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

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.

Calculation:

Given,

Δ  =  \( \begin{vmatrix} k(k+2) & 2k+1 & 1\\ 2k+1 & k+2 & 1\\ 3 & 3 & 1 \end{vmatrix} \)

Simplify the determinant by the column operation  \(C_1 \rightarrow C_1 - C_2\):

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

Expanding along the third row,

\( \Delta = -3 \begin{vmatrix} k^{2}-1 & 1\\ k-1 & 1 \end{vmatrix} \;+\; \begin{vmatrix} k^{2}-1 & 2k+1\\ k-1 & k+2 \end{vmatrix} = (k-1)^{3}. \)

Thus  \( \Delta = (k-1)^{3}\).

Sign analysis

• \(k>0\): if \(0, Δ < 0; if \(k>1\), Δ > 0  ⇒ Statement I is false.
• \(k<0\):  \(\Delta<0\)  ⇒ Statement II is true.
• \(k=0\):  \(\Delta=(-1)^{3}=-1\neq0\)  ⇒ Statement III is false.

∴ Only Statement II is correct  ⇒  exactly one statement is true.

Hence, the correct answer is Option 2.

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:

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:

Property of determinants:

If A and B are two square matrices then |AB| = |A||B|

Calculation:

Given: A = \(\begin{bmatrix} 2 & 5 \\ 2 & 1 \end{bmatrix}\) and B = \(\begin{bmatrix} 4 & -3 \\ 1 & 5 \end{bmatrix}\)

Now,

|A| = 2 × 1 - 5 × 2 = 2 - 10 = -8

|B| = 4 × 5 - (-3 × 1) = 20 + 3 = 23

As we know that, |AB| = |A||B|

= -8 × 23 = -184

Concept:

Elementary row or column transformations do not change the value of the determinant of a matrix.

Calculation:

\(\left| {\begin{array}{*{20}{c}} 1&1&1\\ 1&{1 + {\rm{a}}}&1\\ {1 + {\rm{b}}}&1&1 \end{array}} \right|\)

Applying R₂ → R₂ – R₁, R₃ → R₃ – R₁, we get

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

Now, Expanding along C₃

= 1 (0 – ab) – 0 + 0 = -ab

Concept:

Elementary row or column transformations do not change the value of the determinant of a matrix.

Calculation:

\(\left| {\begin{array}{*{20}{c}} 1&1&1\\ 1&{1 + {\rm{x}}}&1\\ 1&1&{1 + {\rm{y}}} \end{array}} \right|\)

Applying R₂ → R₂ – R₁, R₃ → R₃ – R₁, we get

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

Now, Expanding along C₁

= 1 (xy – 0) – 0 + 0 = xy

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:

\(\begin{vmatrix} 2 & 7 & 37\\ 3& 6 & 33\\ 4 & 5 & 29 \end{vmatrix}\)

Apply C2 → 5C2 + C₁, we get

= \(\begin{vmatrix} 2 & 37 & 37\\ 3& 33 & 33\\ 4 & 29 & 29 \end{vmatrix}\)

As we can see that the second and the third column 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.

∴\(\begin{vmatrix} 2 & 7 & 37\\ 3& 6 & 33\\ 4 & 5 & 29 \end{vmatrix}\) = 0

CONCEPT:

• If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]\) is a square matrix of order 3, then determinant of A is given by |A| = a₁₁ × {(a₂₂ × a₃₃) – (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) – (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) – (a₂₂ × a₃₁)}
• If A is a matrix of order n, then |k ⋅ A| = kn ⋅ |A|, where k ∈ R.

CALCULATION:

Given: \(A = \left[ {\begin{array}{*{20}{c}} 3&4&9\\ {11}&6&7\\ 8&9&5 \end{array}} \right]\) and |2A| = k

⇒ |A| = 3 × (30 - 63) - 4 × (55 - 56) + 9 × (99 - 48)

⇒ |A| = - 99 + 4 + 459 = 364

As we know that, if A is a matrix of order n, then |k ⋅ A| = kn ⋅ |A|, where k ∈ R.

⇒ |2A| = 2³ ⋅ 364 = 2912

Hence, the correct option is 3.

Calculation:

Given \(\begin{vmatrix} \rm x& 2 & 3\\ 1 & \rm x& 1 \\ 3 & 2& \rm x\end{vmatrix}\) = 0

⇒ x(x² - 2) - 2(x - 3) + 3(2 - 3x) = 0

Now, x³ - 2x - 2x + 6 + 6 - 9x = 0

⇒ x³ - 13x + 12 = 0

∵ x = 3 is a root of the equation, ∴ (x - 3) = 0

If we divide (x3 - 13x + 12) from (x - 3) we will get (x2 + 3x - 4)

⇒ (x - 3)(x² + 3x - 4) = 0

⇒ x2 + 3x - 4 = 0

⇒ (x + 4)(x - 1) = 0

⇒ x = 1, -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).

Calculation:

Let, \({\rm{A}} = \left| {\begin{array}{*{20}{c}} {2 + i}&{2 - i}\\ {1 + i}&{i - 1} \end{array}} \right|\)

⇒ |A| = (2 + i) (i – 1) – (2 – i) (1 + i)

= 2i + i² – 2 – i – (2 – i + 2i – i²)

= i – 1 – 2 – (2 + i + 1)                   (∵ i² = -1)

= i – 3 – 2 – i – 1

= -6

∴ |A| is real number.