Evaluation of Determinants MCQ Quiz in தமிழ் - Objective Question with Answer for Evaluation of Determinants - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Three non-zero terms a, b, c are in GP if and only if b² = ac

For a quadratic equation ax² + bx + c = 0,

If α is a root then aα2 + bα + c = 0

Calculation:

Given: Δ = \(\begin{vmatrix}a&b&aα+b\\\ b&c&bα+c\\\ aα+b&bα+c&0\end{vmatrix}\) = 0

C₃ → C3 - C₁α  - C₂

⇒ Δ = \(\begin{vmatrix}a&b&0\\\ b&c&0\\\ aα+b&bα+c&-(aα^2+bα+bα+c)\end{vmatrix}=0\)

⇒ (aα² + 2bα + c) (ac - b²) = 0

⇒ ac - b² = 0 or aα2 + 2bα + c = 0

a, b, c are in GP and (x - α) is the factor ax2 + 2bx + c

∴ Only statement II is correct.

CONCEPT:

• If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\)is a square matrix of order 2, then determinant of A is given by |A| = (a­11 × a22) – (a12 – a21)
• If A is a singular matrix of order n then |A| = 0

CALCULATION:

Given: \(A = \left[ {\begin{array}{*{20}{c}} {3 - 2x}&{x + 1}\\ 2&4 \end{array}} \right]\) is a singular matrix.

As we know that, if \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\) is a square matrix of order 2, then determinant of A is given by |A| = (a­11 × a22) – (a12 – a21)

⇒ |A| = 4(3 - 2x) - 2(x + 1)

⇒ |A| = 12 - 8x - 2x - 2

⇒ |A| = -10x + 10

∵ A is a singular matrix so |A| = 0

⇒ |A| = -10x + 10 = 0

⇒ x = 1

Hence, the correct option is 1.

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, then 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: 

\(\left| {\begin{array}{*{20}{c}} 2&7&65\\ 3&8&75\\ 5&9&86 \end{array}} \right|\)

Apply C2 → 9C2 + C1

\(= \left| {\begin{array}{*{20}{c}} 2&65&65\\ 3&75&75\\ 5&86&86 \end{array}} \right|\)

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.

\(\left| {\begin{array}{*{20}{c}} 2&7&65\\ 3&8&75\\ 5&9&86 \end{array}} \right|=0\)

Concept:

If all the elements of a row or column are zeroes, then the value of the determinant is zero.

To evaluate the determinant row or column operation is done.

Calculation:

The determinant can be written as,

\(\begin{bmatrix} 2(1+1) & 2(1+2) &2(1+3) \\ 2(2+1) &2(2+2) & 2(2+3)\\ 2(3+1)&2(3+2) & 2(3+3) \end{bmatrix}\)

= \(2\times \begin{bmatrix} 1+1 &1+2 & 1+3\\ 2+1 & 2+2 &2+3 \\ 3+1 & 3+2 & 3+3 \end{bmatrix}\)

R₃ → R₃ - R₁ and R2 → R2 - R1

= \(2\times \begin{bmatrix} 1+1 &1+2 & 1+3\\ 1 & 1 &1 \\ 2 & 2 & 2 \end{bmatrix}\)

R₃ → R₃ -2R₂

= \(2\times \begin{bmatrix} 1+1 &1+2 & 1+3\\ 1 & 1 &1 \\ 0 & 0 & 0 \end{bmatrix}\)

= 0

So, the value of the determinant is 0

CONCEPT:

• When the determinant operation is applied to the product of two matrices then the determination will be distributed.

|XY| = |X||Y|

Also, 

|kX| = kⁿ|A| 

Where n is the order of the determinant and k is any constant.

CALCULATION

Given:

|A| = -1, |B| = 3

 ⇒ |3AB| = 3³ |AB| = 27  |A||B| 

⇒ |3AB| =  27 × (-1) × 3 = - 81     [Since |KA| = Kⁿ |A|] 

So, the correct answer is option 2.

CONCEPT:

• If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\)is a square matrix of order 2, then determinant of A is given by |A| = (a­11 × a22) – (a12 – a21)

CALCULATION:

Given: \(\left| {\begin{array}{*{20}{c}} x&4\\ 9&x \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 8&2\\ 16&4 \end{array}} \right|\)

As we know that, \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\)is a square matrix of order 2, then determinant of A is given by |A| = (a­11 × a22) – (a12 – a21)

⇒ x² - 36 = 32 - 32 = 0

⇒ x² = 36

⇒ x = 6

Hence, correct option is 1.

Concept:

Suppose A is a square matrix of 2 rows and 2 columns

\(\rm A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}\)

The determinant is; |A| = [ad - bc]

Calculation:

\(A = \left| {\begin{array}{*{20}{c}} p&q\\ r&s \end{array}} \right|\)

p, q, r and s are any four different prime numbers less than 20.

Here we have to find the maximum value of the determinant so we have to take maximum value for p and s and minimum value for q and r

All prime numbers less than 20 = 2, 3, 5, 7, 11, 13, 17, 19

So, p = 17, q = 2, r = 3 and s = 19

\(\rm A = \left| {\begin{array}{*{20}{c}} 17&2\\3&19 \end{array}} \right|\)

|A| = [17 \(\times\) 19 - 2 \(\times\) 3]

= 323 - 6

= 317

Concept:

If A and B are square matrices of order n, then det (m × AB) = mn × det (A) × det (B) where m ∈ R is a scalar.

If A is a non – singular square matrix of order n, then det (A-1­) = 1/det (A)

Calculation:

Given: Given: A and B be (3 × 3) matrices with det A = 4 and det B = 3

As we know that, if A and B are square matrices of order n, then det (m × AB) = mn × det (A) × det (B) where m ∈ R is a scalar.

⇒ det (3AB-1) = 33 × det (A) × det (B- 1)

As we know that, If A is a non – singular square matrix of order n, then det (A-1­) = 1/det (A)

\(\Rightarrow {\rm{\;det}}\left( {3A{B^{ - 1}}} \right) = 27 \times 4 \times \frac{1}{3} = 36\)

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"

Formula used:

\(\rm A = \begin{vmatrix} a_{1} & a_{2} & a_{1}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3}\ \end{vmatrix}\)

det(A) = \(\rm a_{1}\begin{vmatrix} b_{2} & b_{3}\\ c_{2} & c_{3} \end{vmatrix} - a_{2} \begin{vmatrix} b_{1} & b_{3}\\ c_{1} & c_{3} \end{vmatrix} + a_{3} \begin{vmatrix} b_{1} & b_{2}\\ c_{1} & c_{2} \end{vmatrix}\)

= a1(b2 c3 - b3 c2) - a2(b1c3 - b3c1) + a3 (b1c2 - b2c1)

Calculation:

\(\begin{vmatrix} 1 & -x & x\\ 1 & 1 & -y\\ 1 & z & 1 \end{vmatrix}\)

⇒  \(\begin{vmatrix} 1 & -\frac{a}{b - c} & \frac{a}{b - c}\\ 1 & 1 & -\frac{b}{c - a}\\ 1 & \frac{c}{a -b} & 1 \end{vmatrix}\) = Δ 

⇒ Δ = 1(1 + \(\rm \frac{bc}{(a - b)(c - a)}\)) + \(\rm \frac{a}{b - c}\)(\(\rm 1 + \frac{b}{c - a} \)) + \(\rm \frac{a}{b - c}\) \(\rm (\frac{c}{a - b} - 1)\)

⇒ Δ = 1 + \(\rm \frac{bc}{(a - b)(c - a)} + \frac{ab}{(b - c)(c - a)} + \frac{ac}{(b - c)(a - b)}\)

⇒ Δ = 1 + \( \frac{a}{(b - c)} \times \frac{(b - c)(a - b - c)}{(a - b)(c - a)} + \frac{bc}{(a - b)(c - a)}\)

⇒ Δ = \( \frac{ac - a^2 - bc + ab+ a^2 - ab - ac + bc}{(a - b)(c - a)}\)

∴ Δ =  0

Shortcut TrickPut x = 0, b = 1 and c = 2

⇒ \(x = \frac{a}{b-c}\) = 0, \(y = \frac{b}{c - a}\) = 1/2, \(z = \frac{c}{a - b}\) = -2

Let, \(\begin{vmatrix} 1 & -x & x\\ 1 & 1 & -y\\ 1 & z & 1 \end{vmatrix}\) = Δ 

⇒  Δ = \(\begin{vmatrix} 1 & 0 & 0\\ 1 & 1 & -1/2\\ 1 & -2 & 1 \end{vmatrix}\)

⇒  Δ = 1[1 - (-1/2)(-2)] 

∴   Δ = 0