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

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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(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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(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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(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]

0%

(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]

0%

(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

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(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]

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

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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

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

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

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

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

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

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

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