Matrices JEE Main Questions with Solutions PDF – Chapter-wise Practice Questions

Matrices and Determinants JEE Main Previous Year Questions

Question 1: For $\alpha, \beta \in \mathbb{R}$, suppose the system of linear equations

$ \begin{aligned} x-y+z &= 5,\\ 2x+2y+\alpha z &= 8,\\ 3x-y+4z &= \beta \end{aligned} $

has infinitely many solutions. Then $\alpha$ and $\beta$ are the roots of

1) $x^2+14 x+24=0$

2) $x^2+18 x+56=0$

3) $x^2-18 x+56=0$

4) $x^2-10 x+16=0$

Correct Answer: (C) $x^2-18 x+56=0$

Solution:

$ \begin{aligned} \left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 2 & \alpha \\ 3 & -1 & 4 \end{array}\right| &= 0 \\ 1(8+\alpha)+1(8-3\alpha)+1(-2-6) &= 0 \\ 8+\alpha+8-3\alpha-8 &= 0 \\ -2\alpha &= -8 \\ \alpha &= 4 \end{aligned} $

Since

$D_1=0$, $ \begin{aligned} \left|\begin{array}{ccc} 5 & -1 & 1 \\ 8 & 2 & 4 \\ \beta & -1 & 4 \end{array}\right| &= 0 \\ 5(8+4)+1(32-4\beta)+1(-8-2\beta) &= 0 \\ 60+32-4\beta-8-2\beta &= 0 \\ -6\beta &= -84 \\ \beta &= 14 \end{aligned} $

Therefore, the equation whose roots are $\alpha$ and $\beta$ is $ x^2-(\alpha+\beta)x+\alpha\beta=0 $ $ x^2-18x+56=0 $

Question 2: For the system of linear equations

$ \begin{aligned} x+y+z &= 6 \\ \alpha x+\beta y+7z &= 3 \\ x+2y+3z &= 14 \end{aligned} $

Which of the following is NOT true?

1) If α = β and α ≠ 7, then the system has a unique solution

2) If α = β = 7, then the system has no solution

3) For every point (α, β) ≠ (7,7) on the line x − 2y + 7 = 0, the system has infinitely many solutions

4) There is a unique point (α, β) on the line x + 2y + 18 = 0 for which the system has infinitely many solutions

Correct Answer: For every point (α, β) ≠ (7,7) on the line x − 2y + 7 = 0, the system has infinitely many solutions

Solution:

$ \begin{aligned} x+y+z &= 6 \\ \alpha x+\beta y+7z &= 3 \\ x+2y+3z &= 14 \end{aligned} $

Equation (3) $-$ Equation (1)

$ \begin{aligned} y+2z &= 8 \\ y &= 8-2z \end{aligned} $

From Equation (1),

$ \begin{aligned} x+(8-2z)+z &= 6 \\ x &= -2+z \end{aligned} $

Substituting the values of $x$ and $y$ into Equation (2),

$ \begin{aligned} \alpha(-2+z)+\beta(8-2z)+7z &= 3 \\ -2\alpha+\alpha z+8\beta-2\beta z+7z &= 3 \\ (\alpha-2\beta+7)z &= 2\alpha-8\beta+3 \end{aligned} $

if $\alpha-2 \beta+7 \neq 0$ then system has unique solution

if $(\alpha-2 \beta+7=0)$ and $2 \alpha-8 \beta+3 \neq 0$ then system has no solution if $(\alpha-2 \beta+7=0)$ and $2 \alpha-8 \beta+3=0$ then system has infinite solution

Question 3: Let $P=\left[\begin{array}{lll}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1\end{array}\right]$ and $Q=\left[q_{i j}\right]$ be two $3 \times 3$ matrices such that $Q-P^5=I_3$. Then $\frac{q_{21}+q_{31}}{q_{32}}$ is equal to :

1) 15

2) 9

3) 135

4) 10

Correct Answer: 10

Solution:

$ \begin{aligned} &\left( \begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right) \left( \begin{array}{lll} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array} \right) \\[6pt] &= \left( \begin{array}{lll} a_{11}b_{11}+a_{12}b_{21}+a_{13}b_{31} & a_{11}b_{12}+a_{12}b_{22}+a_{13}b_{32} & a_{11}b_{13}+a_{12}b_{23}+a_{13}b_{33} \\ a_{21}b_{11}+a_{22}b_{21}+a_{23}b_{31} & a_{21}b_{12}+a_{22}b_{22}+a_{23}b_{32} & a_{21}b_{13}+a_{22}b_{23}+a_{23}b_{33} \\ a_{31}b_{11}+a_{32}b_{21}+a_{33}b_{31} & a_{31}b_{12}+a_{32}b_{22}+a_{33}b_{32} & a_{31}b_{13}+a_{32}b_{23}+a_{33}b_{33} \end{array} \right) \end{aligned} $

$ P= \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{array} \right], \qquad P^2= \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 6 & 1 & 0 \\ 27 & 6 & 1 \end{array} \right], \qquad P^3= \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 9 & 1 & 0 \\ 54 & 9 & 1 \end{array} \right] $

$ Q= \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 15 & 1 & 0 \\ 135 & 15 & 1 \end{array} \right] + \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] = \left[ \begin{array}{ccc} 2 & 0 & 0 \\ 15 & 2 & 0 \\ 135 & 15 & 2 \end{array} \right] $

So, ans $\frac{15+135}{15}=10$

Question 4: Let A be any $3 \times 3$ invertible matrix. Then which one of the following is not always true?

1) $\operatorname{adj}(A)=|A| . A^{-1}$

2) $\operatorname{adj}(\operatorname{adj}(A))=|A| . A$

3) $\operatorname{adj}(\operatorname{adj}(A))=|A|^2 \cdot(\operatorname{adj}(A))^{-1}$

4) $\operatorname{adj}(\operatorname{adj}(A))=|A| \cdot(\operatorname{adj}(A))^{-1}$

Correct Answer: (4) $\operatorname{adj}(\operatorname{adj}(A))=|A| \cdot(\operatorname{adj}(A))^{-1}$

Solution:

The inverse of a matrix -

$A^{-1}=\frac{1}{|A|} \cdot \operatorname{adj} A$

Option 1: $A^{-1}=\frac{\operatorname{adj}(A)}{|A|}$ (By formula)

Option 2: $\operatorname{adj}(\operatorname{adj}(\mathrm{A}))==|A|^{n-2} A$

Put $\mathrm{n}=3$

$\therefore \operatorname{adj}(\operatorname{adj}(A))=|A|^{3-2} A=|A| A$

Option 3: and 4

$ \begin{aligned} \because\; A(\operatorname{adj} A) &= |A|I_n \\[4pt] (\operatorname{adj} A)^{-1} &= \frac{A}{|A|} \\[4pt] \operatorname{adj}(\operatorname{adj} A) &= |A|^2(\operatorname{adj} A)^{-1} \\[4pt] &= |A|^2\left(\frac{A}{|A|}\right) \\[4pt] &= |A|A \end{aligned} $

Question 5: Suppose A is any $3 \times 3$ non-singular matrix and $(\mathrm{A}-3 \mathrm{I})(\mathrm{A}-5 \mathrm{I})=\mathrm{O}$, where $\mathrm{I}=\mathrm{I}_3$ and $\mathrm{O}=\mathrm{O}_3$. If $\alpha \mathrm{A}+\beta \mathrm{A}^{-1}=4 \mathrm{I}$, Then $\alpha+\beta$ is equal to:

1) 8

2) 7

3) 13

4) 12

Correct Answer: (1) 8

Solution:

The Inverse of a matrix -

A non-singular square matrix of order $n$ is invertible if there exists a square matrix B of the same order such that $A B=I=B A$

$ \begin{aligned} (A-3I)(A-5I) &= 0 \\ \Rightarrow A^2-8A+15I &= 0 \qquad \cdots (i) \end{aligned} $

Also given that $\alpha A+\beta A^{-1}=4 I$

Multiply both sides by A

$\Rightarrow \alpha A^2-4 A+\beta I=0 \cdots \cdots(i i)$

Compare (i) and (ii)

$ \begin{aligned} A^2-8A+15I &= 0,\\ \alpha A^2-4A+\beta I &= 0. \end{aligned} $

Thus, $\frac{1}{\alpha}=\frac{8}{4}=\frac{15}{\beta}$

Thus, $\alpha=\frac{1}{2}$ and $\beta=\frac{15}{2}$

$\Rightarrow \alpha+\beta=8$

Question 6: For two 3 × 3 matrices A and B, let A + B = 2B' and 3A + 2B = I3, where B' is the transpose of B and I3 is 3 × 3 identity matrix. Then :

1) 5A + 10B = 2I3

2) 10A + 5B = 3I3

3) B + 2A = I3

4) 3A + 6B = 2I3

Correct Answer: (2) 10A + 5B = 3I3

Solution:

Given A + B = 2B' .............(1)

Taking transpose of both the sides

(A + B)' = (2B')'

A' + B' = 2(B')'

A' + B' = 2B ...........(2)

Also given, 3A + 2B = I ....(3)

Taking transpose of both sides

3A' + 2B' = I ....(4)

(Note: Transpose of I is I itself)

Now from these 4 equations, we need to get a relation in A and B by eliminating A' and B'

Let us first eliminate B'

From (4): 3A' + A + B = I (Using (1)) ....(5)

And 2 x (2):

2A' + 2B' = 4B

2A' + A + B = 4B

2A' + A - 3B = 0 .....(6)

From (5) and (6) we can eliminate A' as well

From (5): 3A' = I - A - B .....(7)

From (6): 2A' = 3B - A .....(8)

2x(7) - 3x(8): 0 = 2I + A - 11B .....(9)

From (9) and (3): A = B = I/5

For option (B) : 10A + 5B = 2I + I = 3I

Question 7: If the matrices $A=\left[\begin{array}{ccc}1 & 1 & 2 \\ 1 & 3 & 4 \\ 1 & -1 & 3\end{array}\right], B=\operatorname{adj} A$ and $C=3 A$, then $\frac{|\operatorname{adj} B|}{|C|}$ is equal to:

1) 16

2) 12

3) 8

4) 72

Correct Answer: (3) 8

Solution:

$\begin{aligned} & |A|=\left|\begin{array}{ccc}1 & 1 & 2 \\ 1 & 3 & 4 \\ 1 & -1 & 3\end{array}\right|=6 \\ & |\operatorname{adj} B|=|\operatorname{adj} \operatorname{adj} A|=|A|^{(n-1)^2}=|A|^4=(36)^2 \\ & |C|=|B A|=3^3 \times 6 \frac{|\operatorname{adjB}|}{|C|}=\frac{36 \times 36}{3^3 \times 6}=8\end{aligned}$

Question 8: The following system of linear equations

$ \begin{aligned} 7x+6y-2z &= 0, \\ 3x+4y+2z &= 0, \\ x-2y-6z &= 0. \end{aligned} $

has

1) infinitely many solutions, $(x, y, z)$ satisfying $y=2 \approx$

2) infinitely many solutions, $(x, y, z)$ satisfying $x=2 z$.

3) no solution

4) only the trivial solution.

Correct Answer: (2) infinitely many solutions, $(x, y, z)$ satisfying $x=2 z$.

Solution:

(1) $7 x+6 y-2 z=0$

[2) $3 x+4 y+2 z=0$

(3) $x-2 y-6 z=0$

$ \left| \begin{array}{ccc} 7 & 6 & -2 \\ 3 & 4 & 2 \\ 1 & -2 & -6 \end{array} \right| =7(-20)-6(-20)-2(-10) =-140+120+20=0 $

So infinite non-trivial solutions exist

Now equation (1) +3 equation (3)

$ \begin{aligned} 10x-20z &= 0 \\ x &= 2z \end{aligned} $

Question 9: Let $A=\left\{X=(x, y, z)^2: P x=0\right.$, and,$\left.x^2+y^2+z^2=1\right\}$ where $P=\left(\begin{array}{ccc}1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1\end{array}\right)$ then Set A contains :

1) exactly two elements

2) is an empty set,

3) is a singleton set

4) contains more than 2 elements

Correct Answer: (1) exactly two elements

Solution:

Given $P=\left[\begin{array}{ccc}1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1\end{array}\right],|\mathrm{P}|=1(3 \times(-1)-(-4) \times 9)-2((-1) \times(-2)-(-4) \times 1)+1(9 \times(-2)-3 \times 1)|P|=0$ Here $|P|=0$ also given $P X=0$

$ \left. \begin{aligned} &\begin{bmatrix} 1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} =0 \\[6pt] &x+2y+z=0 \\ &-2x+3y-4z=0 \\ &x+9y-z=0 \end{aligned} \right\} \quad D=0, \text{ so the system has infinitely many solutions.} $

By solving these equations

$l \text { we get } x=\frac{-11 \lambda}{2} ; y=\lambda ; z=\frac{7 \lambda}{2}$

Also given, $x^2+y^2+z^2=1$

$ \begin{aligned} \Rightarrow \left(\frac{-11\lambda}{2}\right)^2+\lambda^2+\left(\frac{7\lambda}{2}\right)^2 &= 1 \\[6pt] \Rightarrow \lambda &= \pm \frac{1}{\sqrt{\frac{121}{4}+1+\frac{49}{4}}} \end{aligned} $

So, there are 2 values of $\lambda$

Therefore, there are 2 solution sets of $(\mathrm{x}, \mathrm{y}, \mathrm{z})$.

Question 10: $a, b, c \in R$ non -zero and satisfies $a^3+b^3+c^3=2$ if the matrix $A=\left\{\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right\}$ satisfies $A^T A=I$ then the value of abc can be

1) $\frac{-1}{3}$

2) $\frac{1}{3}$

3) $\frac{-2}{3}$

4) 3

Correct answer: 2 $\frac{1}{3}$

Solution:

$\begin{aligned} & A=\left[\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right] \\ & A^T=\left[\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right] \\ & A \cdot A^T=I \\ & {\left[\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right] \times\left[\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]} \\ & {\left[\begin{array}{ll}a^2+b^2+c^2 & a b+b c+c a \\ a b+b c+c a & a^2+b^2+c^2 \\ a b+b c+c a & a b+b c+c a \\ a b+c a & a^2+b^2+c^2\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]} \\ & \Rightarrow a^2+b^2+c^2=1 \text { and } a b+b c+c a=0 \\ & (a+b+c)^2=1 \Rightarrow(a+b+c)= \pm 1 \\ & a^3+b^3+c^3=(a+b+c)\left[a^2+b^2+c^2-a b-b c-a c\right]+3 a b c \\ & 2=1 \times[ \pm 1]+3 a b c \\ & a b c=1 \quad \text { or } \quad a b c=\frac{1}{3}\end{aligned}$

Question 11: If the matrix $\mathrm{A}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1\end{array}\right]$ satisfy the equation $A^{20}+\alpha A^{19}+\beta A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{array}\right]$ for some real numbers $\alpha$ and $\beta$, then $\beta-\alpha$ is equal to $\_\_\_\_$

1) 3

2) 2

3) 1

4) 4

Correct Answer: (4) 4

Solution:

Given that

$ \begin{aligned} A&= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1 \end{bmatrix} \\[8pt] A^2&= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \qquad A^3= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 8 & 0 \\ 3 & 0 & -1 \end{bmatrix} \\[8pt] A^4&= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{aligned} $ Hence, $ A^{20}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2^{20} & 0 \\ 0 & 0 & 1 \end{bmatrix}, \qquad A^{19}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2^{19} & 0 \\ 3 & 0 & -1 \end{bmatrix} $

So $A^{20}+\alpha A^{19}+\beta A=\left[\begin{array}{ccc}1+\alpha+\beta & 0 & 0 \\ 0 & 2^{20}+\alpha \cdot 2^{19}+2 \beta & 0 \\ 3 \alpha+3 \beta & 0 & 1-\alpha-\beta\end{array}\right] \quad=\left[\begin{array}{ll}1 & 0 \\ 0 & 0 \\ 0 & 0 \\ 1 & 0\end{array}\right]$

Therefore $\alpha+\beta=0$ and $2^{20}+2^{19} \alpha-2 \alpha=4 \Rightarrow \alpha=\frac{4\left(1-2^{18}\right)}{2\left(2^{18}-1\right)}=-2$ hence $\beta=2$ so $(\beta-\alpha)=4$

Question 12: Let $\theta=\frac{\pi}{5}$ and $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$. If $B=A+A^4$, then $\operatorname{det}(B):$

1) is one

2) Lies in (2,3)

3) is zero

4) Lies in (1,2)

Correct Answer: (4 )Lies in (1,2)

Solution:

$A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right] A^2=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right] A^2=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$

$B=A+A^4=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]+\left[\begin{array}{cc}\cos 4 \theta & \sin 4 \theta \\ -\sin 4 \theta & \cos 4 \theta\end{array}\right] B=\left[\begin{array}{cc}(\cos \theta+\cos 4 \theta) & (\sin \theta+\sin 4 \theta) \\ -(\sin \theta+\sin 4 \theta) & (\cos \theta+\cos 4 \theta)\end{array}\right]$

$|\mathrm{B}|=\left(\cos \theta+\cos ^4 \theta\right)^2+(\sin \theta+\sin 4 \theta)^2|\mathrm{~B}|=2+2 \cos 3 \theta$,

when $\theta=\frac{\pi}{5}|\mathrm{~B}|=2+2 \cos \frac{3 \pi}{5}=2(1-\sin 18)|\mathrm{B}|=2\left(1-\frac{\sqrt{5}-1}{4}\right)=2\left(\frac{5-\sqrt{5}}{4}\right)=\frac{5-\sqrt{5}}{2}$

Question 13: Let $a \in \mathbf{R}$ and $A$ be a matrix of order $3 \times 3$ such that $\operatorname{det}(A)=-4$ and $A+I=\left[\begin{array}{lll}1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2\end{array}\right]$, where $I$ is the identity matrix of order $3 \times 3$.

If $\operatorname{det}((a+1) \operatorname{adj}((a-1) A))$ is $2^{m} 3^{n}, m, n \in$ $\{0,1,2, \ldots . .20\}$, then $m+n$ is equal to :

1) 14

2) 17

3) 15

4) 16

Correct Answer: (4) 16

Solution:

The value of the matrix, $A=\left[\begin{array}{lll}1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2\end{array}\right]-I=\left[\begin{array}{lll}0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1\end{array}\right]$

Use the given determinant,

$|\mathrm{A}|=-4$

$0(0(1)-0(1))-a(2(1)-0(a))+1(2(1)-a(0))=-4$

$\Rightarrow 2-2 \mathrm{a}=-4 \Rightarrow \mathrm{a}=3$

Now, find the value of

$|(\mathrm{a}+1) \operatorname{adj}(\mathrm{a}-1) \mathrm{A}|$ by putting $a=3$

$|(\mathrm{a}+1) \operatorname{adj}(\mathrm{a}-1) \mathrm{A}|=|4 \operatorname{adj} 3 \mathrm{~A}|$

$=4^{3}|\operatorname{adj} 3 \mathrm{~A}|$

$=4^{3} \times|3 \mathrm{~A}|^{3-1}=64|3 \mathrm{~A}|^{2}$

$=64 \times\left(3^{3}\right)^{2}|\mathrm{~A}|^{2}$

Put the value of the determinant $|A|=4$

$=2^{6} \times 3^{6} \times 16$

Compare,

$2^{\mathrm{m}} \times 3^{\mathrm{n}}=2^{10} \times 3^{6}$

For the same bases on both sides, the exponents will be equal on both sides,

$\therefore \mathrm{m}=10, \mathrm{n}=6$

$\Rightarrow \mathrm{m}+\mathrm{n}=16$

Question 14: Let $A$ be a $3 \times 3$ real matrix such that $A^{2}(A-2 I)-$ $4(\mathrm{~A}-\mathrm{I})=\mathrm{O}$, where I and O are the identity and null matrices, respectively. If $\mathrm{A}^{5}=\alpha \mathrm{A}^{2}+\beta \mathrm{A}+\gamma \mathrm{I}$, where $\alpha, \beta$ and $\gamma$ are real constants, then $\alpha+\beta+\gamma$ is equal to:

1) 12

2) 20

3) 76

4) 4

Correct Answer: (1) 12

Solution:

Given matrix equation:

$A^{2}(A - 2I) - 4(A - I) = O$

Expanding,

$A^{3} - 2 A^{2} - 4 A + 4 I = O$

Rearranged as,

$A^{3} = 2 A^{2} + 4 A - 4 I \quad (1)$

To find $A^{5}$, write

$A^{5} = A^{2} \cdot A^{3}$

Using equation $(1)$,

$A^{5} = A^{2}(2 A^{2} + 4 A - 4 I) = 2 A^{4} + 4 A^{3} - 4 A^{2}$

Express $A^{4}$ as

$A^{4} = A \cdot A^{3} = A (2 A^{2} + 4 A - 4 I) = 2 A^{3} + 4 A^{2} - 4 A$

Substitute back,

$A^{5} = 2 (2 A^{3} + 4 A^{2} - 4 A) + 4 A^{3} - 4 A^{2}$

$= 4 A^{3} + 8 A^{2} - 8 A + 4 A^{3} - 4 A^{2}$

$= (4 A^{3} + 4 A^{3}) + (8 A^{2} - 4 A^{2}) - 8 A$

$= 8 A^{3} + 4 A^{2} - 8 A$

Use equation $(1)$ again for $A^{3}$,

$A^{5} = 8 (2 A^{2} + 4 A - 4 I) + 4 A^{2} - 8 A$

$= 16 A^{2} + 32 A - 32 I + 4 A^{2} - 8 A$

$= (16 A^{2} + 4 A^{2}) + (32 A - 8 A) - 32 I$

$= 20 A^{2} + 24 A - 32 I$

Therefore,

$\alpha = 20, \quad \beta = 24, \quad \gamma = -32$

Sum is

$\alpha + \beta + \gamma = 20 + 24 - 32 = 12$

Question 15: If $\mathrm{A}, \mathrm{B}$ and $\left(\operatorname{adj}\left(\mathrm{A}^{-1}\right)+\operatorname{adj}\left(\mathrm{B}^{-1}\right)\right)$ are non-singular matrices of same order, then the inverse of $\mathrm{A}\left(\operatorname{adj}\left(\mathrm{A}^{-1}\right)+\operatorname{adj}\left(\mathrm{B}^{-1}\right)\right)^{-1} \mathrm{~B}$, is equal to

1) $\mathrm{AB}^{-1}+\mathrm{A}^{-1} \mathrm{~B}$

2)$\operatorname{adj}\left(\mathrm{B}^{-1}\right)+\operatorname{adj}\left(\mathrm{A}^{-1}\right)$

3) $\frac{1}{|\mathrm{AB}|}(\operatorname{adj}(\mathrm{B})+\operatorname{adj}(\mathrm{A}))$

4) $\frac{\mathrm{AB}^{-1}}{|\mathrm{~A}|}+\frac{\mathrm{BA}^{-1}}{|\mathrm{~B}|}$

Correct Answer: (3) $\frac{1}{|\mathrm{AB}|}(\operatorname{adj}(\mathrm{B})+\operatorname{adj}(\mathrm{A}))$

Solution:

Given the expression:

$
\left[ A \left( \operatorname{adj}(A^{-1}) + \operatorname{adj}(B^{-1}) \right)^{-1} \cdot B \right]^{-1}
$

First, apply the inverse to the product inside the bracket:

$
= B^{-1} \cdot \left( \operatorname{adj}(A^{-1}) + \operatorname{adj}(B^{-1}) \right) \cdot A^{-1}
$

Distribute $ B^{-1} $ and $ A^{-1} $:

$
= B^{-1} \operatorname{adj}(A^{-1}) A^{-1} + B^{-1} \operatorname{adj}(B^{-1}) A^{-1}
$

Using the property of adjoint and determinant for inverse matrices:

$
\operatorname{adj}(M^{-1}) = |M^{-1}| \cdot M
$

Thus,

$
B^{-1} \operatorname{adj}(A^{-1}) A^{-1} = B^{-1} |A^{-1}| I
$

and

$
B^{-1} \operatorname{adj}(B^{-1}) A^{-1} = |B^{-1}| I A^{-1}
$

Since

$
|A^{-1}| = \frac{1}{|A|} \quad \text{and} \quad |B^{-1}| = \frac{1}{|B|}
$

we get:

$
= \frac{B^{-1}}{|A|} + \frac{A^{-1}}{|B|}
$

Rewrite by expressing inverses in terms of adjoints and determinants:

$
= \frac{\operatorname{adj} B}{|B| |A|} + \frac{\operatorname{adj} A}{|A| |B|}
$

Combine the terms:

$
= \frac{1}{|A||B|} \left( \operatorname{adj} B + \operatorname{adj} A \right)
$

Weightage of Matrices and Determinants in JEE Main 2026

Based on the 2026 exam analysis, matrices and determinants together contribute 6.95% weightage in the JEE Main Mathematics paper. In practical terms, this usually translates to 2 to 3 questions appearing in almost every shift of the exam and around 33 overall in both sessions, worth roughly 8 to 12 marks out of the 100 marks allotted to mathematics.

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Not every sub-topic within Matrices and Determinants carries equal weight. Based on analysis of previous JEE Main papers, here's how the marks typically break down:

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Also Read: JEE Mains 2027: Maths Set of 5 Sample Papers with solutions free PDF

Quick Tips for Solving JEE Main Questions of Matrices and Determinants

Speed matters the most in JEE Main, and this chapter gives that to the students:

• Substitution: When a determinant problem involves variables, try plugging in simple values like a = 1, b = 1 to quickly test which option satisfies the equation. This is often faster than expanding the full determinant symbolically.

• Row/Column Operations: Standard operations such as R1 → R1 + R2 + R3 are used again and again in matrix questions to simplify determinants before solving. Getting comfortable with these operations can turn a two-minute problem into a thirty-second one.