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

Concept:

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

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

Formulas used:

• a³ + b³ + c³ = 3abc, if a + b + c = 0.

Calculation:

Given: p + q + r = a + b + c = 0

To find: determinant \(\left| {\begin{array}{*{20}{c}} {{\rm{pa}}}&{{\rm{qb}}}&{{\rm{rc}}}\\ {{\rm{qc}}}&{{\rm{ra}}}&{{\rm{pb}}}\\ {{\rm{rb}}}&{{\rm{pc}}}&{{\rm{qa}}} \end{array}} \right|\)

\(\left| {\begin{array}{*{20}{c}} {{\rm{pa}}}&{{\rm{qb}}}&{{\rm{rc}}}\\ {{\rm{qc}}}&{{\rm{ra}}}&{{\rm{pb}}}\\ {{\rm{rb}}}&{{\rm{pc}}}&{{\rm{qa}}} \end{array}} \right|\)

Expanding R₁,

= pa (ra × qa - pb × pc) - qb (qc × qa - pb × rb) + rc (qc × pc - ra × rb)

= pa(a²qr - p²bc) - qb(q²ac - b²pr) + rc(c²pq - r²ab)

= a³pqr - p³abc - q³abc + b³pqr + c³pqr - r³abc

= pqr(a³ + b³ + c³) - abc(p³ + q³ + r³)

Now we know that, a³ + b³ + c³ = 3abc, if a + b + c = 0.

= pqr (3abc) - abc (3pqr)

= 0

Calculation:

Given:

x² + y² + z² = 1

then, \(​\begin{vmatrix}1&\rm z&\rm -y\\\ \rm -z&1& \rm x\\\ \rm y& \rm -x&1\end{vmatrix}\)

Expanding along R₁, we get-

1(1 + x²) + z(xy + z) − y(xz − y)

1(1 + x2) - z(-z - xy) − y(xz − y)

= 1 + x² + xyz + z² − xyz + y²

= 1 + x² + y² + z²

= 1 + 1   (∵ x2 + y2 + z2 = 1 (given))

= 2

The correct answer is option "3"

\(\begin{vmatrix}x - 4 & 2x & 2x\\ 2x & x - 4 & 2x\\ 2x & 2x & x - 4\end{vmatrix} = (A + Bx) (x - A)^{2}\)

Consider \(D=\begin{vmatrix}x - 4 & 2x & 2x\\ 2x & x - 4 & 2x\\ 2x & 2x & x - 4\end{vmatrix}\)

Applying \(R_{1}\rightarrow R_{1}+R_{2}+R_{3}\)

\(D=\begin{vmatrix}5x - 4 & 5x-4 & 5x-4\\ 2x & x - 4 & 2x\\ 2x & 2x & x - 4\end{vmatrix}\)

\(D=(5x-4)\begin{vmatrix}1 & 1 & 1\\ 2x & x - 4 & 2x\\ 2x & 2x & x - 4\end{vmatrix}\)

Applying \(R_{2}\rightarrow R_{2}-R_{3}\)

\(D=(5x-4)\begin{vmatrix}1 & 1 & 1\\ 0 & -(x + 4) & x+4\\ 2x & 2x & x - 4\end{vmatrix}\)

\(D=(5x-4)[1(-(x+4)(x-4)-2x(x+4))-1(-2x(x+4))+1(2x(x+4))]\)

\(D=(5x-4)[-x^{2}+16-2x^{2}-8x+2x^{2}+8x+2x^{2}+8x]\)

\(D=(5x-4)[x^{2}+8x+16]\)

\(D=(5x-4)(x+4)^{2}\)

\(D=(-4+5x)(x-(-4))^{2}\)

Hence, \(A=-4, B=5\)

\(\begin{vmatrix}x - 4 & 2x & 2x\\ 2x & x - 4 & 2x\\ 2x & 2x & x - 4\end{vmatrix} = (A + Bx) (x - A)^{2}\)

Consider \(D=\begin{vmatrix}x - 4 & 2x & 2x\\ 2x & x - 4 & 2x\\ 2x & 2x & x - 4\end{vmatrix}\)

Applying \(R_{1}\rightarrow R_{1}+R_{2}+R_{3}\)

\(D=\begin{vmatrix}5x - 4 & 5x-4 & 5x-4\\ 2x & x - 4 & 2x\\ 2x & 2x & x - 4\end{vmatrix}\)

\(D=(5x-4)\begin{vmatrix}1 & 1 & 1\\ 2x & x - 4 & 2x\\ 2x & 2x & x - 4\end{vmatrix}\)

Applying \(R_{2}\rightarrow R_{2}-R_{3}\)

\(D=(5x-4)\begin{vmatrix}1 & 1 & 1\\ 0 & -(x + 4) & x+4\\ 2x & 2x & x - 4\end{vmatrix}\)

\(D=(5x-4)[1(-(x+4)(x-4)-2x(x+4))-1(-2x(x+4))+1(2x(x+4))]\)

\(D=(5x-4)[-x^{2}+16-2x^{2}-8x+2x^{2}+8x+2x^{2}+8x]\)

\(D=(5x-4)[x^{2}+8x+16]\)

\(D=(5x-4)(x+4)^{2}\)

\(D=(-4+5x)(x-(-4))^{2}\)

Hence, \(A=-4, B=5\)

Calculation:

Given:

\(\begin{vmatrix} x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3 \end{vmatrix} = 0\) and \(\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \neq 0\)

\(\begin{vmatrix} x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3 \end{vmatrix} = 0\)

⇒ \(\begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + \begin{vmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{vmatrix} = 0\)

⇒ \(\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} + xyz \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = 0\)

⇒ \((1+xyz)\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = 0\)

Since \(\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \neq 0\), we must have 1 + xyz = 0

⇒ xyz = -1

Hence option 2 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:

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.