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

Q 21 :

If $A = [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}],$ then

$A^{'} A = I$
$A^{'} A = 0$
$A^{'} A = 2 I$
$A^{'} A = - I$

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(1)

Ans. $A^{'} A = I$

Explanation:

Given, $A = [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}]$

$A^{'} = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$

$A^{'} A = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}] [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}]$ $= [\begin{matrix} \cos^{2} α + \sin^{2} α & \sin α \cos α - \cos α \sin α \\ \cos α \sin α - \sin α \cos α & \sin^{2} α + \cos^{2} α \end{matrix}]$

$= [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I$

$∴$ $A^{'} A = I$

Q 22 :

If $A = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}],$ then $A^{2} - 5 A + 7 I =$

$0$
$I$
$2 I$
$3 I$

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(1)

Ans. 0

Explanation:

$A = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}]$

$A^{2} - 5 A + 7 I = ?$

$A^{2} = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] = [\begin{matrix} 8 & 5 \\ - 5 & 3 \end{matrix}]$

$- 5 A = - 5 [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] = [\begin{matrix} - 15 & - 5 \\ 5 & - 10 \end{matrix}]$

$7 I = 7 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 7 & 0 \\ 0 & 7 \end{matrix}]$

$A^{2} - 5 A + 7 I = [\begin{matrix} 8 & 5 \\ - 5 & 3 \end{matrix}] + [\begin{matrix} - 15 & - 5 \\ 5 & - 10 \end{matrix}] + [\begin{matrix} 7 & 0 \\ 0 & 7 \end{matrix}]$

$= [\begin{matrix} 8 - 15 + 7 & 5 - 5 + 0 \\ - 5 + 5 + 0 & 3 - 10 + 7 \end{matrix}] = 0$

Q 23 :

The set of natural numbers is divided into array of rows and columns in the form of matrices $A_{1} = [1],$ $A_{2} = [\begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix}],$ $A_{3} = [\begin{matrix} 6 & 7 & 8 \\ 9 & 10 & 11 \\ 12 & 13 & 14 \end{matrix}]$ and so on. Let the trace of $A_{10}$ be $λ$ . Find unit digit of $λ$ ?

(5)

$First element of matrix A_{10} = 286$ $(10^{th} of sequence 1, 2, 6, 15, \dots)$

$Trace of A_{10} = 286 + 297 + 308 + 319 + \dots + 385 = 3055$

Q 24 :

$A = [\begin{matrix} 1 & - 2 \\ 0 & 1 \end{matrix}],$ $B = [\begin{matrix} 5 & 2 \\ 2 & 1 \end{matrix}] then {(A B A^{T})}^{5} {(A B^{T} A^{T})}^{10} = X then trace of matrix X is$ ______

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8
2

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4

(4)

$A B A^{T} = I \Rightarrow A B B^{T} A^{T} = I$

Q 25 :

Let $A = [\begin{matrix} 1 & 2 \\ 1 & α \end{matrix}]$ and $B = [\begin{matrix} 3 & 3 \\ β & 2 \end{matrix}]$ . If $A^{2} - 4 A + I = O$ and $B^{2} - 5 B - 6 I = O$ , then among the two statements :

(S1): ${[(B - A) (B + A)]}^{T} = [\begin{matrix} 13 & 15 \\ 7 & 10 \end{matrix}]$ and

(S2): det (adj (A + B)) = –5 [2026]

Only (S1) is correct
Only (S2) is correct
Both (S1) and (S2) are correct
Both (S1) and (S2) are wrong

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(2)

Given: $A = [\begin{matrix} 1 & 2 \\ 1 & α \end{matrix}]$ & $B = [\begin{matrix} 3 & 3 \\ β & 2 \end{matrix}]$

Now, $A^{2} - 4 A + I = O$

$\Rightarrow [\begin{matrix} 1 & 2 \\ 1 & α \end{matrix}] [\begin{matrix} 1 & 2 \\ 1 & α \end{matrix}] - [\begin{matrix} 4 & 8 \\ 4 & 4 α \end{matrix}] + [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

$\Rightarrow [\begin{matrix} 3 & 2 + 2 α \\ 1 + α & 2 + α^{2} \end{matrix}] - [\begin{matrix} 4 & 8 \\ 4 & 4 α \end{matrix}] + [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

$\Rightarrow 2 + 2 α - 8 = 0 \Rightarrow 1 + α - 4 = 0$

$\Rightarrow α = 3$

Also, $B^{2} - 5 B - 6 I = O$

$\Rightarrow [\begin{matrix} 3 & 3 \\ β & 2 \end{matrix}] [\begin{matrix} 3 & 3 \\ β & 2 \end{matrix}] - [\begin{matrix} 15 & 15 \\ 5 β & 10 \end{matrix}] + [\begin{matrix} 6 & 0 \\ 0 & 6 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

$\Rightarrow [\begin{matrix} 9 + 3 β & 15 \\ 5 β & 3 β + 4 \end{matrix}] - [\begin{matrix} 15 & 15 \\ 5 β & 10 \end{matrix}] + [\begin{matrix} 6 & 0 \\ 0 & 6 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

$\Rightarrow 9 + 3 β - 15 - 6 = 0 \Rightarrow β = 4$

Thus, $A = [\begin{matrix} 1 & 2 \\ 1 & 3 \end{matrix}]$ & $B = [\begin{matrix} 3 & 3 \\ 4 & 2 \end{matrix}]$

(S1): L.H.S. = ${[(B - A) (B + A)]}^{T}$

$[[\begin{matrix} 2 & 1 \\ 3 & - 1 \end{matrix}] {[\begin{matrix} 4 & 5 \\ 5 & 5 \end{matrix}]]}^{T} = [\begin{matrix} 13 & 7 \\ 15 & 10 \end{matrix}]$

R.H.S. = $[\begin{matrix} 13 & 15 \\ 7 & 10 \end{matrix}]$

$∴$ L.H.S. $\neq$ R.H.S.

Hence, (S1) is wrong.

(S2): $A + B = [\begin{matrix} 4 & 5 \\ 5 & 5 \end{matrix}]$

Adj(A + B) = $[\begin{matrix} 5 & - 5 \\ - 5 & 4 \end{matrix}]$

|adj . (A + B)| = 20 – 25 = –5 = R.H.S.

Hence, (S2) is correct.

Q 26 :

Let M be a 3 x 3 matrix such that $M = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) = (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix})$ , $M = (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) = (\begin{matrix} 0 \\ 1 \\ 2 \end{matrix})$ and $M = (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) = (\begin{matrix} - 1 \\ 1 \\ 1 \end{matrix})$ .

If $M = (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 1 \\ 7 \\ 11 \end{matrix})$ , then x + y + z equals : [2026]

4
5
7
11

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(2)

Let $M = [\begin{matrix} a & b & c \\ p & q & r \\ l & m & n \end{matrix}]$

Given, $M = [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}] = [\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}] \Rightarrow [\begin{matrix} a & b & c \\ p & q & r \\ l & m & n \end{matrix}] [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}] = [\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]$

$\Rightarrow$ a = 1, p = 2, $l$ = 3

Also, $[\begin{matrix} a & b & c \\ p & q & r \\ l & m & n \end{matrix}] [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}]$

$\Rightarrow$ b = 0, q = 1, m = 2

And $[\begin{matrix} a & b & c \\ p & q & r \\ l & m & n \end{matrix}] [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}] = [\begin{matrix} - 1 \\ 1 \\ 1 \end{matrix}]$

$\Rightarrow$ c = –1, r = 1, n = 1

$∴ M = [\begin{matrix} 1 & 0 & - 1 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{matrix}]$

Now, $[\begin{matrix} 1 & 0 & - 1 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 1 \\ 7 \\ 11 \end{matrix}]$

$\Rightarrow$ x – z = 1 ... (i)

2x + y + z = 7 ... (ii)

3x + 2y + z = 11 ... (iii)

From (i), (ii) and (iii), we get

x = 2, y = 2, z = 1

$∴$ x + y + z = 5

Q 27 :

Let $A = [\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{matrix}]$ and $B = [b_{i j}]$ , $1 \leq i, j \leq 3$ . If $B = A^{99} - I$ , then the value of $\frac{b_{31} - b_{21}}{b_{32}}$ is : [2026]

99
199
149
159

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(3)

Given, $A = [\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{matrix}]$

$\Rightarrow A = I + N$ , where

$N = [\begin{matrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{matrix}]$ and $I = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

$\Rightarrow N^{2} = [\begin{matrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{matrix}] [\begin{matrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{matrix}] = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{matrix}]$

$\Rightarrow N^{2} = N^{2} \cdot N = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$

$\Rightarrow A^{99} = {(I + N)}^{99} = I + 99 N + \frac{9998}{2} N^{2} + . . .$

Since, $N^{k} = 0$ , for $k \geq 3$

$∴ A^{99} = I + 99 N + 4851 N^{2}$

$B = I + 99 N + 4851 N^{2}$

$B = 99 N + 4851 N^{2}$

$= [\begin{matrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{matrix}] + 4851 [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{matrix}]$

$= [\begin{matrix} 0 & 0 & 0 \\ 297 & 0 & 0 \\ 891 + 43659 & 297 & 0 \end{matrix}]$

$= [\begin{matrix} 0 & 0 & 0 \\ 297 & 0 & 0 \\ 44550 & 297 & 0 \end{matrix}]$

$∴ \frac{b_{31} - b_{21}}{b_{32}} = \frac{44550 - 297}{297} = 149$

Q 28 :

Let $A = [\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}]$ and $B = [\begin{matrix} a & b \\ c & d \end{matrix}]$ are two matrices such that $A B = B A$ and $c \neq 0$ . Then the value of $\frac{a - d}{3 b - c} = \frac{- 14}{K}$ then $K =$

(14)

$A B = [\begin{matrix} a + 2 c & b + 2 d \\ 3 a + 4 c & 3 b + 4 d \end{matrix}]$

$B A = [\begin{matrix} a + 3 b & 2 a + 4 b \\ c + 3 d & 2 c + 4 d \end{matrix}]$

$A B = B A \Rightarrow 2 a - 2 d = - 3 b,$ $\frac{a - d}{3 b - c} = - 1$

Topic Question Set

Q 21 :

If $A = [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}],$ then

Q 22 :

If $A = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}],$ then $A^{2} - 5 A + 7 I =$

Q 24 :

$A = [\begin{matrix} 1 & - 2 \\ 0 & 1 \end{matrix}],$ $B = [\begin{matrix} 5 & 2 \\ 2 & 1 \end{matrix}] then {(A B A^{T})}^{5} {(A B^{T} A^{T})}^{10} = X then trace of matrix X is$ ______

Q 27 :

Let $A = [\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{matrix}]$ and $B = [b_{i j}]$ , $1 \leq i, j \leq 3$ . If $B = A^{99} - I$ , then the value of $\frac{b_{31} - b_{21}}{b_{32}}$ is : [2026]

Q 28 :

Let $A = [\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}]$ and $B = [\begin{matrix} a & b \\ c & d \end{matrix}]$ are two matrices such that $A B = B A$ and $c \neq 0$ . Then the value of $\frac{a - d}{3 b - c} = \frac{- 14}{K}$ then $K =$

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Topic Question Set

Q 21 : If A=[cosαsinα-sinαcosα], then

Q 22 : If A=[31-12], then A2-5A+7I=

Q 23 : The set of natural numbers is divided into array of rows and columns in the form of matrices A1=[1], A2=[2345], A3=[67891011121314] and so on. Let the trace of A10 be λ. Find unit digit of λ?

Q 24 : A=[1-201], B=[5221] then (ABAT)5(ABTAT)10=X then trace of matrix X is ______

Q 25 : Let A=[121α] and B=[33β2]. If A2–4A+I=O and B2–5B–6I=O, then among the two statements : (S1): [(B–A)(B+A)]T=[1315710] and (S2): det (adj (A + B)) = –5 [2026]

Q 26 : Let M be a 3 x 3 matrix such that M=(100)=(123), M=(010)=(012) and M=(001)=(–111). If M=(xyz)=(1711), then x + y + z equals : [2026]

Q 27 : Let A = [100310931] and B=[bij], 1≤i, j≤3. If B=A99–I, then the value of b31–b21b32 is : [2026]

Q 28 : Let A=[1234] and B=[abcd] are two matrices such that AB=BA and c≠0. Then the value of a-d3b-c=-14K then K=

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Q 21 :

If $A = [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}],$ then

Q 22 :

If $A = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}],$ then $A^{2} - 5 A + 7 I =$

Q 24 :

$A = [\begin{matrix} 1 & - 2 \\ 0 & 1 \end{matrix}],$ $B = [\begin{matrix} 5 & 2 \\ 2 & 1 \end{matrix}] then {(A B A^{T})}^{5} {(A B^{T} A^{T})}^{10} = X then trace of matrix X is$ ______

Q 27 :

Let $A = [\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{matrix}]$ and $B = [b_{i j}]$ , $1 \leq i, j \leq 3$ . If $B = A^{99} - I$ , then the value of $\frac{b_{31} - b_{21}}{b_{32}}$ is : [2026]

Q 28 :

Let $A = [\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}]$ and $B = [\begin{matrix} a & b \\ c & d \end{matrix}]$ are two matrices such that $A B = B A$ and $c \neq 0$ . Then the value of $\frac{a - d}{3 b - c} = \frac{- 14}{K}$ then $K =$