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JEE • MATHEMATICS • MATRICES

Matrices practice · Medium

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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 Medium
301 of 408
Analyzing +4 | -1
If a2+b2+c2=−2a^2+b^2+c^2=-2 and
f(x)=∣1+a2x(1+b2)x(1+c2)x(1+a2)x1+b2x(1+c2)x(1+a2)x(1+b2)x1+c2x∣\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc} 1+a^2 x & \left(1+b^2\right) x & \left(1+c^2\right) x \\ \left(1+a^2\right) x & 1+b^2 x & \left(1+c^2\right) x \\ \left(1+a^2\right) x & \left(1+b^2\right) x & 1+c^2 x \end{array}\right|
then f(x)f(x) is a polynomial of degree
View answer and explanation
Correct option: D
2
The correct answer is 2 because, after simplification using the given condition, the function f(x) becomes a polynomial of degree 2.
302 Medium
302 of 408
Application +4 | -1
If 0≤[x]<2;−1≤[y]<10 \leq[{x}] < 2 ;-1 \leq[{y}] < 1 and 1≤[z]<31 \leq[{z}] < 3. where [∗][*] denotes the greatest integer function,
View answer and explanation
Correct option: A
4
Option A (4) is correct because by substituting the possible values of [x][x], [y][y], and [z][z] into the determinant and simplifying, we obtain the value 4.
303 Medium
303 of 408
Analyzing +4 | -1
The value of xx obtained from the equation ∣x+αβγγx+βααβx+γ∣=0\left| {\begin{array}{cccccccccccccccccccc} {x + \alpha }&\beta &\gamma \\ \gamma &{x + \beta }&\alpha \\ \alpha &\beta &{x + \gamma } \end{array}} \right| = 0 will be
View answer and explanation
Correct option: A
0 and −(α+β+γ)- \left( {\alpha + \beta + \gamma } \right)
Option A (0 and −(α+β+γ)- \left( {\alpha + \beta + \gamma } \right)) is correct because these are the values of x that satisfy the given determinant equation.
304 Medium
304 of 408
Application +4 | -1
If AA is square matrix of order n,a=n, a= maximum number of distinct entries. If AA is triangular matrix, b=b= maximum number of distinct entries. If AA is a diagonal matrix. c=c= minimum number of zeros. If AA is a triangular matrix if a+5=c+2ba+5=c+2 b
View answer and explanation
Correct option: B
4
Option B, 4, is correct because substituting the expressions for a, b, and c in terms of n into the given equation a+5=c+2ba+5 = c+2b yields n=4.
305 Medium
305 of 408
Understanding +4 | -1
If A=[3457]A = \left[ {\begin{array}{cccccccccccccccccccc} 3&4\\ 5&7 \end{array}} \right] then A.(adj  A)A.\left( {\mathrm{adj}\;A} \right) is equal to
View answer and explanation
Correct option: C
∣A∣I\left| A \right|I
Option C (∣A∣I\left| A \right|I) is correct because A.(adj;A)A.\left( {\mathrm{adj\\;}A} \right) is equal to the determinant of A multiplied by the identity matrix.
306 Medium
306 of 408
Application +4 | -1
If a,b,ca, b, c are positive integers such that a>b>ca > b > c and ∣111abca2b2c2∣=−2\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array}\right|=-2 then 3a+7b−10c3 a+7 b-10 c equals
View answer and explanation
Correct option: D
13
Option D (13) is correct because by analyzing the given determinant condition, we can determine the values of a, b, and c, and then calculate 3a + 7b - 10c.
307 Medium
307 of 408
Application +4 | -1
Let A[12−51]A\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right] and A−1=xA+yIA^{-1}=x A+y I, then the value of (2x+3y)(2 x+3 y) is______
View answer and explanation
Correct option: A
0.36
Option A (0.36) is correct because by calculating the inverse of matrix A and comparing it with the given equation, we can find the values of x and y, and then calculate 2x + 3y.
308 Medium
308 of 408
Application +4 | -1
If [12x4−1724−6]\left[ {\begin{array}{cccccccccccccccccccc} 1&2&x\\ 4&{ - 1}&7\\ 2&4&{ - 6} \end{array}} \right] is a singular matrix, then xx is equal to
View answer and explanation
Correct option: C
−3- 3
Option C (-3) is correct because when the determinant of the given matrix is set to zero (since it's a singular matrix), solving for x yields x = -3.
309 Medium
309 of 408
Analyzing +4 | -1
The value of the determinant ∣xab+cxbc+axca+b∣=0\left| {\begin{array}{cccccccccccccccccccc} x&a&{b + c}\\ x&b&{c + a}\\ x&c&{a + b} \end{array}} \right| = 0, if
View answer and explanation
Correct option: D
xx has any value
The correct answer is D (xx has any value). The determinant simplifies to 0 regardless of the value of x.
310 Medium
310 of 408
Analyzing +4 | -1
A factor of Δ(x)=∣x3+12x4+3x23x5+4x25731419∣\Delta \left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} {{x^3} + 1}&{2{x^4} + 3{x^2}}&{3{x^5} + 4x}\\ 2&5&7\\ 3&{14}&{19} \end{array}} \right| is
View answer and explanation
Correct option: B
(x−1)2{\left( {x - 1} \right)^2}
Option B ((x−1)2{\left( {x - 1} \right)^2}) is correct because both Δ(1)=0\Delta(1) = 0 and Δ′(1)=0\Delta'(1) = 0, indicating that (x−1)2(x-1)^2 is a factor.
311 Medium
311 of 408
Application +4 | -1
If A=[010001pqr]A=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r\end{array}\right] then A3−rA2−qA=A^3-r A^2-q A=
View answer and explanation
Correct option: A
pIp I
Option A (pIpI) is correct because by calculating A2 and A3, and substituting them into the expression A3 - rA2 - qA, we find that it simplifies to pI.
312 Medium
312 of 408
Analyzing +4 | -1
If Dr=∣2r−13r−14r−1xyz2n−1(3n−1)/2(4n−1)/3∣{D_r} = \left| {\begin{array}{cccccccccccccccccccc} {{2^{r - 1}}}&{{3^{r - 1}}}&{{4^{r - 1}}}\\ x&y&z\\ {{2^n} - 1}&{\left( {{3^n} - 1} \right)/2}&{\left( {{4^n} - 1} \right)/3} \end{array}} \right|, then the value of ∑r=1nDr\sum\nolimits_{r = 1}^n {{D_r}} is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because when the summation is applied, the first row becomes identical to the third row, resulting in a determinant value of zero.
313 Medium
313 of 408
Application +4 | -1
One root of the equation ∣x−6−12−3xx−3−32xx+2∣=0\left| {\begin{array}{cccccccccccccccccccc} x&{ - 6}&{ - 1}\\ 2&{ - 3x}&{x - 3}\\ { - 3}&{2x}&{x + 2} \end{array}} \right| = 0 is
View answer and explanation
Correct option: B
1
Option B (1) is correct because substituting x = 1 into the determinant equation results in 0, indicating that 1 is a root.
314 Medium
314 of 408
Application +4 | -1
[0    ab    O]4=I{\left[ {\begin{array}{cccccccccccccccccccc} {0\;\;a}\\ {b\;\;O} \end{array}} \right]^4} = I, then
View answer and explanation
Correct option: D
ab=1ab = 1
Option D (ab=1ab = 1) is correct because after calculating the fourth power of the matrix and equating it to the identity matrix, we find that ab=1ab = 1.
315 Medium
315 of 408
Application +4 | -1
The number of integers xx satisfying
−3x4+det⁡[1xx21x2x41x3x6]=0 is equal to -3 x^4+\operatorname{det}\left[\begin{array}{ccc} 1 & x & x^2 \\ 1 & x^2 & x^4 \\ 1 & x^3 & x^6 \end{array}\right]=0 \text { is equal to }
View answer and explanation
Correct option: A
2
Option C (1, -4) is correct because by finding the values of p, c, b, and q from the given roots, we can form the equation x2+bx+c=0x^2 + bx + c = 0 and solve for its roots.
316 Medium
316 of 408
Application +4 | -1
Let A  be  a 3×3 A\,\,\mathrm{be}\,\,\mathrm{a}\,3 \times 3\, matrix such that adj A = [2−11−1021−2−1]  and B=  adj  (adj A) adj\,A\, = \,\left[ {\begin{array}{cccccccccccccccccccc} 2&{ - 1}&1\\ { - 1}&0&2\\ 1&{ - 2}&{ - 1} \end{array}} \right]\,\,\mathrm{and}\,B = \,\,\mathrm{adj}\,\,\mathrm{(adj}\,A\mathrm{)}\, If ∣A∣ = λ and ∣(B−1)T∣ = μ,\mathrm{If}\,\left| A \right|\,\mathrm{ = }\,\lambda \,\mathrm{and}\,\left| {{{({B^{ - 1}})}^T}} \right|\,\mathrm{ = }\,\mu, then the ordered pair,  (∣λ∣, μ)\,\left( {\left| \lambda \right|,\,\mu } \right) is equal to
View answer and explanation
Correct option: D
(3, 181)\left( {3,\,\frac{1}{{81}}} \right)
Option D (3,181)\left( {3,\frac{{1}}{{81}}} \right) is correct because by using the properties of adjoint and determinant of a matrix, we can find the values of |λ| and μ.
317 Medium
317 of 408
Application +4 | -1
The matrix XX for which [1−43−2]X=[−16−672]\left[\begin{array}{cc}1 & -4 \\ 3 & -2\end{array}\right] X=\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right] is
View answer and explanation
Correct option: C
[621122]\left[\begin{array}{cc}6 & 2 \\ \frac{11}{2} & 2\end{array}\right]
Option C (\left[\begin{{array}}{{cc}}6 & 2 \\ \frac{{11}}{{2}} & 2\end{{array}}\right]\left[\begin{{array}}{{cc}}6 & 2 \\ \frac{{11}}{{2}} & 2\end{{array}}\right]) is correct because it is the result of multiplying the inverse of the matrix \left[\begin{{array}}{{cc}}1 & -4 \\ 3 & -2\end{{array}}\right]\left[\begin{{array}}{{cc}}1 & -4 \\ 3 & -2\end{{array}}\right] by the matrix \left[\begin{{array}}{{cc}}-16 & -6 \\ 7 & 2\end{{array}}\right]\left[\begin{{array}}{{cc}}-16 & -6 \\ 7 & 2\end{{array}}\right].
318 Medium
318 of 408
Analyzing +4 | -1
Let A=∣2b1bb2+1b1b2∣A = \left| {\begin{array}{cccccccccccccccccccc} 2&b&1\\ b&{{b^2} + 1}&b\\ 1&b&2 \end{array}} \right|, where b>0b > 0. Then, the minimum value of det⁡(A)b\frac{{\det \left( A \right)}}{b} is
View answer and explanation
Correct option: C
232\sqrt 3
Option C (232\sqrt 3) is correct because the minimum value of the expression is found using AM-GM inequality, resulting in 232\sqrt 3.
319 Medium
319 of 408
Analyzing +4 | -1
The determinant Δ=∣a2+x2abacabb2+x2bcacbcc2+x2∣\Delta = \left| {\begin{array}{cccccccccccccccccccc} {{a^2} + {x^2}}&{ab}&{ac}\\ {ab}&{{b^2} + {x^2}}&{bc}\\ {ac}&{bc}&{{c^2} + {x^2}} \end{array}} \right| is divisible by
View answer and explanation
Correct option: B
x4{x^4}
Option B (x4{x^4}) is correct because after applying column operations and simplifying, the determinant is shown to be divisible by x4{x^4}.
320 Medium
320 of 408
Analyzing +4 | -1
The value of k∈Rk \in R, for which the following system of linear equations 3x−y+4z=3,x+2y−3z=−2,6x+5y+kz=−33 x-y+4 z=3, x+2 y-3 z=-2,6 x+5 y+k z=-3, has infinitely many solutions, is:
View answer and explanation
Correct option: A
-5
Option A (-5) is the correct value of k because it makes the system of linear equations have infinitely many solutions by making the equations linearly dependent.
321 Medium
321 of 408
Application +4 | -1
If MM is square matrix such that M2=MM^2=M, then (I+M)3−7M(I+M)^3-7 M is equal to
View answer and explanation
Correct option: C
I
Option C (I) is correct because after simplifying the expression (I+M)3−7M(I+M)^3-7M using the property M2=MM^2=M, the result is the identity matrix I.
322 Medium
322 of 408
Application +4 | -1
If x,y,zx,y,z are in AP, then the value of the det AA is, where A=∣456x567y678zxyz0∣A = \left| {\begin{array}{cccccccccccccccccccc} 4&5&6&x\\ 5&6&7&y\\ 6&7&8&z\\ x&y&z&0 \end{array}} \right|
View answer and explanation
Correct option: A
0
Option A (0) is correct because since x,y,zx, y, z are in AP, 2y=x+z2y = x + z. Applying row operations makes the second row zero, resulting in a zero determinant.
323 Medium
323 of 408
Analyzing +4 | -1
Let PP be a square matrix such that P2=I−PP^{2}=I-P. For α,β,γ,δ∈N,\alpha, \beta, \gamma, \delta \in \mathbb{N}, if Pα+Pβ=γI−29PP^{\alpha}+P^{\beta}=\gamma I-29 P and Pα−Pβ=δI−13PP^{\alpha}-P^{\beta}=\delta I-13 P, then α+β+γ−δ\alpha+\beta+\gamma-\delta is equal to
View answer and explanation
Correct option: D
24
The correct answer is D (24) because by repeatedly applying the given relation P2 = I - P, we can find expressions for P6 and P8 in terms of I and P. Then, solving for alpha, beta, gamma, and delta and summing them gives 24.
324 Medium
324 of 408
Application +4 | -1
The number of 2×22 \times 2 matrices XX satistying the matrix equation X2=IX^2=I ( I\mathrm{I} is 2×22 \times 2 unit matrix) is
View answer and explanation
Correct option: D
infinite
Option D (infinite) is correct because there are infinitely many 2x2 matrices that satisfy the equation X2 = I, as shown by the case where a + d = 0.
325 Medium
325 of 408
Understanding +4 | -1
The matrix [cos⁡αsin⁡α−sin⁡αcos⁡α]\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right] is -
View answer and explanation
Correct option: C
orthogonal
Option C (orthogonal) is correct because the matrix satisfies the condition A*A' = I, where A' is the transpose of A and I is the identity matrix.
326 Medium
326 of 408
Analyzing +4 | -1
Let A=(1234)A=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right) and B=(a00b),a,b∈NB=\left(\begin{array}{ll}a & 0 \\ 0 & b\end{array}\right), a, b \in N. Then
View answer and explanation
Correct option: C
there exist infinitely many B′\mathrm{B}^{\prime} such that AB=BA\mathrm{AB}=\mathrm{BA}
The correct answer is that there exist infinitely many BB's such that AB=BAAB = BA because AB=BAAB = BA only when a=ba = b, and a,b∈Na, b \in N.
327 Medium
327 of 408
Analyzing +4 | -1
If A=(3−342−340−11)A=\left(\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right) then value of trace of the matrix adj⁡(adj⁡A)\operatorname{adj}(\operatorname{adj} A) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the adj(adj A) simplifies to A, and the trace of A is the sum of its diagonal elements.
328 Medium
328 of 408
Analyzing +4 | -1
The determinant Δ=∣abax+bbcbx+cax+bbx+c0∣\Delta = \left| {\begin{array}{cccccccccccccccccccc} a&b&{ax + b}\\ b&c&{bx + c}\\ {ax + b}&{bx + c}&0 \end{array}} \right| is equal to zero, if
View answer and explanation
Correct option: B
a,b,c,a,b,c, are in G.P.
The correct answer is B (a,b,c,a,b,c, are in G.P.). The determinant simplifies to zero if a,b,ca, b, c are in G.P., i.e., b² = ac.
329 Medium
329 of 408
Application +4 | -1
If [1x1][1322511532][12x]=[0]\left[\begin{array}{lll}1 & x & 1\end{array}\right]\left[\begin{array}{rrr}1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2\end{array}\right]\left[\begin{array}{l}1 \\ 2 \\ x\end{array}\right]=[0] then x=x =
View answer and explanation
Correct option: C
−2,−14-2,-14
Option C (−2,−14-2, -14) is correct because solving the matrix equation leads to a quadratic equation whose roots are -2 and -14.
330 Medium
330 of 408
Analyzing +4 | -1
Let ∣x2+x2x−1x+33x+12+x2x3−3x−3x2+42x∣\left|\begin{array}{ccc}x^2+x & 2 x-1 & x+3 \\ 3 x+1 & 2+x^2 & x^3-3 \\ x-3 & x^2+4 & 2 x\end{array}\right|
=px7+qx6+rx5+sx4+tx3+ux2+vx+w, =p x^7+q x^6+r x^5+s x^4+t x^3+u x^2+v x+w \text {, }
then which of the following is not true?
View answer and explanation
Correct option: B
q=0,s=−4q=0, s=-4
Option B ((∼p)∨(∼q)(\sim p) \vee (\sim q)) is correct because the given expression simplifies to this using logical equivalences.
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