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 in 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, 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 from this topic in JEE Examination.

JEE Main Maths Matrices And Determinants Previous Year Questions With Solutions

Question 1:

[100011024];I=[100010001]\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]

A1=16(A2+cA+dI)A^{-1}=\frac{1}{6}(A^{2}+cA+dI) where c, d ∈ R, then pair of values (c, d) are __________.

Solution:

Given

A=[100011024]A=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \\ \end{matrix} \right] A1=16[600041021]A^{-1}=\frac{1}{6}\left[ \begin{matrix} 6 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 2 & 1 \\ \end{matrix} \right] A2=[100011024][100011024]=[10001501014]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] cA=[c000cc02c4c] cA=\left[ \begin{matrix} c & 0 & 0 \\ 0 & c & c \\ 0 & -2c & 4c \\ \end{matrix} \right] dI=[d000d000d] dI=\left[ \begin{matrix} d & 0 & 0 \\ 0 & d & 0 \\ 0 & 0 & d \\ \end{matrix} \right]

Therefore, by A1=16(A2+cA+dI)A^{-1}=\frac{1}{6}(A^2+cA+dI)

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

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

Question 2: If P=[32121232],A=[1101] 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] 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

A2005 = [1200501]\left[ \begin{matrix} 1 & 2005 \\ 0 & 1 \\ \end{matrix} \right]

Question 3: If A=[α011]and B=[1051]A=\left[ \begin{matrix} \alpha & 0 \\ 1 & 1 \\ \end{matrix} \right] and \ B=\left[ \begin{matrix} 1 & 0 \\ 5 & 1 \\ \end{matrix} \right] then value of α for which A2 = B, is

A) 1

B) -1

C) 4

D) No real values

Solution:

A2=[α011][α011]=[α20α+11]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]

Clearly, no real value of a.

Question 4: 1aa2bc1bb2ac1cc2ab=\left| \,\begin{matrix} 1 & a & {{a}^{2}}-bc \\ 1 & b & {{b}^{2}}-ac \\ 1 & c & {{c}^{2}}-ab \\ \end{matrix}\, \right|= _________.

Solution:

1aa2bc1bb2ac1cc2ab=0ab(ab)(a+b+c)0bc(bc)(a+b+c)1cc2ab\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|\\

by R1 → R1 − R2 & R2 → R2 − R3

(ab)(bc)01a+b+c01a+b+c1cc2ab=0(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|=0\,

= 0, {Because R1 = R2}.

Question 5: The roots of the equation 142012512x5x2=0\left| \,\begin{matrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & 5{{x}^{2}} \\ \end{matrix}\, \right|=0 = 0 are _________.

Solution:

142012512x5x2=0\left| \begin{matrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & 5{{x}^{2}} \\ \end{matrix}\,\, \right|=0  0615022x5(1x2)12x5x2=0 \left| \begin{matrix} 0 & 6 & 15 \\ 0 & -2-2x & 5(1-{{x}^{2}}) \\ 1 & 2x & 5{{x}^{2}} \\ \end{matrix}\, \right|\,=0 [R1 → R1 − R2 and R2 → R2 − R3]

0110(1+x)1x21xx2=0\begin{vmatrix} 0 &1 &1 \\ 0 & -(1+x) &1-x^2 \\ 1&x & x^2 \end{vmatrix}=0

=> Rightarrow x+1=0 or x-2=0
=> x=-1

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

At x = −1, 1420125125=0\left| \,\begin{matrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & -2 & 5 \\ \end{matrix}\, \right|\,=0 as R2 = R3

At x = 2,

\begin{vmatrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 4 & 20 \\ \end{vmatrix}\, \right|=0 as R1 = R3

Question 6: a+ba+2ba+3ba+2ba+3ba+4ba+4ba+5ba+6b=\left| \,\begin{matrix} a+b & a+2b & a+3b \\ a+2b & a+3b & a+4b \\ a+4b & a+5b & a+6b \\ \end{matrix}\, \right|= __________.

Solution:

a+ba+2ba+3ba+2ba+3ba+4ba+4ba+5ba+6b=a+ba+2ba+3bbbb2b2b2b=0\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

by R2 → R2 − R1 and R3 → R3 − R2

Trick: Putting a = 1 = b.

The determinant will be 234345567=0\left| \,\begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 5 & 6 & 7 \\ \end{matrix}\, \right|=0

Obviously the answer is 0.

Question 7: b2+c2a2a2b2c2+a2b2c2c2a2+b2=\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|= _________.

Solution:

Δ=b2+c2a2a2b2c2+a2b2c2c2a2+b2=20c2b2b2c2+a2b2c2c2a2+b2\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} =-2 \begin{vmatrix} 0 & {{c}^{2}} & {{b}^{2}} \\ {{b}^{2}} & {{c}^{2}}+{{a}^{2}} & {{b}^{2}} \\ {{c}^{2}} & {{c}^{2}} & {{a}^{2}}+{{b}^{2}} \\ \end{vmatrix}

by R2 → R2 − R1 and R3 → R3 − R1

=>2{c2(b2a2)+b2(c2a2)}=4a2b2c2-2\{-{{c}^{2}}({{b}^{2}}{{a}^{2}})+{{b}^{2}}(-{{c}^{2}}{{a}^{2}})\}=4{{a}^{2}}{{b}^{2}}{{c}^{2}}

Question 8: If – 9 is a root of the equation x372x276x=0\left| \,\begin{matrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \\ \end{matrix}\, \right|=0 = 0, then the other two roots are ________.

Solution:

x372x276x=0  (x+9)1112x276x=0,\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,

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 Δ=aa2a3+1bb2b3+1cc2c3+1\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|?

Solution:

Splitting the determinant into two determinants, we get

Δ=1aa21bb21cc2+abc1aa21bb21cc2=0=(1+abc)[(ab)(bc)(ca)]=0\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

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

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

Question 10: If pλ4+qλ3+rλ2+sλ+t=λ2+3λλ1λ+3λ+12λλ4λ3λ+43λ,p{{\lambda }^{4}}+q{{\lambda }^{3}}+r{{\lambda }^{2}}+s\lambda +t=\left| \,\begin{matrix} {{\lambda }^{2}}+3\lambda & \lambda -1 & \lambda +3 \\ \lambda +1 & 2-\lambda & \lambda -4 \\ \lambda -3 & \lambda +4 & 3\lambda \\ \end{matrix}\, \right|, then the value of t is ______.

Solution:

Since it is an identity in a(1+w)bw2awb(w+w2)cbw2c(w2+1)awc\left| \,\begin{matrix} a(1+w) & b{{w}^{2}} & aw \\ b(w+{{w}^{2}}) & c & b{{w}^{2}} \\ c({{w}^{2}}+1) & aw & c \\ \end{matrix}\, \right|

is satisfied by every value of λ.

Now put BC = [ba][-b\,\,-a]\, (aa)\begin{pmatrix} a\\ -a \end{pmatrix}

= [a2 −ab] in the given equation, we have

013124340\begin{vmatrix} 0 &-1 &3 \\ 1& 2 & -4\\ -3& 4 & 0 \end{vmatrix}

= −12 + 30

= 18

Question 11: The determinant abaα+bbcbα+caα+bbα+c0=0\left| \,\begin{matrix} a & b & a\alpha +b \\ b & c & b\alpha +c \\ a\alpha +b & b\alpha +c & 0 \\ \end{matrix}\, \right|=0 if a, b, c are in

A) A. P.

B) G. P.

C) H. P.

D) None of these

Solution:

Δabaα+bbcbα+caα+bbα+c0=abaα+bbcbα+c00(aα2+2bα+c),\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|,

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=0or aα2+2bα+c=0 i.e., a,b,c in G.P. or aα2+2bα+c=0.

Trick: Put α = 0, then the determinant

abbbccbc0=ab0bc0bcc=c(acb2)=0.\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.

Question 12: If x2+xx+1x22x2+3x13x3x3x2+2x+32x12x1=Ax12,\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, then what is the value of A?

Solution:

Put x = 1, then we have

221430611=A12021130511=A12\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

Apply C1 → C1 − C2

⇒ −2 + (−1) (−14) = A − 12

⇒ A = 24

Question 13: Let 6i3i143i1203i=x+iy,\left| \,\begin{matrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \\ \end{matrix}\, \right|=x+iy, then

A) x = 3, y = 1

B) x = 0, y = 0

C) x = 0, y = 3

D) x = 1, y = 3

Solution:

(a+b+c)22bc2c2bb2c+ab0c20a+bc{{(a+b+c)}^{2}} \left|\begin{matrix} 2bc & -2c & -2b \\ {{b}^{2}} & c+a-b & 0 \\ {{c}^{2}} & 0 & a+b-c \\ \end{matrix}\, \right|

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

⇒ (C1 → C1 + C2 + C3)

⇒ x = 0, y = 0

Question 14: Let ω=12+i32\omega =-\frac{1}{2}+i\frac{\sqrt{3}}{2}. Then the value of the determinant 11111ω2ω21ω2ω4\left| \,\begin{matrix} 1 & 1 & 1 \\ 1 & -1-{{\omega }^{2}} & {{\omega }^{2}} \\ 1 & {{\omega }^{2}} & {{\omega }^{4}} \\ \end{matrix}\, \right| is ______.

Solution:

Δ=31101ω2ω20ω2ω\Delta =\left| \,\begin{matrix} 3 & 1 & 1 \\ 0 & -1-{{\omega }^{2}} & {{\omega }^{2}} \\ 0 & {{\omega }^{2}} & \omega \\ \end{matrix}\, \right|

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

= 3 [ω . ω − ω4]

= 3 (ω2 − ω)

= 3 ω (ω − 1)

Question 15: 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

By cross multiplication,

[k2] / [−8 + 9] = [k] / [3 − 2] = [1] / [−3 + 4]

k2 = 1 and k = 1;

k = 1

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