Matrices and Determinants jee mains questions | JEE Main Maths Matrices and Determinants Previous Year Question

Q 1 :

For $α, β \in R$ and a natural number $n$ , let $A_{r} =$ $| \begin{matrix} r & 1 & \frac{n^{2}}{2} + α \\ 2 r & 2 & n^{2} - β \\ 3 r - 2 & 3 & \frac{n (3 n - 1)}{2} \end{matrix} |$ . Then $2 A_{10} - A_{8}$ is [2024]

$0$
$4 α + 2 β$
$2 n$
$2 α + 4 β$

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(2)

We have, $2 A_{10} - A_{8} =$ $| \begin{matrix} 20 & 1 & \frac{n^{2}}{2} + α \\ 40 & 2 & n^{2} - β \\ 56 & 3 & \frac{n (3 n - 1)}{2} \end{matrix} |$ $-$ $| \begin{matrix} 8 & 1 & \frac{n^{2}}{2} + α \\ 16 & 2 & n^{2} - β \\ 22 & 3 & \frac{n (3 n - 1)}{2} \end{matrix} |$

$=$ $| \begin{matrix} 20 - 8 & 1 & \frac{n^{2}}{2} + α \\ 40 - 16 & 2 & n^{2} - β \\ 56 - 22 & 3 & \frac{n (3 n - 1)}{2} \end{matrix} |$ $=$ $| \begin{matrix} 12 & 1 & \frac{n^{2}}{2} + α \\ 24 & 2 & n^{2} - β \\ 34 & 3 & \frac{n (3 n - 1)}{2} \end{matrix} |$

$=$ $| \begin{matrix} 0 & 1 & \frac{n^{2}}{2} + α \\ 0 & 2 & n^{2} - β \\ - 2 & 3 & \frac{n (3 n - 1)}{2} \end{matrix} |$ $(Applying C_{1} \to C_{1} - 12 C_{2})$

$= - 2 (n^{2} - β - n^{2} - 2 α)$

$= - 2 (- β - 2 α) = 4 α + 2 β$

Q 2 :

Let $A = [\begin{matrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{matrix}] .$ If $A^{3} = 4 A^{2} - A - 21 I,$ where $I$ is the identity matrix of order $3 \times 3$ , then $2 a + 3 b$ is equal to [2024]

$- 12$
$- 13$
$- 10$
$- 9$

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(2)

$A = [\begin{matrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{matrix}],$ $A^{3} = 4 A^{2} - A - 21 I$

characteristic equation of $A$ is given by

$x^{3} - 4 x^{2} + x + 21 = 0$

Sum of roots = 4 = trace of A

$\Rightarrow 2 + 3 + b = 4 \Rightarrow b = - 1$

Also, product of roots = $- 21$ = det A

$\Rightarrow 2 (3 b - 5) - a (b) = - 21 \Rightarrow - 6 - 10 + a = - 21$

$\Rightarrow a = - 21 + 16 = - 5$ $∴$ $2 a + 3 b = - 10 - 3 = - 13$

Q 3 :

If $α \neq a,$ $β \neq b,$ $γ \neq c$ and $| \begin{matrix} α & b & c \\ a & β & c \\ a & b & γ \end{matrix} |$ $= 0,$ then $\frac{a}{α - a} + \frac{b}{β - b} + \frac{γ}{γ - c}$ is equal to: [2024]

1
2
0
3

1

2

3

4

(3)

$| \begin{matrix} α & b & c \\ a & β & c \\ a & b & γ \end{matrix} |$ $= 0$

$R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}$

$\Rightarrow$ $| \begin{matrix} α - a & b - β & 0 \\ 0 & β - b & c - γ \\ a & b & γ \end{matrix} |$ $= 0$

Taking out $α - a, β - b$ and $γ - c$ common from column-1, 2 and 3 respectively,

$\Rightarrow (α - a) (β - b) (γ - c)$ $| \begin{matrix} 1 & - 1 & 0 \\ 0 & 1 & - 1 \\ \frac{a}{α - a} & \frac{b}{β - b} & \frac{γ}{γ - c} \end{matrix} |$ $= 0$

$\Rightarrow (α - a) (β - b) (γ - c) [1 (\frac{γ}{γ - c} + \frac{b}{β - b}) + 1 (0 + \frac{a}{α - a})] = 0$

$\Rightarrow (α - a) (β - b) (γ - c) [\frac{γ}{γ - c} + \frac{b}{β - b} + \frac{a}{α - a}] = 0$

$\Rightarrow [\frac{γ}{γ - c} + \frac{b}{β - b} + \frac{a}{α - a}] = 0$ $(∵ α \neq a, β \neq b, γ \neq c)$

$∴$ $\frac{γ}{γ - c} + \frac{b}{β - b} + \frac{a}{α - a} = 0$

Q 4 :

If $A = [\begin{matrix} \sqrt{2} & 1 \\ - 1 & \sqrt{2} \end{matrix}], B = [\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}], C = A B A^{T}$ and $X = A^{T} C^{2} A,$ then det $X$ is equal to [2024]

729
243
891
27

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2

3

4

(1)

Given, $X = A^{T} C^{2} A$ and $C = A B A^{T}$

$\Rightarrow | X | = | A^{T} C^{2} A | \Rightarrow | X | = | A | | C |^{2} | A^{T} |$

$\Rightarrow | C | = | A B A^{T} | = | A | | B | | A^{T} |$

As $| A | = 3, | A^{T} | = 3, | B | = 1, | C | = 3 \times 1 \times 3 = 9$

$∴$ $| X | = 3 \times 9^{2} \times 3 = 9 \times 81 = 729$

Q 5 :

The values of $α,$ for which $| \begin{matrix} 1 & \frac{3}{2} & α + \frac{3}{2} \\ 1 & \frac{1}{3} & α + \frac{1}{3} \\ 2 α + 3 & 3 α + 1 & 0 \end{matrix} |$ $= 0,$ lie in the interval [2024]

$(- \frac{3}{2}, \frac{3}{2})$
$(- 2, 1)$
$(- 3, 0)$
$(0, 3)$

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(3)

Given, $| \begin{matrix} 1 & \frac{3}{2} & α + \frac{3}{2} \\ 1 & \frac{1}{3} & α + \frac{1}{3} \\ 2 α + 3 & 3 α + 1 & 0 \end{matrix} |$ $= 0$

$\Rightarrow (2 α + 3) (\frac{7 α}{6}) - (3 α + 1) (\frac{- 7}{6}) = 0$

$\Rightarrow 14 α^{2} + 42 α + 7 = 0 \Rightarrow 2 α^{2} + 6 α + 1 = 0$

$Roots of this equation are α = \frac{- 3 \pm \sqrt{7}}{2}$

$\frac{- 3 + \sqrt{7}}{2} \in (- 3, 0) and \frac{- 3 - \sqrt{7}}{2} \in (- 3, 0)$

Q 6 :

Let $A = [\begin{matrix} 1 & 0 & 0 \\ 0 & α & β \\ 0 & β & α \end{matrix}]$ and $| 2 A |^{3} = 2^{21}$ where $α, β \in Z .$ Then a value of $α$ is [2024]

5
9
17
3

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2

3

4

(1)

Given, $A = [\begin{matrix} 1 & 0 & 0 \\ 0 & α & β \\ 0 & β & α \end{matrix}]$

$\Rightarrow | A | = α^{2} - β^{2}$ ...(i)

Also, $| 2 A |^{3} = 2^{21}$

$\Rightarrow {(2^{3} | A |)}^{3} = 2^{21} \Rightarrow | A |^{3} = 2^{12} \Rightarrow | A | = 2^{4}$

From (i), we get, $α^{2} - β^{2} = 2^{4}$

$\Rightarrow α^{2} - β^{2} = 16 \Rightarrow (α + β) (α - β) = 8 \times 2$

So, $α + β = 8 and α - β = 2$

Solving, we get, $α = 5$ and $β = 3$

Q 7 :

Let $A = [\begin{matrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{matrix}]$ and $P = [\begin{matrix} 1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5 \end{matrix}] .$ The sum of the prime factors of $| P^{- 1} A P - 2 I |$ is equal to [2024]

66
27
26
23

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(3)

$P^{- 1} A P - 2 I = P^{- 1} (A - 2 I) P$

$\Rightarrow | P^{- 1} A P - 2 I | = | P^{- 1} | | A - 2 I | | P |$

$= | A - 2 I |$ $(∵ | P^{- 1} | \cdot | P | = 1)$

$∴$ $\det (A - 2 I) =$ $| \begin{matrix} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{matrix} |$

$= - 1 (- 33) + 2 (18) = 69 = 3 \times 23$

Sum of prime factors = 23 + 3 = 26

Q 8 :

Let $A$ be a $3 \times 3$ matrix of non-negative real elements such that $A [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = 3 [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] .$ Then the maximum value of $d e t (A)$ is _______. [2024]

(27)

Let $A = [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}]$

Now, $A [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = 3 [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] \Rightarrow [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}] [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = [\begin{matrix} 3 \\ 3 \\ 3 \end{matrix}]$

$a_{11} + a_{12} + a_{13} = 3$

$a_{21} + a_{22} + a_{23} = 3$

$a_{31} + a_{32} + a_{33} = 3$

Now, for maximum value of $d e t (A) = a_{i j} = {\begin{matrix} 0, & i \neq j \\ 3, & i = j \end{matrix}$

$∴$ $| A | = 27$

Q 9 :

Let $A = [\begin{matrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}],$ $B = [\begin{matrix} B_{1}, & B_{2}, & B_{3} \end{matrix}]$ , where $B_{1}, B_{2}, B_{3}$ are column matrices, and $A B_{1} = [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}], A B_{2} = [\begin{matrix} 2 \\ 3 \\ 0 \end{matrix}], A B_{3} = [\begin{matrix} 3 \\ 2 \\ 1 \end{matrix}] .$ If $α = | B |$ and $β$ is the sum of all the diagonal elements of B, then $α^{3} + β^{3}$ is equal to _____ . [2024]

(28)

Let $B_{1} = [\begin{matrix} a_{1} \\ b_{1} \\ c_{1} \end{matrix}],$ $A B_{1} = [\begin{matrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}] [\begin{matrix} a_{1} \\ b_{1} \\ c_{1} \end{matrix}] = [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}]$

$\Rightarrow 2 a_{1} + c_{1} = 1, a_{1} + b_{1} = 0, a_{1} + c_{1} = 0$

On solving, we get $a_{1} = 1, b_{1} = - 1, c_{1} = - 1$ ...(i)

Similarly, let $B_{2} = [\begin{matrix} a_{2} \\ b_{2} \\ c_{2} \end{matrix}], A B_{2} = [\begin{matrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}] [\begin{matrix} a_{2} \\ b_{2} \\ c_{2} \end{matrix}] = [\begin{matrix} 2 \\ 3 \\ 0 \end{matrix}]$

$\Rightarrow 2 a_{2} + c_{2} = 2, a_{2} + b_{2} = 3$ and $a_{2} + c_{2} = 0$

On solving, we get $a_{2} = 2, b_{2} = 1$ and $c_{2} = - 2$ ...(ii)

Similarly, let $B_{3} = [\begin{matrix} a_{3} \\ b_{3} \\ c_{3} \end{matrix}], A B_{3} = [\begin{matrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}] [\begin{matrix} a_{3} \\ b_{3} \\ c_{3} \end{matrix}] = [\begin{matrix} 3 \\ 2 \\ 1 \end{matrix}]$

$\Rightarrow 2 a_{3} + c_{3} = 3, a_{3} + b_{3} = 2$ and $a_{3} + c_{3} = 1$

$\Rightarrow a_{3} = 2, b_{3} = 0, c_{3} = - 1$ ...(iii)

Thus, $B = [\begin{matrix} 1 & 2 & 2 \\ - 1 & 1 & 0 \\ - 1 & - 2 & - 1 \end{matrix}]$ [By using (i), (ii) & (iii)]

$α = | B | = 1 (- 1 + 0) - 2 (1 - 0) + 2 (2 + 1) = - 1 - 2 + 6 = 3$

and $β =$ sum of diagonal elements of $B = 1 + 1 + (- 1) = 1$

Hence, $α^{3} + β^{3} = {(3)}^{3} + {(1)}^{3} = 28$

Q 10 :

Let $A$ be a $2 \times 2$ real matrix and $I$ be the identity matrix of order 2. If the roots of the equation $| A - x I | = 0$ be $- 1$ and $3$ , then the sum of the diagonal elements of the matrix $A^{2}$ is _______. [2024]

(10)

Let $A = {[\begin{matrix} a & b \\ c & d \end{matrix}]}_{2 \times 2}, I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

Roots are $- 1, 3$

Sum of roots $= - 1 + 3 = 2 = t r (A)$

$a + d = 2$ ...(i)

Product of roots $= (- 1) \times 3 = - 3 = \det (A)$

$a d - b c = - 3$ ...(ii)

$A^{2} = [\begin{matrix} a & b \\ c & d \end{matrix}] [\begin{matrix} a & b \\ c & d \end{matrix}] = [\begin{matrix} a^{2} + b c & a b + b d \\ a c + d c & b c + d^{2} \end{matrix}]$

$tr (A^{2}) = a^{2} + 2 b c + d^{2} = {(a + d)}^{2} - 2 a d + 2 b c$

$= 2^{2} - 2 (- 3) = 4 + 6 = 10$

Q 11 :

Let $A = [\begin{matrix} α & - 1 \\ 6 & β \end{matrix}], α > 0$ , such that det(A) = 0 and $α + β = 1 .$ If $I$ denotes $2 \times 2$ identity matrix, then the matrix ${(I + A)}^{8}$ is: [2025]

$[\begin{matrix} 257 & - 64 \\ 514 & - 127 \end{matrix}]$
$[\begin{matrix} 1025 & - 511 \\ 2024 & - 1024 \end{matrix}]$
$[\begin{matrix} 4 & - 1 \\ 6 & - 1 \end{matrix}]$
$[\begin{matrix} 766 & - 255 \\ 1530 & - 509 \end{matrix}]$

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(4)

We have, $A = [\begin{matrix} α & - 1 \\ 6 & β \end{matrix}], | A | = 0$

$\Rightarrow α β + 6 = 0 \Rightarrow α β = - 6$

Also, $α + β = 1$ [Given]

$\Rightarrow α = 3 and β = - 2$ ( $∵ α > 0$ )

Now, $A = [\begin{matrix} 3 & - 1 \\ 6 & - 2 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 3 & - 1 \\ 6 & - 2 \end{matrix}] [\begin{matrix} 3 & - 1 \\ 6 & - 2 \end{matrix}] = [\begin{matrix} 3 & - 1 \\ 6 & - 2 \end{matrix}]$

$\Rightarrow A^{2} = A$

So, $A = A^{2} = A^{3} = A^{4} = A^{5} = A^{6} = A^{7}$

$∴ {(I + A)}^{8} = I + {}^{8}C_{1} A^{7} + {}^{8}C_{2} A^{6} + . . . + {}^{8}C_{8} A^{8}$

$= I + A ({}^{8}C_{1} + {}^{8}C_{2} + . . . + {}^{8}C_{8}) = I + A (2^{8} - 1)$

$= [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + [\begin{matrix} 765 & - 255 \\ 1530 & - 510 \end{matrix}] = [\begin{matrix} 766 & - 255 \\ 1530 & - 509 \end{matrix}]$ .

Q 12 :

For some, a, b, let $f (x) =$ $| \begin{matrix} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{matrix} |$ $, x \neq 0, \lim_{x \to 0} f (x) = λ + μ a + ν b$ . Then ${(λ + μ + ν)}^{2}$ is equal to : [2025]

25
9
16
36

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2

3

4

(3)

$f (x) =$ $| \begin{matrix} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{matrix} |$ $, x \neq 0$

Since, $\lim_{x \to 0} f (x) = λ + μ a + ν b$

$\Rightarrow λ + μ a + ν b = \lim_{x \to 0}$ $| \begin{matrix} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{matrix} |$

$=$ $| \begin{matrix} a + 1 & 1 & b \\ a & 1 + 1 & b \\ a & 1 & b + 1 \end{matrix} |$ $=$ $| \begin{matrix} a + 1 & 1 & b \\ a & 2 & b \\ a & 1 & b + 1 \end{matrix} |$

$= (a + 1) [2 (b + 1) - b] - 1 [a (b + 1) - a b] + b \times (a - 2 a)$

$= (a + 1) (b + 2) - a - a b = a + b + 2$

On comparing, we get

$λ = 2, μ = 1 and ν = 1$

Hence, ${(λ + μ + ν)}^{2} = {(2 + 1 + 1)}^{2} = 16$ .

Q 13 :

Let $A = [a_{i j}] = [\begin{matrix} \log_{5} 128 & \log_{4} 5 \\ \log_{5} 8 & \log_{4} 25 \end{matrix}]$ . If $A_{i j}$ is the cofactor of $a_{i j}, C_{i j} = \sum_{k = 1}^{2} a_{i k} A_{j k}, 1 \leq i, j \leq 2$ , and $C = [C_{i j}]$ , then $8$ $| C |$ is equal to : [2025]

242
288
262
222

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2

3

4

(1)

From the given matrix A, $| A |$ $= \frac{11}{2}$

Here, $C_{i j} = \sum_{k = 1}^{2} a_{i k} A_{j k}, 1 \leq i, j \leq 2$

$C_{11} = \sum_{k = 1}^{2} a_{1 k} A_{1 k} = a_{11} A_{11} + a_{12} A_{12} = | A | = \frac{11}{2}$

$C_{12} = \sum_{k = 1}^{2} a_{1 k} A_{2 k} = 0$

$C_{21} = \sum_{k = 1}^{2} a_{2 k} A_{1 k} = 0$

$C_{22} = \sum_{k = 1}^{2} a_{2 k} A_{2 k} = | A | = \frac{11}{2}$

Now, $C = [\begin{matrix} 11 / 2 & 0 \\ 0 & 11 / 2 \end{matrix}] \Rightarrow$ $| C |$ $= \frac{121}{4}$

Hence, $8$ $| C |$ $= 242$ .

Q 14 :

Let M and m respectively be the maximum and the minimum values of $f (x) =$ $| \begin{matrix} 1 + \sin^{2} x & \cos^{2} x & 4 \sin 4 x \\ \sin^{2} x & 1 + \cos^{2} x & 4 \sin 4 x \\ \sin^{2} x & \cos^{2} x & 1 + 4 \sin 4 x \end{matrix} |$ $, x \in R$ . Then $M^{4} - m^{4}$ is equal to : [2025]

1295
1040
1215
1280

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2

3

4

(4)

$f (x) =$ $| \begin{matrix} 1 + \sin^{2} x & \cos^{2} x & 4 \sin 4 x \\ \sin^{2} x & 1 + \cos^{2} x & 4 \sin 4 x \\ \sin^{2} x & \cos^{2} x & 1 + 4 \sin 4 x \end{matrix} |$

$=$ $| \begin{matrix} 2 & \cos^{2} x & 4 \sin 4 x \\ 2 & 1 + \cos^{2} x & 4 \sin 4 x \\ 1 & \cos^{2} x & 1 + 4 \sin 4 x \end{matrix} |$ $(Applying C_{1} \to C_{1} + C_{2})$

$=$ $| \begin{matrix} 2 & \cos^{2} x & 4 \sin 4 x \\ 0 & 1 & 0 \\ 1 & \cos^{2} x & 1 + 4 \sin 4 x \end{matrix} |$ $(Applying R_{2} \to R_{2} - R_{1})$

Expanding along $R_{2}$ , we get f(x) = 2(1 + 4 sin 4x) – 4 sin 4x

$\Rightarrow f (x) = 2 + 4 \sin 4 x$

Maximum value of f(x), M = 6

Minimum value of f(x), m = –2

$∴ M^{4} - m^{4} = 1280$ .

Q 15 :

Let $I$ be the identity matrix of order $3 \times 3$ and for the matrix $A = [\begin{matrix} λ & 2 & 3 \\ 4 & 5 & 6 \\ 7 & - 1 & 2 \end{matrix}], | A | = - 1$ . Let B be the inverse of the matrix adj $(A a d j (A^{2}))$ . Then $| (λ B + I) |$ is equal to __________. [2025]

(38)

Given, $A = [\begin{matrix} λ & 2 & 3 \\ 4 & 5 & 6 \\ 7 & - 1 & 2 \end{matrix}]$

$\Rightarrow | A | = [\begin{matrix} λ & 2 & 3 \\ 4 & 5 & 6 \\ 7 & - 1 & 2 \end{matrix}] = - 1$

$\Rightarrow λ (16) - 2 (- 34) + 3 (- 39) = - 1$

$\Rightarrow 16 λ = 48 \Rightarrow λ = 3$

Given, $B^{- 1} = adj (A \cdot adj (A^{2}))$

Let $C = A \cdot adj (A^{2})$

$A C = A^{2} adj (A^{2}) = | A |^{2} \cdot I = I$

$\Rightarrow C = A^{- 1}$ [ $∵$ |A| = – 1]

Now, $B^{- 1} = adj (A^{- 1}) \Rightarrow B = adj (A)$

Now, $λ B + I = 3 B + I$

Let P = 3B + $I$

$\Rightarrow$ P = 3 adj(A) + $I$

$\Rightarrow$ AP = A 3 adj(A)+ (A)

$\Rightarrow$ AP = 3 $| A |$ $\cdot$ $I$ + A $\Rightarrow$ AP = A – 3 $I$

$\Rightarrow$ $| A P |$ $=$ $| A - 3 I |$

$| A |$ $\cdot$ $| P |$ $=$ $[\begin{matrix} 0 & 2 & 3 \\ 4 & 2 & 6 \\ 7 & - 1 & - 1 \end{matrix}]$

= 0 – 2(–46) + 3(–18)

= 92 – 54 = 38

$\Rightarrow$ $| P |$ = 38.

Q 16 :

Let integers $a, b \in [- 3, 3]$ be such that $a + b \neq 0$ . Then the number of all possible ordered pairs (a, b) for which $| \frac{z - a}{z + b} |$ $= 1$ and $| \begin{matrix} z + 1 & ω & ω^{2} \\ ω & z + ω^{2} & 1 \\ ω^{2} & 1 & z + ω \end{matrix} |$ $= 1, z \in C$ , where $ω$ and $ω^{2}$ are the roots of $x^{2} + x + 1 = 0$ , is equal to __________. [2025]

(10)

We have, $a, b \in z, - 3 \leq a, b \leq 3$ and $a + b \neq 0$ .

Also, $| \frac{z - a}{z + b} |$ $= 1$ and $| \begin{matrix} z + 1 & ω & ω^{2} \\ ω & z + ω^{2} & 1 \\ ω^{2} & 1 & z + ω \end{matrix} |$ $= 1$

Applying $C_{1} \to C_{1} + C_{2} + C_{3}$

$z$ $| \begin{matrix} 1 & ω & ω^{2} \\ 1 & z + ω^{2} & 1 \\ 1 & 1 & z + ω \end{matrix} |$ $= 1$ [ $∵ 1 + ω + ω^{2} = 0$ ]

On expanding, we get

$z [(z^{2} + (ω^{2} + ω) z + ω^{3} - 1) - ω z - ω^{2} - ω^{2} + ω - ω^{2} z - ω] = 1$

$\Rightarrow z (z^{2}) = 1 \Rightarrow z^{3} = 1 \Rightarrow z = 1, ω, ω^{2}$

Case 1 : $z = ω$ , then $a + b \neq 0$ and a – b = –1

a = –3, b = –2; a = –2; b = –1;

a = –1, b = 0; a = 0, b = 1

a = 1, b = 2; a = 2, b = 3

Case 2 : z = 1; then a – b = 2 and $a + b \neq 0$

a = –1, b = –3; a = 0, b = –2; a = 2, b = 0; a = 3, b = 1

Total pairs = 10.

Q 17 :

Let A be a $2 \times 2$ matrix with real entries such that $A' = α A + I,$ where $α \in R - {- 1, 1}$ . If $\det (A^{2} - A) = 4$ , then the sum of all possible values of $α$ is equal to [2023]

0
$\frac{3}{2}$
$\frac{5}{2}$
2

1

2

3

4

(3)

We have, $A^{T} = α A + I$

$∴$ $A = α A^{T} + I \Rightarrow A = α (α A + I) + I$

$\Rightarrow A = α^{2} A + (α + 1) I \Rightarrow A (1 - α^{2}) = (α + 1) I$

$\Rightarrow A = \frac{I}{1 - α},$ Now, $| A |$ $= \frac{1}{{(1 - α)}^{2}} ...(i)$

$A - I = \frac{I}{1 - α} - I = \frac{α}{1 - α} I$

We know that

$| A^{2} - A |$ $= | A | | A - I |$ ...(ii)

$∴$ $| A - I |$ $= {(\frac{α}{1 - α})}^{2} ...(iii)$

Now, $| A^{2} - A |$ $= 4$ [ $∵$ Given]

$\Rightarrow \frac{1}{{(1 - α)}^{2}} \times \frac{α^{2}}{{(1 - α)}^{2}} = 4$ [Using (i), (ii) and (iii)]

$\Rightarrow \frac{α}{{(1 - α)}^{2}} = \pm 2 \Rightarrow 2 {(1 - α)}^{2} = \pm α$

Case I: $2 {(1 - α)}^{2} = α$ $\Rightarrow 2 α^{2} - 5 α + 2 = 0$

$∴$ $α_{1} + α_{2} = \frac{5}{2}$

Case II: $2 {(1 - α)}^{2} = - α$

$\Rightarrow 2 α^{2} - 3 α + 2 = 0 \Rightarrow α \notin R$

$∴$ $Sum of values of α = \frac{5}{2}$

Q 18 :

$If$ $| \begin{matrix} x + 1 & x & x \\ x & x + λ & x \\ x & x & x + λ^{2} \end{matrix} |$ $= \frac{9}{8} (103 x + 81), then λ, \frac{λ}{3} are the roots of the equation$ [2023]

$4 x^{2} - 24 x + 27 = 0$
$4 x^{2} + 24 x + 27 = 0$
$4 x^{2} - 24 x - 27 = 0$
$4 x^{2} + 24 x - 27 = 0$

1

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(1)

$| \begin{matrix} x + 1 & x & x \\ x & x + λ & x \\ x & x & x + λ^{2} \end{matrix} |$

$=$ $| \begin{matrix} 1 & - λ & 0 \\ 0 & λ & - λ^{2} \\ x & x & x + λ^{2} \end{matrix} |$ $[Applying R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}]$

$= (1) (x λ + λ^{3} + x λ^{2}) + λ (x λ^{2})$

$= λ^{3} + x (λ + λ^{2} + λ^{3}) = \frac{9}{8} \times 103 x + {(\frac{9}{2})}^{3} (Given)$

$\Rightarrow λ = \frac{9}{2};$ $Roots of the equation are \frac{9}{2} and \frac{3}{2}$

$Required quadratic equation is (x - \frac{9}{2}) (x - \frac{3}{2}) = 0$

$\Rightarrow (2 x - 9) (2 x - 3) = 0 \Rightarrow 4 x^{2} - 24 x + 27 = 0$

Q 19 :

Let $α$ be a root of the equation $(a - c) x^{2} + (b - a) x + (c - b) = 0$ where $a, b, c$ are distinct real numbers such that the matrix $[\begin{matrix} α^{2} & α & 1 \\ 1 & 1 & 1 \\ a & b & c \end{matrix}]$ is singular. Then, the value of $\frac{{(a - c)}^{2}}{(b - a) (c - b)} + \frac{{(b - a)}^{2}}{(a - c) (c - b)} + \frac{{(c - b)}^{2}}{(a - c) (b - a)}$ is [2023]

6
3
9
12

1

2

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(2)

$Given matrix is singular,$

$∴$ $| \begin{matrix} α^{2} & α & 1 \\ 1 & 1 & 1 \\ a & b & c \end{matrix} |$ $= 0$

$\Rightarrow α^{2} (c - b) - α (c - a) + 1 (b - a) = 0$

$It is singular if α = 1 .$

$\Rightarrow c - b - c + a + b - a = 0$

$Now,$ $\frac{{(a - c)}^{2}}{(b - a) (c - b)} + \frac{{(b - a)}^{2}}{(a - c) (c - b)} + \frac{{(c - b)}^{2}}{(a - c) (b - a)}$

$= \frac{{(a - c)}^{3}}{(a - c) (b - a) (c - b)} + \frac{{(b - a)}^{3}}{(a - c) (b - a) (c - b)} + \frac{{(c - b)}^{3}}{(a - c) (b - a) (c - b)}$

$= \frac{{(a - c)}^{3} + {(b - a)}^{3} + {(c - b)}^{3}}{(a - c) (b - a) (c - b)}$

$= \frac{3 (a - c) (b - a) (c - b)}{(a - c) (b - a) (c - b)} = 3$ $(∵ a - c + b - a + c - b = 0)$

Q 20 :

The set of all values of $t \in ℝ$ , for which the matrix $[\begin{matrix} e^{t} & e^{- t} (\sin t - 2 \cos t) & e^{- t} (- 2 \sin t - \cos t) \\ e^{t} & e^{- t} (2 \sin t + \cos t) & e^{- t} (\sin t - 2 \cos t) \\ e^{t} & e^{- t} \cos t & e^{- t} \sin t \end{matrix}]$ is invertible, is [2023]

${(2 k + 1) \frac{π}{2}, k \in ℤ}$
$ℝ$
${k π + \frac{π}{4}, k \in ℤ}$
${k π, k \in ℤ}$

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(2)

$If the matrix is invertible, then its determinant should not be zero.$

So, $| \begin{matrix} e^{t} & e^{- t} (\sin t - 2 \cos t) & e^{- t} (- 2 \sin t - \cos t) \\ e^{t} & e^{- t} (2 \sin t + \cos t) & e^{- t} (\sin t - 2 \cos t) \\ e^{t} & e^{- t} \cos t & e^{- t} \sin t \end{matrix} |$ $\neq 0$

$\Rightarrow e^{t} \times e^{- t} \times e^{- t}$ $| \begin{matrix} 1 & \sin t - 2 \cos t & - 2 \sin t - \cos t \\ 1 & 2 \sin t + \cos t & \sin t - 2 \cos t \\ 1 & \cos t & \sin t \end{matrix} |$ $\neq 0$

$Applying R_{1} \to R_{1} - R_{2} then R_{2} \to R_{2} - R_{3}, we get:$

$e^{- t}$ $| \begin{matrix} 0 & - \sin t - 3 \cos t & - 3 \sin t + \cos t \\ 0 & 2 \sin t & - 2 \cos t \\ 1 & \cos t & \sin t \end{matrix} |$ $\neq 0$

$\Rightarrow e^{- t} (2 \sin t \cos t + 6 \cos^{2} t + 6 \sin^{2} t - 2 \sin t \cos t) \neq 0$

$\Rightarrow e^{- t} \times 6 \neq 0, \forall t \in ℝ .$

Q 21 :

$Let D_{k} =$ $| \begin{matrix} 1 & 2 k & 2 k - 1 \\ n & n^{2} + n + 2 & n^{2} \\ n & n^{2} + n & n^{2} + n + 2 \end{matrix} |$ . $If \sum_{k = 1}^{n} D_{k} = 96, then n is equal to$ _________ . [2023]

(6)

$Given, D_{k} =$ $| \begin{matrix} 1 & 2 k & 2 k - 1 \\ n & n^{2} + n + 2 & n^{2} \\ n & n^{2} + n & n^{2} + n + 2 \end{matrix} |$

$∴$ $\sum_{k = 1}^{n} D_{k} =$ $| \begin{matrix} \sum_{k = 1}^{n} 1 & \sum_{k = 1}^{n} 2 k & \sum_{k = 1}^{n} (2 k - 1) \\ n & n^{2} + n + 2 & n^{2} \\ n & n^{2} + n & n^{2} + n + 2 \end{matrix} |$

$Apply \sum_{k = 1}^{n} 2 k = \frac{2 n (n + 1)}{2} = n (n + 1)$

$\sum_{k = 1}^{n} (2 k - 1) = \sum_{k = 1}^{n} 2 k - \sum_{k = 1}^{n} 1 = \frac{2 n (n + 1)}{2} - n = n^{2} + n - n = n^{2}$

$∴$ $\sum_{k = 1}^{n} D_{k} =$ $| \begin{matrix} n & n^{2} + n & n^{2} \\ n & n^{2} + n + 2 & n^{2} \\ n & n^{2} + n & n^{2} + n + 2 \end{matrix} |$

$Apply R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}$

$\Rightarrow$ $| \begin{matrix} n & n^{2} + n & n^{2} \\ 0 & 2 & 0 \\ 0 & 0 & n + 2 \end{matrix} |$ $= 96$

$\Rightarrow n (2 (n + 2) - 0) = 96 \Rightarrow n (n + 2) = 48$

$\Rightarrow n^{2} + 2 n - 48 = 0 \Rightarrow n^{2} + 8 n - 6 n - 48 = 0 \Rightarrow n (n + 8) - 6 (n + 8) = 0 \Rightarrow (n - 6) (n + 8) = 0$

$\Rightarrow n = 6, - 8$

$Since n cannot be negative, ∴$ $n = 6$

Q 22 :

Among the statements:

$I: If$ $| \begin{matrix} 1 & \cos α & \cos β \\ \cos α & 1 & \cos γ \\ \cos β & \cos γ & 1 \end{matrix} |$ $=$ $| \begin{matrix} 0 & \cos α & \cos β \\ \cos α & 0 & \cos γ \\ \cos β & \cos γ & 0 \end{matrix} |$ $, then \cos^{2} α + \cos^{2} β + \cos^{2} γ = \frac{3}{2},$ and

$II: If$ $| \begin{matrix} x^{2} + x & x + 1 & x - 2 \\ 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 \\ x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} |$ $= p x + q, then p^{2} = 196 q^{2} .$ [2026]

only I is true
both are false
only II is true
both are true

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(2)

$Let \cos α = x$

$\cos β = y$

$\cos γ = z$

$| \begin{matrix} 0 & x & y \\ x & 0 & z \\ y & z & 0 \end{matrix} |$ $=$ $| \begin{matrix} 1 & x & y \\ x & 1 & z \\ y & z & 1 \end{matrix} |$

Expanding both sides, we get

$x^{2} + y^{2} + z^{2} = 1$

$i.e. \cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$

Statement 1 is false.

Now,

$| \begin{matrix} x^{2} + x & 1 + x & x - 2 \\ 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 \\ x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} |$ $= p x + q$

Put $x = 0$ both sides

$q =$ $| \begin{matrix} 0 & 1 & - 2 \\ - 1 & 0 & - 3 \\ 3 & - 1 & - 1 \end{matrix} |$

$\Rightarrow q = - 12$

Now put $x = 1$ both sides

$p + q =$ $| \begin{matrix} 2 & 2 & - 1 \\ 4 & 3 & 3 \\ 6 & 1 & 1 \end{matrix} |$ $= 42$

$\Rightarrow p = 54$

Now,

$\frac{p^{2}}{q^{2}} = {(\frac{54}{- 12})}^{2} + 196$

$\Rightarrow p^{2} \neq 196 q^{2}$

Statement (2) is false.

Correct option (2).

Q 23 :

Let $| A |$ $= 6$ , where A is a $3 \times 3 matrix.$ If $| adj (3 adj (A^{2} \cdot adj (2 A))) |$ $= 2^{m} \cdot 3^{n}, m, n \in N,$ then $m + n$ is equal to _________ . [2026]

(62)

$adj (2 A) = 2^{2} adj A ∵ adj (k A) = k^{n - 1} adj (A)$

$= 4 adj A$

Now, $A^{2} (adj (2 A)) = 4 A (adj A)$

$= 4 A$ $| A |$ $I_{3}$

$= 24 A$

Now, $3 adj (A^{2} (adj (2 A))) = 3 adj (24 A)$

$= 3 {(24)}^{2} adj A$

Now, $| adj (3 adj (A^{2} (adj (2 A)))) |$

$=$ $| adj (3 . {(24)}^{2} adj A) |$

$=$ $| (3 . {{(24)}^{2})}^{2} adj (adj A) |$

$=$ $| 3^{6} \cdot 2^{12} adj (adj A) |$

$= {(3^{6} \cdot 2^{12})}^{3}$ $| adj (adj A) |$

$= 3^{18} \cdot 2^{36} \cdot | A |^{4}$

$= 3^{22} \cdot 2^{40}$

$∴$ $m + n = 62$

Q 24 :

For some $α, β \in ℝ$ , let $A = [\begin{matrix} α & 2 \\ 1 & 2 \end{matrix}] and B = [\begin{matrix} 1 & 1 \\ 1 & β \end{matrix}]$ be such that $A^{2} - 4 A + 2 I = B^{2} - 3 B + I = O .$ Then $(\det (adj {(A^{3} - B^{3})))}^{2} is equal to$ ________ [2026]

(225)

$Tr (A) = 4 \Rightarrow α + 2 = 4 \Rightarrow α = 2$

$Tr (B) = 3 \Rightarrow β + 1 = 3 \Rightarrow β = 2$

$A^{2} - 4 A + 2 I = 0$

$A^{3} = 4 A^{2} - 2 A = 16 A - 8 I - 2 A = 14 A - 8 I$

$= [\begin{matrix} 28 & 28 \\ 14 & 28 \end{matrix}] + [\begin{matrix} - 8 & 0 \\ 0 & - 8 \end{matrix}]$

$= A^{3} = [\begin{matrix} 20 & 28 \\ 14 & 20 \end{matrix}]$

$B^{2} - 3 B + I = 0$

$B^{2} = 3 B - I$

$\Rightarrow B^{3} = 3 B^{2} - B = 3 (3 B - I) - B = 8 B - 3 I$

$B^{3} = [\begin{matrix} 8 & 8 \\ 8 & 16 \end{matrix}] + [\begin{matrix} - 3 & 0 \\ 0 & - 3 \end{matrix}] = [\begin{matrix} 5 & 8 \\ 8 & 13 \end{matrix}]$

$A^{3} - B^{3} = [\begin{matrix} 20 & 28 \\ 14 & 20 \end{matrix}] - [\begin{matrix} 5 & 8 \\ 8 & 13 \end{matrix}] = [\begin{matrix} 15 & 20 \\ 6 & 7 \end{matrix}]$

$\Rightarrow$ $| A^{3} - B^{3} |$ $= 105 - 120 = - 15$

$\Rightarrow$ $| adj (A^{3} - B^{3}) |$ $=$ $| A^{3} - B^{3} |$ $= - 15$

$\Rightarrow | adj (A^{3} - B^{3}) |^{2} = 225$

Q 25 :

If $A = [\begin{matrix} 2 & 3 \\ 3 & 5 \end{matrix}],$ then the determinant of the matrix $(A^{2025} - 3 A^{2024} + A^{2023})$ is [2026]

16
12
24
28

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(1)

$A = [\begin{matrix} 2 & 3 \\ 3 & 5 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 13 & 21 \\ 21 & 34 \end{matrix}]$

$| A^{2025} - 3 A^{2024} + A^{2023} |$

$=$ $| A^{2023} (A^{2} - 3 A + I) |$

$= | A |^{2023}$ $| A^{2} - 3 A + I |$

$= 1 \cdot$ $| \begin{matrix} 8 & 12 \\ 12 & 20 \end{matrix} |$ $= 160 - 144 = 16$

Q 26 :

Let $P = [p_{i j}]$ and $Q = [q_{i j}]$ be two square matrices of order 3 such that $q_{i j} = 2^{(i + j - 1)} p_{i j}$ and $\det (Q) = 2^{10} .$ Then the value of $\det (adj (adj P))$ is: [2026]

124
16
81
32

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(2)

$| \begin{matrix} 2 p_{11} & 2^{2} p_{12} & 2^{3} p_{13} \\ 2^{2} p_{21} & 2^{3} p_{22} & 2^{4} p_{23} \\ 2^{3} p_{31} & 2^{4} p_{32} & 2^{5} p_{33} \end{matrix} |$ $= 2^{10}$

$= 2^{2} \cdot 2 \cdot 2^{3}$ $| \begin{matrix} p_{11} & p_{12} & p_{13} \\ 2 p_{21} & 2 p_{22} & 2 p_{23} \\ 2^{2} p_{31} & 2^{2} p_{32} & 2^{2} p_{33} \end{matrix} |$ $= 2^{10}$

$= 2^{9}$ $| \begin{matrix} p_{11} & p_{12} & p_{13} \\ p_{21} & p_{22} & p_{23} \\ p_{31} & p_{32} & p_{33} \end{matrix} |$ $= 2^{10}$ $\Rightarrow$ $| P |$ $= 2$

$| adj (adj (P)) |$ $=$ $| P |^{{(n - 1)}^{2}}$ $=$ $| P |^{4} = 2^{4} = 16$

Q 27 :

The system of linear equations

$\begin{matrix} x + y + z = 6 \\ 2 x + 5 y + a z = 36 \\ x + 2 y + 3 z = b \end{matrix}$ has [2026]

unique solution for a = 8 and b = 14
infinitely many solutions for a = 8 and b = 16
infinitely many solutions for a = 8 and b = 14
unique solution for a = 8 and b = 16

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(3)

$If D =$ $| \begin{matrix} 1 & 1 & 1 \\ 2 & 5 & a \\ 1 & 2 & 3 \end{matrix} |$ $= 0 \Rightarrow a = 8$

$If D_{1} =$ $| \begin{matrix} 6 & 1 & 1 \\ 36 & 5 & a \\ b & 2 & 3 \end{matrix} |$ $= 0 \Rightarrow a b - 5 b - 12 a + 54 = 0$

$If D_{2} =$ $| \begin{matrix} 1 & 6 & 1 \\ 2 & 36 & a \\ 1 & b & 3 \end{matrix} |$ $= 0 \Rightarrow a b - 6 a - 2 b - 36 = 0$

$If D_{3} =$ $| \begin{matrix} 1 & 1 & 6 \\ 2 & 5 & 36 \\ 1 & 2 & b \end{matrix} |$ $= 0 \Rightarrow b = 14$

$For a = 8 and b = 14 \Rightarrow D_{1}, D_{2} are also zero$

$For a = 8 and b = 14 \Rightarrow D = D_{1} = D_{2} = D_{3} = 0$

$\Rightarrow infinitely many solutions.$

Q 28 :

If the system of equations

$\begin{matrix} 3 x + y + 4 z = 3 \\ 2 x + α y - z = - 3 \\ x + 2 y + z = 4 \end{matrix}$

has no solution, then the value of $α$ is equal to : [2026]

19
23
13
4

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(1)

$for no solution Δ = 0$

$| \begin{matrix} 3 & 1 & 4 \\ 2 & α & - 1 \\ 1 & 2 & 1 \end{matrix} |$ $= 0$

$\Rightarrow 3 (α + 2) + 1 (- 1 - 2) + 4 (4 - α) = 0$

$\Rightarrow 19 - α = 0 \Rightarrow α = 19$

$For α = 19$

$Δ_{x} =$ $| \begin{matrix} 3 & 1 & 4 \\ - 3 & 19 & - 1 \\ 4 & 2 & 1 \end{matrix} |$ $= 3 (21) + 1 (- 1) + 4 (- 82)$

$\neq 0$

$∴$ $no solution for α = 19$

Q 29 :

If $X = [[\begin{matrix} x \\ y \\ z \end{matrix}]]$ is a solution of the system of equations AX=B, where adj $A = [[\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{matrix}]] and B = [[\begin{matrix} 4 \\ 0 \\ 2 \end{matrix}]]$ then $| x + y + z |$ is equal to: [2026]

3/2
2
1
3

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(2)

$X = A^{- 1} B = (\frac{adj A}{| A |}) B$

$= \pm \frac{1}{10} (\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{matrix}) (\begin{matrix} 4 \\ 0 \\ 2 \end{matrix})$

$= \pm \frac{1}{10} (\begin{matrix} 20 \\ - 10 \\ 10 \end{matrix}) = \pm (\begin{matrix} 2 \\ - 1 \\ 1 \end{matrix})$

$∴$ $| x + y + z |$ $= 2$

Topic Question Set

Q 1 :

For $α, β \in R$ and a natural number $n$ , let $A_{r} =$ $| \begin{matrix} r & 1 & \frac{n^{2}}{2} + α \\ 2 r & 2 & n^{2} - β \\ 3 r - 2 & 3 & \frac{n (3 n - 1)}{2} \end{matrix} |$ . Then $2 A_{10} - A_{8}$ is [2024]

Q 2 :

Let $A = [\begin{matrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{matrix}] .$ If $A^{3} = 4 A^{2} - A - 21 I,$ where $I$ is the identity matrix of order $3 \times 3$ , then $2 a + 3 b$ is equal to [2024]

Q 3 :

If $α \neq a,$ $β \neq b,$ $γ \neq c$ and $| \begin{matrix} α & b & c \\ a & β & c \\ a & b & γ \end{matrix} |$ $= 0,$ then $\frac{a}{α - a} + \frac{b}{β - b} + \frac{γ}{γ - c}$ is equal to: [2024]

Q 4 :

If $A = [\begin{matrix} \sqrt{2} & 1 \\ - 1 & \sqrt{2} \end{matrix}], B = [\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}], C = A B A^{T}$ and $X = A^{T} C^{2} A,$ then det $X$ is equal to [2024]

Q 5 :

The values of $α,$ for which $| \begin{matrix} 1 & \frac{3}{2} & α + \frac{3}{2} \\ 1 & \frac{1}{3} & α + \frac{1}{3} \\ 2 α + 3 & 3 α + 1 & 0 \end{matrix} |$ $= 0,$ lie in the interval [2024]

Q 6 :

Let $A = [\begin{matrix} 1 & 0 & 0 \\ 0 & α & β \\ 0 & β & α \end{matrix}]$ and $| 2 A |^{3} = 2^{21}$ where $α, β \in Z .$ Then a value of $α$ is [2024]

Q 7 :

Let $A = [\begin{matrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{matrix}]$ and $P = [\begin{matrix} 1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5 \end{matrix}] .$ The sum of the prime factors of $| P^{- 1} A P - 2 I |$ is equal to [2024]

Q 8 :

Let $A$ be a $3 \times 3$ matrix of non-negative real elements such that $A [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = 3 [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] .$ Then the maximum value of $d e t (A)$ is _______. [2024]

Q 10 :

Let $A$ be a $2 \times 2$ real matrix and $I$ be the identity matrix of order 2. If the roots of the equation $| A - x I | = 0$ be $- 1$ and $3$ , then the sum of the diagonal elements of the matrix $A^{2}$ is _______. [2024]

Q 11 :

Let $A = [\begin{matrix} α & - 1 \\ 6 & β \end{matrix}], α > 0$ , such that det(A) = 0 and $α + β = 1 .$ If $I$ denotes $2 \times 2$ identity matrix, then the matrix ${(I + A)}^{8}$ is: [2025]

Q 12 :

For some, a, b, let $f (x) =$ $| \begin{matrix} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{matrix} |$ $, x \neq 0, \lim_{x \to 0} f (x) = λ + μ a + ν b$ . Then ${(λ + μ + ν)}^{2}$ is equal to : [2025]

Q 13 :

Let $A = [a_{i j}] = [\begin{matrix} \log_{5} 128 & \log_{4} 5 \\ \log_{5} 8 & \log_{4} 25 \end{matrix}]$ . If $A_{i j}$ is the cofactor of $a_{i j}, C_{i j} = \sum_{k = 1}^{2} a_{i k} A_{j k}, 1 \leq i, j \leq 2$ , and $C = [C_{i j}]$ , then $8$ $| C |$ is equal to : [2025]

Q 15 :

Let $I$ be the identity matrix of order $3 \times 3$ and for the matrix $A = [\begin{matrix} λ & 2 & 3 \\ 4 & 5 & 6 \\ 7 & - 1 & 2 \end{matrix}], | A | = - 1$ . Let B be the inverse of the matrix adj $(A a d j (A^{2}))$ . Then $| (λ B + I) |$ is equal to __________. [2025]

Q 17 :

Let A be a $2 \times 2$ matrix with real entries such that $A' = α A + I,$ where $α \in R - {- 1, 1}$ . If $\det (A^{2} - A) = 4$ , then the sum of all possible values of $α$ is equal to [2023]

Q 18 :

$If$ $| \begin{matrix} x + 1 & x & x \\ x & x + λ & x \\ x & x & x + λ^{2} \end{matrix} |$ $= \frac{9}{8} (103 x + 81), then λ, \frac{λ}{3} are the roots of the equation$ [2023]

Q 21 :

$Let D_{k} =$ $| \begin{matrix} 1 & 2 k & 2 k - 1 \\ n & n^{2} + n + 2 & n^{2} \\ n & n^{2} + n & n^{2} + n + 2 \end{matrix} |$ . $If \sum_{k = 1}^{n} D_{k} = 96, then n is equal to$ _________ . [2023]

Q 23 :

Let $| A |$ $= 6$ , where A is a $3 \times 3 matrix.$ If $| adj (3 adj (A^{2} \cdot adj (2 A))) |$ $= 2^{m} \cdot 3^{n}, m, n \in N,$ then $m + n$ is equal to _________ . [2026]

Q 24 :

For some $α, β \in ℝ$ , let $A = [\begin{matrix} α & 2 \\ 1 & 2 \end{matrix}] and B = [\begin{matrix} 1 & 1 \\ 1 & β \end{matrix}]$ be such that $A^{2} - 4 A + 2 I = B^{2} - 3 B + I = O .$ Then $(\det (adj {(A^{3} - B^{3})))}^{2} is equal to$ ________ [2026]

Q 25 :

If $A = [\begin{matrix} 2 & 3 \\ 3 & 5 \end{matrix}],$ then the determinant of the matrix $(A^{2025} - 3 A^{2024} + A^{2023})$ is [2026]

Q 26 :

Let $P = [p_{i j}]$ and $Q = [q_{i j}]$ be two square matrices of order 3 such that $q_{i j} = 2^{(i + j - 1)} p_{i j}$ and $\det (Q) = 2^{10} .$ Then the value of $\det (adj (adj P))$ is: [2026]

Q 27 :

The system of linear equations

$\begin{matrix} x + y + z = 6 \\ 2 x + 5 y + a z = 36 \\ x + 2 y + 3 z = b \end{matrix}$ has [2026]

Q 28 :

If the system of equations

$\begin{matrix} 3 x + y + 4 z = 3 \\ 2 x + α y - z = - 3 \\ x + 2 y + z = 4 \end{matrix}$

has no solution, then the value of $α$ is equal to : [2026]

Q 29 :

If $X = [[\begin{matrix} x \\ y \\ z \end{matrix}]]$ is a solution of the system of equations AX=B, where adj $A = [[\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{matrix}]] and B = [[\begin{matrix} 4 \\ 0 \\ 2 \end{matrix}]]$ then $| x + y + z |$ is equal to: [2026]

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Topic Question Set

Q 1 : For α,β∈R and a natural number n, let Ar=|r1n22+α2r2n2-β3r-23n(3n-1)2|. Then 2A10-A8 is [2024]

Q 2 : Let A=[2a013105b]. If A3=4A2-A-21I, where I is the identity matrix of order 3×3, then 2a+3b is equal to [2024]

Q 3 : If α≠a, β≠b, γ≠c and |αbcaβcabγ|=0, then aα-a+bβ-b+γγ-c is equal to: [2024]

Q 4 : If A=[21-12], B=[1011], C=ABATand X=ATC2A, then det X is equal to [2024]

Q 5 : The values of α, for which |132α+32113α+132α+33α+10|=0, lie in the interval [2024]

Q 6 : Let A=[1000αβ0βα] and |2A|3=221 where α,β∈Z. Then a value of α is [2024]

Q 7 : Let A=[2126211332] and P=[120502715]. The sum of the prime factors of|P-1AP-2I| is equal to [2024]

Q 8 : Let A be a 3×3 matrix of non-negative real elements such that A[111]=3[111]. Then the maximum value of det(A) is _______. [2024]

Q 9 : Let A=[201110101], B=[B1,B2,B3], where B1,B2,B3 are column matrices, and AB1=[100], AB2=[230], AB3=[321]. If α=|B| and β is the sum of all the diagonal elements of B, then α3+β3 is equal to _____ . [2024]

Q 10 : Let A be a 2×2 real matrix and I be the identity matrix of order 2. If the roots of the equation |A-xI|=0 be -1 and 3, then the sum of the diagonal elements of the matrix A2 is _______. [2024]

Q 11 : Let A=[α–16β], α>0, such that det(A) = 0 and α+β=1. If I denotes 2×2 identity matrix, then the matrix (I+A)8 is: [2025]

Q 12 : For some, a, b, let f(x)=|a+sin xx1ba1+sin xxba1b+sin xx|, x≠0, limx→0f(x)=λ+μa+νb. Then (λ+μ+ν)2 is equal to : [2025]

Q 13 : Let A=[aij]=[log5128log45log58log425]. If Aij is the cofactor of aij,Cij=∑k=12aik Ajk, 1≤i, j≤2, and C=[Cij], then 8|C| is equal to : [2025]

Q 14 : Let M and m respectively be the maximum and the minimum values of f(x)=|1+sin2xcos2x4sin4xsin2x1+cos2x4sin4xsin2xcos2x1+4sin4x|, x∈R. Then M4–m4 is equal to : [2025]

Q 15 : Let I be the identity matrix of order 3×3 and for the matrix A=[λ234567–12], |A|=–1. Let B be the inverse of the matrix adj(A adj (A2)). Then |(λB+I)| is equal to __________. [2025]

Q 16 : Let integers a, b∈[–3,3] be such that a+b≠0. Then the number of all possible ordered pairs (a, b) for which |z–az+b|=1 and |z+1ωω2ωz+ω21ω21z+ω|=1, z∈C, where ω and ω2 are the roots of x2+x+1=0, is equal to __________. [2025]

Q 17 : Let A be a 2×2 matrix with real entries such that A'=αA+I, where α∈R-{-1,1}. If det(A2-A)=4, then the sum of all possible values of α is equal to [2023]

Q 18 : If |x+1xxxx+λxxxx+λ2|=98(103x+81), then λ,λ3 are the roots of the equation [2023]

Q 19 : Let α be a root of the equation (a-c)x2+(b-a)x+(c-b)=0 where a,b,c are distinct real numbers such that the matrix [α2α1111abc] is singular. Then, the value of (a-c)2(b-a)(c-b)+(b-a)2(a-c)(c-b)+(c-b)2(a-c)(b-a) is [2023]

Q 20 : The set of all values of t∈ℝ, for which the matrix [ete-t(sint-2cost)e-t(-2sint-cost)ete-t(2sint+cost)e-t(sint-2cost)ete-tcoste-tsint] is invertible, is [2023]

Q 21 : Let Dk=|12k2k-1nn2+n+2n2nn2+nn2+n+2|. If ∑k=1nDk=96, then n is equal to _________ . [2023]

Q 22 : Among the statements: I: If |1cosαcosβcosα1cosγcosβcosγ1|=|0cosαcosβcosα0cosγcosβcosγ0|, then cos2α+cos2β+cos2γ=32, and II: If |x2+xx+1x-22x2+3x-13x3x-3x2+2x+32x-12x-1|=px+q, then p2=196q2. [2026]

Q 23 : Let |A|=6, where A is a 3×3 matrix. If |adj(3adj(A2·adj(2A)))|=2m·3n, m,n∈N, then m+n is equal to _________ . [2026]

Q 24 : For some α,β∈ℝ, let A=[α212] and B=[111β] be such that A2-4A+2I=B2-3B+I=O. Then (det(adj(A3-B3)))2 is equal to ________ [2026]

Q 25 : If A=[2335], then the determinant of the matrix (A2025-3A2024+A2023) is [2026]

Q 26 : Let P=[pij] and Q=[qij] be two square matrices of order 3 such that qij=2(i+j-1)pij and det(Q)=210. Then the value of det(adj(adjP)) is: [2026]

Q 27 : The system of linear equations x+y+z=62x+5y+az=36x+2y+3z=b has [2026]

Q 28 : If the system of equations 3x+y+4z=32x+αy−z=−3 x+2y+z=4 has no solution, then the value of α is equal to : [2026]

Q 29 : If X=[xyz] is a solution of the system of equations AX=B, where adj A=[422-5051-23] and B=[402] then |x+y+z| is equal to: [2026]

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Q 1 :

For $α, β \in R$ and a natural number $n$ , let $A_{r} =$ $| \begin{matrix} r & 1 & \frac{n^{2}}{2} + α \\ 2 r & 2 & n^{2} - β \\ 3 r - 2 & 3 & \frac{n (3 n - 1)}{2} \end{matrix} |$ . Then $2 A_{10} - A_{8}$ is [2024]

Q 2 :

Let $A = [\begin{matrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{matrix}] .$ If $A^{3} = 4 A^{2} - A - 21 I,$ where $I$ is the identity matrix of order $3 \times 3$ , then $2 a + 3 b$ is equal to [2024]

Q 3 :

If $α \neq a,$ $β \neq b,$ $γ \neq c$ and $| \begin{matrix} α & b & c \\ a & β & c \\ a & b & γ \end{matrix} |$ $= 0,$ then $\frac{a}{α - a} + \frac{b}{β - b} + \frac{γ}{γ - c}$ is equal to: [2024]

Q 4 :

If $A = [\begin{matrix} \sqrt{2} & 1 \\ - 1 & \sqrt{2} \end{matrix}], B = [\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}], C = A B A^{T}$ and $X = A^{T} C^{2} A,$ then det $X$ is equal to [2024]

Q 5 :

The values of $α,$ for which $| \begin{matrix} 1 & \frac{3}{2} & α + \frac{3}{2} \\ 1 & \frac{1}{3} & α + \frac{1}{3} \\ 2 α + 3 & 3 α + 1 & 0 \end{matrix} |$ $= 0,$ lie in the interval [2024]

Q 6 :

Let $A = [\begin{matrix} 1 & 0 & 0 \\ 0 & α & β \\ 0 & β & α \end{matrix}]$ and $| 2 A |^{3} = 2^{21}$ where $α, β \in Z .$ Then a value of $α$ is [2024]

Q 7 :

Let $A = [\begin{matrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{matrix}]$ and $P = [\begin{matrix} 1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5 \end{matrix}] .$ The sum of the prime factors of $| P^{- 1} A P - 2 I |$ is equal to [2024]

Q 8 :

Let $A$ be a $3 \times 3$ matrix of non-negative real elements such that $A [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = 3 [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] .$ Then the maximum value of $d e t (A)$ is _______. [2024]

Q 10 :

Let $A$ be a $2 \times 2$ real matrix and $I$ be the identity matrix of order 2. If the roots of the equation $| A - x I | = 0$ be $- 1$ and $3$ , then the sum of the diagonal elements of the matrix $A^{2}$ is _______. [2024]

Q 11 :

Let $A = [\begin{matrix} α & - 1 \\ 6 & β \end{matrix}], α > 0$ , such that det(A) = 0 and $α + β = 1 .$ If $I$ denotes $2 \times 2$ identity matrix, then the matrix ${(I + A)}^{8}$ is: [2025]

Q 12 :

For some, a, b, let $f (x) =$ $| \begin{matrix} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{matrix} |$ $, x \neq 0, \lim_{x \to 0} f (x) = λ + μ a + ν b$ . Then ${(λ + μ + ν)}^{2}$ is equal to : [2025]

Q 13 :

Let $A = [a_{i j}] = [\begin{matrix} \log_{5} 128 & \log_{4} 5 \\ \log_{5} 8 & \log_{4} 25 \end{matrix}]$ . If $A_{i j}$ is the cofactor of $a_{i j}, C_{i j} = \sum_{k = 1}^{2} a_{i k} A_{j k}, 1 \leq i, j \leq 2$ , and $C = [C_{i j}]$ , then $8$ $| C |$ is equal to : [2025]

Q 15 :

Let $I$ be the identity matrix of order $3 \times 3$ and for the matrix $A = [\begin{matrix} λ & 2 & 3 \\ 4 & 5 & 6 \\ 7 & - 1 & 2 \end{matrix}], | A | = - 1$ . Let B be the inverse of the matrix adj $(A a d j (A^{2}))$ . Then $| (λ B + I) |$ is equal to __________. [2025]

Q 17 :

Let A be a $2 \times 2$ matrix with real entries such that $A' = α A + I,$ where $α \in R - {- 1, 1}$ . If $\det (A^{2} - A) = 4$ , then the sum of all possible values of $α$ is equal to [2023]

Q 18 :

$If$ $| \begin{matrix} x + 1 & x & x \\ x & x + λ & x \\ x & x & x + λ^{2} \end{matrix} |$ $= \frac{9}{8} (103 x + 81), then λ, \frac{λ}{3} are the roots of the equation$ [2023]

Q 21 :

$Let D_{k} =$ $| \begin{matrix} 1 & 2 k & 2 k - 1 \\ n & n^{2} + n + 2 & n^{2} \\ n & n^{2} + n & n^{2} + n + 2 \end{matrix} |$ . $If \sum_{k = 1}^{n} D_{k} = 96, then n is equal to$ _________ . [2023]

Q 23 :

Let $| A |$ $= 6$ , where A is a $3 \times 3 matrix.$ If $| adj (3 adj (A^{2} \cdot adj (2 A))) |$ $= 2^{m} \cdot 3^{n}, m, n \in N,$ then $m + n$ is equal to _________ . [2026]

Q 24 :

For some $α, β \in ℝ$ , let $A = [\begin{matrix} α & 2 \\ 1 & 2 \end{matrix}] and B = [\begin{matrix} 1 & 1 \\ 1 & β \end{matrix}]$ be such that $A^{2} - 4 A + 2 I = B^{2} - 3 B + I = O .$ Then $(\det (adj {(A^{3} - B^{3})))}^{2} is equal to$ ________ [2026]

Q 25 :

If $A = [\begin{matrix} 2 & 3 \\ 3 & 5 \end{matrix}],$ then the determinant of the matrix $(A^{2025} - 3 A^{2024} + A^{2023})$ is [2026]

Q 26 :

Let $P = [p_{i j}]$ and $Q = [q_{i j}]$ be two square matrices of order 3 such that $q_{i j} = 2^{(i + j - 1)} p_{i j}$ and $\det (Q) = 2^{10} .$ Then the value of $\det (adj (adj P))$ is: [2026]

Q 27 :

The system of linear equations

$\begin{matrix} x + y + z = 6 \\ 2 x + 5 y + a z = 36 \\ x + 2 y + 3 z = b \end{matrix}$ has [2026]

Q 28 :

If the system of equations

$\begin{matrix} 3 x + y + 4 z = 3 \\ 2 x + α y - z = - 3 \\ x + 2 y + z = 4 \end{matrix}$

has no solution, then the value of $α$ is equal to : [2026]

Q 29 :

If $X = [[\begin{matrix} x \\ y \\ z \end{matrix}]]$ is a solution of the system of equations AX=B, where adj $A = [[\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{matrix}]] and B = [[\begin{matrix} 4 \\ 0 \\ 2 \end{matrix}]]$ then $| x + y + z |$ is equal to: [2026]