JEE Main 2024 Question Paper Solution Discussion Live JEE Main 2024 Question Paper Solution Discussion Live

# JEE Main Maths Matrices And Determinants Previous Year Questions With Solutions

The matrices and determinants questions from the previous years of JEE Main are present on this page, along with the detailed solution for each question. This article covers the matrices and their types, operations on matrices, transpose of a matrix, adjoint of a matrix, determinant of a matrix, the inverse of a matrix, and cofactors of a matrix to find the solution for a system of equations through matrix method, types of solutions and cube roots of unity.  These questions include all the important topics and formulae. About 2-3 questions are asked on this topic in JEE Examination.

## JEE Main Maths Matrices And Determinants Previous Year Questions With Solutions

Question 1:

$$\begin{array}{l}If A = \ \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \\ \end{matrix} \right];\,\,I=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right],\ A^{-1}=\frac{1}{6}(A^{2}+cA+dI)\end{array}$$
, where c, d ∈ R, then pair of values (c, d) are __________.

Solution:

Given

$$\begin{array}{l}A=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \\ \end{matrix} \right]\end{array}$$
$$\begin{array}{l}A^{-1}=\frac{1}{6}\left[ \begin{matrix} 6 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 2 & 1 \\ \end{matrix} \right]\end{array}$$
$$\begin{array}{l}A^2=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \\ \end{matrix} \right]\,\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \\ \end{matrix} \right]\\=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 5 \\ 0 & -10 & 14 \\ \end{matrix} \right]\end{array}$$
$$\begin{array}{l} cA=\left[ \begin{matrix} c & 0 & 0 \\ 0 & c & c \\ 0 & -2c & 4c \\ \end{matrix} \right]\end{array}$$
$$\begin{array}{l} dI=\left[ \begin{matrix} d & 0 & 0 \\ 0 & d & 0 \\ 0 & 0 & d \\ \end{matrix} \right]\end{array}$$

Therefore, by

$$\begin{array}{l}A^{-1}=\frac{1}{6}(A^2+cA+dI)\end{array}$$

⇒ 6 = 1 + c + d, (by equality of matrices)

So, (-6, 11) satisfy the relation.

Question 2:

$$\begin{array}{l}If\ P=\left[ \begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \end{matrix} \right],\,A=\left[ \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right]\end{array}$$
and Q = PAPT, then P (Q2005)PT equal to ________.

Solution:

If Q = PAPT, PT Q = APT,

(as PPT = I) PT Q2005 P = A PT Q2004 P

= A2 PT Q2003 P

= A3 PT Q2002 P

= A2004 PT (QP)

= A2004 PT (PA) (Q = PAPT ⇒ QP = PA)

= A2005

$$\begin{array}{l}A^{2005} = \left[ \begin{matrix} 1 & 2005 \\ 0 & 1 \\ \end{matrix} \right]\end{array}$$

Question 3:

$$\begin{array}{l}If\ A=\left[ \begin{matrix} \alpha & 0 \\ 1 & 1 \\ \end{matrix} \right] and \ B=\left[ \begin{matrix} 1 & 0 \\ 5 & 1 \\ \end{matrix} \right]\end{array}$$
then value of α for which A2 = B, is

A) 1

B) -1

C) 4

D) No real values

Solution:

$$\begin{array}{l}A^2=\left[ \,\begin{matrix} \alpha & 0 \\ 1 & 1 \\ \end{matrix}\, \right]\,\left[ \,\begin{matrix} \alpha & 0 \\ 1 & 1 \\ \end{matrix}\, \right]=\left[ \,\begin{matrix} {{\alpha }^{2}} & 0 \\ \alpha +1 & 1 \\ \end{matrix}\, \right]\end{array}$$

Clearly, no real value of a.

Question 4:

$$\begin{array}{l}\begin{vmatrix} 1 & a & {{a}^{2}}-bc \\ 1 & b & {{b}^{2}}-ac \\ 1 & c & {{c}^{2}}-ab \\ \end{vmatrix} =\end{array}$$

Solution:

$$\begin{array}{l}\left| \,\begin{matrix} 1 & a & {{a}^{2}}-bc \\ 1 & b & {{b}^{2}}-ac \\ 1 & c & {{c}^{2}}-ab \\ \end{matrix}\, \right|=\left| \,\begin{matrix} 0 & a-b & (a-b)\,(a+b+c) \\ 0 & b-c & (b-c)\,\,(a+b+c) \\ 1 & c & {{c}^{2}}-ab \\ \end{matrix}\, \right|\end{array}$$

by R1 → R1 − R2 & R2 → R2 − R3

$$\begin{array}{l}(a-b)\,(b-c)\,\left| \,\begin{matrix} 0 & 1 & a+b+c \\ 0 & 1 & a+b+c \\ 1 & c & {{c}^{2}}-ab \\ \end{matrix}\, \right|\end{array}$$

= 0, {Because R1 = R2}.

Question 5: The roots of the equation

$$\begin{array}{l}\left| \,\begin{matrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & 5{{x}^{2}} \\ \end{matrix}\, \right|=0\end{array}$$
are _________.

Solution:

$$\begin{array}{l}\left| \begin{matrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & 5{{x}^{2}} \\ \end{matrix}\,\, \right|=0\end{array}$$
$$\begin{array}{l} \left| \begin{matrix} 0 & 6 & 15 \\ 0 & -2-2x & 5(1-{{x}^{2}}) \\ 1 & 2x & 5{{x}^{2}} \\ \end{matrix}\, \right|\,=0\end{array}$$
[R1 → R1 − R2 and R2 → R2 − R3]
$$\begin{array}{l}\begin{vmatrix} 0 &1 &1 \\ 0 & -(1+x) &1-x^2 \\ 1&x & x^2 \end{vmatrix}=0\end{array}$$

Rightarrow x + 1 = 0 or x – 2 = 0
x = -1, x = 2

Trick: Obviously, by inspection, x = −1, 2 satisfies the equation.

At x = −1,

$$\begin{array}{l}\left| \,\begin{matrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & -2 & 5 \\ \end{matrix}\, \right|\,=0\end{array}$$
as R2 = R3

At x = 2,

$$\begin{array}{l}\left| \,\begin{matrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & -2 & 5 \\ \end{matrix}\, \right|\,=0\end{array}$$
as R1 = R3

Question 6:

$$\begin{array}{l}\left| \,\begin{matrix} a+b & a+2b & a+3b \\ a+2b & a+3b & a+4b \\ a+4b & a+5b & a+6b \\ \end{matrix}\, \right|=——-.\end{array}$$

Solution:

$$\begin{array}{l}\left| \,\begin{matrix} a+b & a+2b & a+3b \\ a+2b & a+3b & a+4b \\ a+4b & a+5b & a+6b \\ \end{matrix}\, \right|\,=\,\left| \,\begin{matrix} a+b & a+2b & a+3b \\ b & b & b \\ 2b & 2b & 2b \\ \end{matrix}\, \right| = 0\end{array}$$

by R2 → R2 − R1 and R3 → R3 − R2

Trick: Putting a = 1 = b.

The determinant will be

$$\begin{array}{l}\left| \,\begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 5 & 6 & 7 \\ \end{matrix}\, \right|=0\end{array}$$

Question 7:

$$\begin{array}{l}\left| \,\begin{matrix} {{b}^{2}}+{{c}^{2}} & {{a}^{2}} & {{a}^{2}} \\ {{b}^{2}} & {{c}^{2}}+{{a}^{2}} & {{b}^{2}} \\ {{c}^{2}} & {{c}^{2}} & {{a}^{2}}+{{b}^{2}} \\ \end{matrix}\, \right|=——–.\end{array}$$

Solution:

$$\begin{array}{l}\Delta =\begin{vmatrix} {{b}^{2}}+{{c}^{2}} & {{a}^{2}} & {{a}^{2}} \\ {{b}^{2}} & {{c}^{2}}+{{a}^{2}} & {{b}^{2}} \\ {{c}^{2}} & {{c}^{2}} & {{a}^{2}}+{{b}^{2}} \\ \end{vmatrix} = \begin{vmatrix} 0 & {{-2c}^{2}} & {{-2b}^{2}} \\ {{b}^{2}} & {{c}^{2}}+{{a}^{2}} & {{b}^{2}} \\ {{c}^{2}} & {{c}^{2}} & {{a}^{2}}+{{b}^{2}} \\ \end{vmatrix}\end{array}$$

by R1 → R1 − R2 − R

Solving the determinant, we get ∆ = 4a2b2c2.

Question 8: If – 9 is a root of the equation

$$\begin{array}{l}\left| \,\begin{matrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \\ \end{matrix}\, \right|=0\end{array}$$
, then the other two roots are ________.

Solution:

$$\begin{array}{l}\left| \,\begin{matrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \\ \end{matrix}\, \right|\,=0\\\ \Rightarrow \ (x+9)\,\left| \,\begin{matrix} 1 & 1 & 1 \\ 2 & x & 2 \\ 7 & 6 & x \\ \end{matrix}\, \right|=0,\end{array}$$

by R1 → R1 + R2 + R3

⇒ (x + 9) {(x2 − 12) − (2x − 14) + (12 − 7x)} = 0

⇒ (x + 9) (x2 − 9x + 14) = 0

⇒ (x + 9) (x − 2) (x − 7) = 0

Hence, the other two roots are x = 2, 7.

Question 9: If a, b, c are unequal what is the condition that the value of the following determinant is zero

$$\begin{array}{l}\Delta =\left| \,\begin{matrix} a & {{a}^{2}} & {{a}^{3}}+1 \\ b & {{b}^{2}} & {{b}^{3}}+1 \\ c & {{c}^{2}} & {{c}^{3}}+1 \\ \end{matrix}\, \right|?\end{array}$$

Solution:

Splitting the determinant into two determinants, we get

$$\begin{array}{l}\Delta =\left| \,\begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix}\, \right|+abc\,\left| \,\begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix}\, \right|\,=0 \\= (1+abc)\,[(a-b)\,(b-c)\,(c-a)]=0\end{array}$$

Because a, b, c are different, the second factor cannot be zero.

Hence, 1 + abc = 0, is the condition.

Question 10: The determinant

$$\begin{array}{l}\left| \,\begin{matrix} a & b & a\alpha +b \\ b & c & b\alpha +c \\ a\alpha +b & b\alpha +c & 0 \\ \end{matrix}\, \right|=0\end{array}$$
if a, b, c are in

A) A. P.

B) G. P.

C) H. P.

D) None of these

Solution:

$$\begin{array}{l}\Delta \equiv \left| \,\begin{matrix} a & b & a\alpha +b \\ b & c & b\alpha +c \\ a\alpha +b & b\alpha +c & 0 \\ \end{matrix}\, \right| = \left| \,\begin{matrix} a & b & a\alpha +b \\ b & c & b\alpha +c \\ 0 & 0 & -(a{{\alpha }^{2}}+2b\alpha +c) \\ \end{matrix}\, \right|,\end{array}$$

R3 → R3 − αR1 − R2

= a {−c(aα2 + 2bα + c) − 0} − b{−b(aα2 + 2bα + c) − 0} by expanding along

C1 = (b2 − ac)(aα2 + 2bα + c)

Thus, Δ = 0, if either b2 − ac = 0 or aα2 + 2bα + c = 0 i.e., a, b, c in G.P. or aα2 + 2bα + c = 0.

Trick: Put α = 0, then the determinant

$$\begin{array}{l}\left| \,\begin{matrix} a & b & b \\ b & c & c \\ b & c & 0 \\ \end{matrix}\, \right|\,=\,\left| \,\begin{matrix} a & b & 0 \\ b & c & 0 \\ b & c & -c \\ \end{matrix}\, \right|\,\\=\,-c(ac-{{b}^{2}})\\=0.\end{array}$$

Question 11:

$$\begin{array}{l}If\ \left| \,\begin{matrix} {{x}^{2}}+x & x+1 & x-2 \\ 2{{x}^{2}}+3x-1 & 3x & 3x-3 \\ {{x}^{2}}+2x+3 & 2x-1 & 2x-1 \\ \end{matrix}\, \right|=Ax-12,\end{array}$$
then what is the value of A?

Solution:

Put x = 1, then we have

$$\begin{array}{l}\left| \,\begin{matrix} 2 & 2 & -1 \\ 4 & 3 & 0 \\ 6 & 1 & 1 \\ \end{matrix}\, \right|=A-12\\\Rightarrow \left| \,\begin{matrix} 0 & 2 & -1 \\ 1 & 3 & 0 \\ 5 & 1 & 1 \\ \end{matrix}\, \right|=A-12\end{array}$$

Apply C1 → C1 − C2

⇒ −2 + (−1) (−14) = A − 12

⇒ A = 24

Question 12:

$$\begin{array}{l}If\ \left| \,\begin{matrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \\ \end{matrix}\, \right|=x+iy,\end{array}$$
then

A) x = 3, y = 1

B) x = 0, y = 0

C) x = 0, y = 3

D) x = 1, y = 3

Solution:

=> [6i(-3 + 3) + 3i(4i + 20) + 1(12 – 60i) = x + iy

⇒ x = 0, y = 0

Hence option B is the answer.

Question 13:

$$\begin{array}{l}If\ \omega =-\frac{1}{2}+i\frac{\sqrt{3}}{2}\end{array}$$
, then the value of the determinant
$$\begin{array}{l}\left| \,\begin{matrix} 1 & 1 & 1 \\ 1 & -1-{{\omega }^{2}} & {{\omega }^{2}} \\ 1 & {{\omega }^{2}} & {{\omega }^{4}} \\ \end{matrix}\, \right|\ is——-.\end{array}$$

Solution:

$$\begin{array}{l}\Delta =\left| \,\begin{matrix} 3 & 1 & 1 \\ 0 & -1-{{\omega }^{2}} & {{\omega }^{2}} \\ 0 & {{\omega }^{2}} & \omega \\ \end{matrix}\, \right|\end{array}$$

(C1 → C1 + C2 + C3) (Because 1 + ω + ω2 = 0)

= 3 [ω . ω − ω4]

= 3 (ω2 − ω)

= 3 ω (ω − 1)

Question 14: The number of values of k for which the system of equations (k + 1) x + 8y = 4k, kx + (k + 3) y = 3k − 1 has infinitely many solutions, is _______.

Solution:

For infinitely many solutions, the two equations must be identical.

⇒ [k + 1] / [k] = [8] / [k + 3] = [4k] / [3k − 1]

⇒ (k + 1) (k + 3) = 8k and 8 (3k − 1) = 4k (k + 3)

⇒ k2 − 4k + 3 = 0 and k2 −3k + 2 = 0

We get k = 1, 3 and k = 1, 2

Common is k = 1

So k = 1