Determinants MCQ Quiz - Objective Question with Answer for 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, \(A=-4, B=5\)

Concept

1. Let a be the first term of the GP and r is the common ratio.

So, n^th term of a GP = a r^{n - 1}

2. Product rule: The log of a product equals the sum of two logs.

\({\log _a}\left( {mn} \right) = \;{\log _a}m + \;{\log _a}n\)

3. Quotient rule: The log of a quotient equals the difference of two logs.

\({\log _a}\frac{m}{n} = \;{\log _a}m - \;{\log _a}n\)

4. Power rule: In the log of a power the exponent becomes a coefficient.

\({\log _a}{m^n} = n{\log _a}m\)

Calculation:

Let a is the first term of the GP and R is the common ratio, then we are given,

Given: u, v and w (all positive) are the p^th, q^th and r^th terms of a GP

So, u = aR^{p - 1}, v = aR^{q - 1} and w = aR^{r - 1}

\(\left[ {\begin{array}{*{20}{c}} {\ln {\rm{u}}}&{\rm{p}}&1\\ {\ln {\rm{v}}}&{\rm{q}}&1\\ {\ln {\rm{w}}}&{\rm{r}}&1 \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} {\ln \;(a{R^{p\; - \;1}})}&{\rm{p}}&1\\ {\ln \;(a{R^{q\; - \;1}})}&{\rm{q}}&1\\ {\ln (a{R^{r\; - \;1}})}&{\rm{r}}&1 \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} {\ln a + \left( {p - 1} \right)\ln R}&{\rm{p}}&1\\ {\ln a + \left( {{\rm{q}} - 1} \right)\ln R}&{\rm{q}}&1\\ {\ln a + \left( {{\rm{r}} - 1} \right)\ln R}&{\rm{r}}&1 \end{array}} \right]\)

Now, Applying R₁ → R₁ - R₃ and R₂ → R₂ - R₃

\(= \left[ {\begin{array}{*{20}{c}} {\left( {{\rm{p}} - {\rm{r}}} \right){\rm{ln\;R}}}&{{\rm{p}} - {\rm{r}}}&0\\ {\left( {{\rm{q}} - {\rm{r\;}}} \right)\ln {\rm{R}}}&{{\rm{q}} - {\rm{r\;}}}&0\\ {\ln a{\rm{\;}} + {\rm{\;}}\left( {{\rm{r}} - 1} \right)\ln R}&{\rm{r}}&1 \end{array}} \right]\)

Now, Expanding along C₃

= 1 × {(p - r) ln R × (q - r) - (p - r) × (q - r) ln R}

= (p - r) (q - r) (ln R - ln R)

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:

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:

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:

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