Determinants MCQ Quiz - Objective Question with Answer for Determinants - Download Free PDF

Given:

M = \(\begin{vmatrix} -1 & 0 \\ 2 & 3 \end{vmatrix} \)

N = \(\begin{vmatrix} 0 & -2 \\ -2 & 3 \end{vmatrix}\)

Calculation:

First, calculate 2M:

2M = \(\begin{vmatrix} -2 & 0 \\ 4 & 6 \end{vmatrix} \)

Now, calculate 2M + N :

2M + N = \(\begin{vmatrix} -2 & 0 \\ 4 & 6 \end{vmatrix} \) + \(\begin{vmatrix} 0 & -2 \\ -2 & 3 \end{vmatrix}\)

\(\begin{vmatrix} -2+0 & 0 +(-2)\\ 4+(-2) & 6+3 \end{vmatrix} \)

\(\begin{vmatrix} -2 & -2\\ 2 & 9 \end{vmatrix} \)

∴ Option 4 is the correct answer.

Calculation:

Statement I

\( |M^2| = |M \times M| = |M| \cdot |M| = |M|^2 \)

⇒ Statement I is correct.

Statement II

For a non-singular matrix, \( M \times M^{-1} = I \), where \( I \) is the identity matrix.

\( |M| \cdot |M^{-1}| = |I| = 1 \)

⇒ Statement II is incorrect unless \( |M| = \pm 1 \).

Statement III

The determinant of a matrix is equal to the determinant of its transpose:

\( |M| = |M^T| \)

⇒ Statement III is correct.

Out of the three statements, two are correct: I and III.

Calculation:

Given,

Let ω be a non-real cube root of unity, so \( \omega^{3}=1 \) and \( 1+\omega+\omega^{2}=0 \).

Consider the determinant

\( \Delta(x)= \begin{vmatrix} x+1 & \omega & \omega^{2}\\ \omega & x+\omega^{2} & 1\\ \omega^{2} & 1 & x+\omega \end{vmatrix}=0. \)

Step 1 — Column operation:  Replace the first column by \(C_{1}-C_{2}\):

\( \Delta(x)= \begin{vmatrix} x^{2}-1 & \,2\!k\!+\!1\, & 1\\ k-1 & k+2 & 1\\ 0 & 3 & 1 \end{vmatrix}. \)

Step 2 — Expansion along the third row:

\( \Delta(x)= -3\! \begin{vmatrix} k^{2}-1 & 1\\ k-1 & 1 \end{vmatrix} + \begin{vmatrix} k^{2}-1 & 2k+1\\ k-1 & k+2 \end{vmatrix}, \)

which simplifies to

\( \Delta(x)=x(x^{2}-1)-x\bigl(\omega+\omega^{2}\bigr) =x(x^{2}-1)+x =x^{3}. \)

Step 3 — Equate to zero:

\( \Delta(x)=0 \;\Longrightarrow\; x^{3}=0 \;\Longrightarrow\; x=0. \)

∴ The root of the equation is  \( x = 0 \).

Hence, the correct answer is Option 1.

Concept:

When A² + B² + C² = 0, it implies A = B = C = 0 (since the squares of real numbers are non-negative).

Substitute the values of A, B, and C for determinant calculation into the matrix

Calculation:

\(\begin{vmatrix} 1& cos0& cos0\\ cos0&1&cos0 \\ cos0&cos0&1 \end{vmatrix} \)

Since, Cos0 =1

Thus Matrix becomes 

\(\begin{vmatrix} 1& 1& 1\\ 1&1&1 \\ 1&1&1 \end{vmatrix} \)

Now determinant = 1[(1×1 - 1×1)] - 1[(1×1 - 1×1)] + 1[(1×1 - 1×1)]

= = 1(0) - 1(0) + 1(0) = 0

∴ The value of the determinant is 0.

Hence, the correct answer is Option 2.

Calculation:

Determinant Δ = \(a(ei−fh)−b(di−fg)+c(dh−eg)\)

Now, For our matrix, 

\(a=2,b=3+i,c=−1,d=3−i,e=0,f=i,g=−1,h=−i,i=1\)

calculate the subdeterminants

⇒ \( ei−fh=(0)(1)−(i)(−i)=0−(−1)=1\)

⇒ \(di−fg=(3−i)(1)−(i)(−1)=3−i+i=3\)

⇒ \(dh−eg=(3−i)(−i)−(0)(−1)=−3i+i 2=−3i−1=−1−3i\)

⇒ Δ = \(2(1)−(3+i)(3)+(−1)(−1−3i)\)

⇒ Δ = \(2−9−3i+1+3i\)

⇒ \(Δ=−6+0i\)

Since we are given that $\backslash (\Delta =A+iB\backslash )$ comparing the real and imaginary parts, we find:

A  = -6 and B = 0

Thus A + B = -6 + 0 = - 6

Hence, the Correct answer is Option 2.

Concept:

If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_{11}}}&{{{\rm{a}}_{12}}}\\ {{{\rm{a}}_{21}}}&{{{\rm{a}}_{22}}} \end{array}} \right]\) then determinant of A is given by:

|A| = (a­11 × a22) - (a12 × a21)

Calculation:

Given: \({\rm{f}}\left( {\rm{x}} \right) = {\rm{\;}}\left| {\begin{array}{*{20}{c}} 1&2\\ {\rm{x}}&{\rm{a}} \end{array}} \right|\)

To find: 2f(x) – f(2x) =?

\({\rm{f}}\left( {\rm{x}} \right) = {\rm{\;}}\left| {\begin{array}{*{20}{c}} 1&2\\ {\rm{x}}&{\rm{a}} \end{array}} \right| = \left( {1 \times {\rm{a}}} \right) - \left( {2 \times {\rm{x}}} \right)\)

\( \Rightarrow {\rm{f}}\left( {\rm{x}} \right) = {\rm{a}} - 2{\rm{x\;\;}}\)

So, \({\rm{f}}\left( {2{\rm{x}}} \right) = {\rm{a}} - 2\left( {2{\rm{x}}} \right) = {\rm{a}} - 4{\rm{x}}\)                 (put x = 2x)

\(\therefore 2{\rm{f}}\left( {\rm{x}} \right) - {\rm{\;f}}\left( {2{\rm{x}}} \right) = 2\left( {{\rm{a}} - 2{\rm{x}}} \right) - \left( {{\rm{a}} - 4{\rm{x}}} \right)\)

\(\Rightarrow 2{\rm{a}} - 4{\rm{x}} - {\rm{a}} + 4{\rm{x}} = {\rm{a}}\)

Hence, option (4) is correct.

Concept:

Properties of Determinant of a Matrix:

• If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.
• For any square matrix say A, |A| = |AT|.
• If we interchange any two rows (columns) of a matrix, the determinant is multiplied by -1.
• If any two rows (columns) of a matrix are same then the value of the determinant is zero.

Calculation:

\(\begin{vmatrix} \rm x-a & \rm y-b & \rm z-c\\ \rm a & \rm b & \rm c \\ \rm x & \rm y & \rm z \end{vmatrix}\)

Apply R3 → R3 - R2

= \(\begin{vmatrix} \rm x-a & \rm y-b & \rm z-c\\ \rm a & \rm b & \rm c \\ \rm x-a & \rm y-b & \rm z-c \end{vmatrix}\)

As we can see that the first and the third row of the given matrix are equal. 

We know that, if any two rows (columns) of a matrix are same then the value of the determinant is zero.

\(\begin{vmatrix} \rm x-a & \rm y-b & \rm z-c\\ \rm a & \rm b & \rm c \\ \rm x & \rm y & \rm z \end{vmatrix}\) = 0

Concept:

\(\rm i = \sqrt {-1}\)

i² = -1 , i³ = - i, i⁴ = 1, i⁶ = - 1 , i⁸ = 1 , i⁹ = i, i ¹² = 1, and i¹⁵ = - i

Calculations: 

Given determinant is \(\left| {\begin{array}{*{20}{c}} {\rm{i}}&{{{\rm{i}}^2}}&{{{\rm{i}}^3}}\\ {{{\rm{i}}^4}}&{{{\rm{i}}^6}}&{{{\rm{i}}^8}}\\ {{{\rm{i}}^9}}&{{{\rm{i}}^{12}}}&{{{\rm{i}}^{15}}} \end{array}} \right|\)

Since, we have, 

\(\rm i = \sqrt {-1}\)

i2 = -1 , i3 = - i, i4 = 1, i6 = - 1 , i8 = 1 , i9 = i, i 12 = 1, and i15 = - i

=\(\left| {\begin{array}{*{20}{c}} {\rm{i}}&{{{\rm{-1}}}}&{{{\rm{-i}}}}\\ {{{\rm{1}}}}&{{{\rm{-1}}}}&{{{\rm{1}}}}\\ {{{\rm{i}}}}&{{{\rm{1}}}}&{{{\rm{-i}}}} \end{array}} \right|\)

=i(i - 1) + 1(-i - i) - i (1 + i)

= i² - i - 2i - i - i²

= - 4i

Concept

Let the system of equations be,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

\(\; \Rightarrow \left[ {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right]\;\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} {{d_1}}\\ {{d_2}}\\ {{d_3}} \end{array}} \right]\)

⇒ AX = B

⇒ X = A^-1 B = \(\frac{{{\rm{adj\;}}\left( {\rm{A}} \right)}}{{\det {\rm{\;}}({\rm{A}})}}\;B\)

⇒ If det (A) ≠ 0, the system is consistent having unique solution.

⇒ If det (A) = 0 and (adj A). B = 0, system is consistent, with infinitely many solutions.

⇒ If det (A) = 0 and (adj A). B ≠ 0, system is inconsistent (no solution)

Calculation:

Given: 

kx + y + z = 1, x + ky + z = k and x + y + kz = k²

\( \Rightarrow {\rm{A}} = {\rm{}}\left[ {\begin{array}{*{20}{c}} {\rm{k}}&1&1\\ 1&{\rm{k}}&1\\ 1&1&{\rm{k}} \end{array}} \right],{\rm{B}} = {\rm{}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right]{\rm{and\;C}} = \left[ {\begin{array}{*{20}{c}} 1\\ {\rm{k}}\\ {{{\rm{k}}^2}} \end{array}} \right]\)

⇒ For the given equations to have no solution, |A| = 0

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {\rm{k}}&1&1\\ 1&{\rm{k}}&1\\ 1&1&{\rm{k}} \end{array}} \right| = 0\)

⇒ k(k² – 1) - 1(k – 1) + 1(1 – k) = 0

⇒ k³ – k – k + 1 + 1 – k = 0

⇒ k³ -3k +2 = 0

⇒ (k – 1) (k – 1) (k + 2) = 0

⇒ k = 1, -2

If we put k = 1 in the above given equations, then all the equations will become the same.

Hence, the given equations have no solution if k = - 2. 

Concept:

1. Determinant of a 3 × 3 matrix:

• Let A be a 3 × 3 matrix given by:

\({\rm{A}} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{b}}&{\rm{c}}\\ {\rm{f}}&{\rm{e}}&{\rm{d}}\\ {\rm{g}}&{\rm{h}}&{\rm{i}} \end{array}} \right]\)

then the value of |A| also written as det(A) is:

det (A) = a (ei - dh) – b (fi - dg) + c (fh - eg)

2. Property of determinant of a matrix:

• Let A be a matrix of order n × n and det(A) = k. Then for a scaler c, the following property holds:

          det(cA) = cⁿ det(A)

Calculation:

First evaluate the determinant of the given matrix:

det(A) = 4(15 - 0) – 7(-5 + 4) + 1(0 + 6)

= 4(15) -7(-1) + 1(6)

= 60 + 7 + 6

= 73

Now using the property the value of det(3A) is:

det(3A) = 3³ det(A)

= 27 × 73

= 1971

Concept:

If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_{11}}}&{{{\rm{a}}_{12}}}\\ {{{\rm{a}}_{21}}}&{{{\rm{a}}_{22}}} \end{array}} \right]\) then determinant of A is given by:

|A| = a11 × a22 – a21 × a12

|An| = |A|n 

Calculation:

Given that,

\({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {\rm{x}}&2\\ 4&{\rm{x }} \end{array}} \right]\) and |A2| = 64

⇒ |A| = x2 - 8           .... (1)

Given |A2| = 64

⇒ |A|2 = 64          [∵ |An| = |A|n]

⇒ |A| = (64)1/2 = 8       ....(2)

From equation 1 and 2

⇒ x2 - 8 = 8

⇒ x2 = 16

Concept:

Area of a triangle with vertices (x1, y1) , (x2, y2), (x3, y3) is given by

​Area = \(\rm \dfrac 12 \rm \begin{vmatrix} \rm x_1&\rm y_1 &1 \\ x_2& \rm y_2&1 \\ \rm\rm x_3 &\rm y_3&1 \end{vmatrix}\)

Calculations:

Given that, vertices of triangle are (-3, 0), (3, 0) and (0, k)

By using the above formula,

​⇒ Area = \(\rm \dfrac 12 \rm \begin{vmatrix} \rm -3&\rm 0 &1 \\ 3& 0&1 \\ 0 &k&1 \end{vmatrix}\)

⇒ Area = \(\dfrac 12\)[-3(0 - k) - 0 + 1(3k)]

⇒ Area = 3k

According to the question, area of triangle is 9 square unit,

⇒ 3k = 9

⇒ k = 3

∴ Required value of k is 3 unit.

Concept:

Area of equilateral triangle = \(\frac{\sqrt{3}}{4}\)×a2

Calculation:

Given: The co-ordinates of its vertices are (x1, y1); (x2, y2): (x3, y3) then the square of the determinant \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}\)

⇒ (△ ABC ) = \(\dfrac{1}{2}\) \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}\) = ( \(\frac{\sqrt{3}}{4}\)) ×a²

On squaring both side, 

⇒ \(\dfrac{1}{4}\)  \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}^2\) = \(\dfrac{3a^4}{16}\)

⇒ \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}^2\) = \(\dfrac{3a^4}{4}\)

Concept:

If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\) then determinant of A is given by: |A| = (a­₁₁ × a₂₂) - (a₁₂ - a₂₁).

If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]\) then determinant of A is given by:

|A| = a₁₁ × {(a₂₂ × a₃₃) - (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) - (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) - (a₂₂ × a₃₁)}

Calculation:

\(\Rightarrow \left| {\begin{array}{*{20}{c}} x&0&2\\ {2x}&2&1\\ 1&1&1 \end{array}} \right| = x \times \left( {2 - 1} \right) - 0 \times \left( {2x - 1} \right) + 2 \times \left( {2x - 2} \right) = 5x - 4\)

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {3x}&0&2\\ {{x^2}}&2&1\\ 0&1&1 \end{array}} \right| = 3x \times \left( {2 - 1} \right) - 0 \times \left( {{x^2} - 0} \right) + 2 \times \left( {{x^2} - 0} \right) = 2{x^2} + 3x\)

\(\Rightarrow \left| {\begin{array}{*{20}{c}} x&0&2\\ {2x}&2&1\\ 1&1&1 \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {3x}&0&2\\ {{x^2}}&2&1\\ 0&1&1 \end{array}} \right| = \left( {5x - 4} \right) + \left( {2{x^2} + 3x} \right) = 2{x^2} + 8x - 4 = 0\)

⇒ 2x² + 8x - 4 = 0

By comparing the equation 2x² + 8x - 4 = 0 with ax² + bx + c = 0, we get a = 2, b = 8 and c = - 4.

Concept:

Properties of determinants:​

For a n×n matrix A, det(kA) = kn det(A).

Calculation:

Given:

|A| = 5

k = 5

From the properties of the determinants, we know that |KA| = Kn |A|, where n is the order of the determinant.

Here, n = 2, therefore, the answer is K2 |A|.

|5A| = 52|A|

|5A| = 52 × 5 = 125