Evaluation of Determinants MCQ Quiz in தமிழ் - Objective Question with Answer for Evaluation of Determinants - இலவச PDF ஐப் பதிவிறக்கவும்
Last updated on Mar 11, 2025
Latest Evaluation of Determinants MCQ Objective Questions
Top Evaluation of Determinants MCQ Objective Questions
Evaluation of Determinants Question 1:
If the determinant \(\begin{vmatrix}a&b&aα+b\\\ b&c&bα+c\\\ aα+b&bα+c&0\end{vmatrix}\) is equal to zero, then consider the following statements
I. a, b, c are in A.P.
II. α is a root of the equation ax2 + 2bx + c = 0.
Which of the above statements is/are correct?
Answer (Detailed Solution Below)
Evaluation of Determinants Question 1 Detailed Solution
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.
Evaluation of Determinants Question 2:
For what value of x is the matrix \(A = \left[ {\begin{array}{*{20}{c}} {3 - 2x}&{x + 1}\\ 2&4 \end{array}} \right]\)is singular ?
Answer (Detailed Solution Below)
Evaluation of Determinants Question 2 Detailed Solution
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.
Evaluation of Determinants Question 3:
Find the determinant of the matrix \(\left| {\begin{array}{*{20}{c}} 2&7&65\\ 3&8&75\\ 5&9&86 \end{array}} \right|\) ?
Answer (Detailed Solution Below)
Evaluation of Determinants Question 3 Detailed Solution
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.
\(\left| {\begin{array}{*{20}{c}} 2&7&65\\ 3&8&75\\ 5&9&86 \end{array}} \right|=0\)
Evaluation of Determinants Question 4:
The element in the ith row and the jth column of a determinant of third order is equal to 2(i + j). What is the value of the determinant?
Answer (Detailed Solution Below)
Evaluation of Determinants Question 4 Detailed Solution
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
Evaluation of Determinants Question 5:
If A and B are square matrices of order 3 such that |A| = -1, |B| = 3, then |3AB| =
Answer (Detailed Solution Below)
Evaluation of Determinants Question 5 Detailed Solution
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.
Evaluation of Determinants Question 6:
Find the value of x, if \(\left| {\begin{array}{*{20}{c}} x&4\\ 9&x \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 8&2\\ 16&4 \end{array}} \right|\) ?
Answer (Detailed Solution Below)
Evaluation of Determinants Question 6 Detailed Solution
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.
Evaluation of Determinants Question 7:
Let \(A = \left| {\begin{array}{*{20}{c}} p&q\\ r&s \end{array}} \right|\)
where p, q, r and s are any four different prime numbers less than 20. What is the maximum value of the determinant?
Answer (Detailed Solution Below)
Evaluation of Determinants Question 7 Detailed Solution
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
Evaluation of Determinants Question 8:
Comprehension:
Direction: Consider the following for the next 02 (two) items:
Let A and B be (3 × 3) matrices with det A = 4 and det B = 3What is det (3AB-1) equal to?
Answer (Detailed Solution Below)
Evaluation of Determinants Question 8 Detailed Solution
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)
\(\Rightarrow {\rm{\;det}}\left( {3A{B^{ - 1}}} \right) = 27 \times 4 \times \frac{1}{3} = 36\)
Evaluation of Determinants Question 9:
If x2 + y2 + z2 = 1, then what is the value
\(\begin{vmatrix}1&z&-y\\\ -z&1&x\\\ y&-x&1\end{vmatrix}=?\)
Answer (Detailed Solution Below)
Evaluation of Determinants Question 9 Detailed Solution
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"
Evaluation of Determinants Question 10:
If \(x = \frac{a}{b-c}\), \(y = \frac{b}{c - a}\), \(z = \frac{c}{a - b}\) then what is the value of the following?
\(\begin{vmatrix} 1 & -x & x\\ 1 & 1 & -y\\ 1 & z & 1 \end{vmatrix}\)
Answer (Detailed Solution Below)
Evaluation of Determinants Question 10 Detailed Solution
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