3I5A + 3I) is correct because repeated application of the given relation A2=A+I{A^2} = A + I simplifies A5{A^5} to 5A+3I5A + 3I.
325 Medium
325 of 995
Application +4 | -1
Let P=[3212−1232],A=[1101]P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] and Q=PAPTQ=P A P^{T}. If PTP^{T} Q2007P=[abcd]Q^{2007} P=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right], then 2a+b−3c−4d2 a+b-3 c-4 d equal to
View answer and explanation
Correct option: B
2005
The correct answer is 2005 because after simplifying the matrix operations and using the properties of matrix exponentiation, the final expression evaluates to 2005.
326 Medium
326 of 995
Application +4 | -1
A root of the equation ∣3−x−63−63−x333−6−x∣=0,\left| {\begin{array}{cccccccccccccccccccc} {3 - x}&{ - 6}&3\\ { - 6}&{3 - x}&3\\ 3&3&{ - 6 - x} \end{array}} \right| = 0, is
View answer and explanation
Correct option: C
0
Option C (0) is correct because x=0 is a root of the equation after applying column and row operations and simplifying the determinant.
327 Easy
327 of 995
Analyzing +4 | -1
If SS is the set of distinct values of ′b′'b' for which the following system of linear equations
x+y+z=1x+y+z=1

x+ay+z=1x+a y+z=1

ax+by+z=0a x+b y+z=0
has no solution, then SS is
View answer and explanation
Correct option: D
a singleton set
The correct answer is a singleton set because the system has no solution only when a=b=1, resulting in a single value for 'b'.
328 Easy
328 of 995
Application +4 | -1
If a1,a2,a3,…,an,…a_1, a_2, a_3, \ldots, a_n, \ldots are in G.P., then the value of the determinant ∣log⁡anlog⁡an+1log⁡an+2log⁡an+1log⁡an+2log⁡an+3log⁡an+2log⁡an+3log⁡an+4∣\left|\begin{array}{ccc} \log a_n & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+1} & \log a_{n+2} & \log a_{n+3} \\ \log a_{n+2} & \log a_{n+3} & \log a_{n+4} \end{array}\right| is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the rows of the determinant are linearly dependent due to the properties of geometric progressions.
329 Easy
329 of 995
Remembering +4 | -1
If A=[100010ab−1]A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1\end{array}\right], then the trace of the matrix AA is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because the trace of a matrix is the sum of its diagonal elements, which in this case is 1 + 1 + (-1) = 1.
330 Easy
330 of 995
Application +4 | -1
If A=[abba]A=\left[\begin{array}{ll}a & b \\ b & a\end{array}\right] and A2=[αββα]A^2=\left[\begin{array}{ll}\alpha & \beta \\ \beta & \alpha\end{array}\right], then
View answer and explanation
Correct option: C
α=a2+b2,β=2ab\alpha=a^2+b^2, \beta=2 a b
Option C (α=a2+b2,β=2ab\alpha=a^2+b^2, \beta=2 a b) is correct because when A is multiplied by itself, the resulting matrix elements are α=a2+b2\alpha = a^2 + b^2 and β=2ab\beta = 2ab.
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