Numerical elements MCQ Quiz in తెలుగు - Objective Question with Answer for Numerical elements - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Last updated on Mar 17, 2025

పొందండి Numerical elements సమాధానాలు మరియు వివరణాత్మక పరిష్కారాలతో బహుళ ఎంపిక ప్రశ్నలు (MCQ క్విజ్). వీటిని ఉచితంగా డౌన్‌లోడ్ చేసుకోండి Numerical elements MCQ క్విజ్ Pdf మరియు బ్యాంకింగ్, SSC, రైల్వే, UPSC, స్టేట్ PSC వంటి మీ రాబోయే పరీక్షల కోసం సిద్ధం చేయండి.

Latest Numerical elements MCQ Objective Questions

Top Numerical elements MCQ Objective Questions

Numerical elements Question 1:

Find the determinant of the matrix \(\begin{vmatrix} 2 & 7 & 37\\ 3& 6 & 33\\ 4 & 5 & 29 \end{vmatrix}\)

  1. 234
  2. 132
  3. 83
  4. 0
  5. None of these

Answer (Detailed Solution Below)

Option 4 : 0

Numerical elements Question 1 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, 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

Numerical elements Question 2:

If \(P=\begin{bmatrix} 1 & \alpha & 3\ 1 & 3 & 3\ 2 & 4 & 4 \end{bmatrix}\) is the adjoint of a \(3\times 3\) matrix \(A\) and \(|A| =4\), then \(\alpha\) is equal to:

  1. \(11\)
  2. \(5\)
  3. \(0\)
  4. \(4\)

Answer (Detailed Solution Below)

Option 1 : \(11\)

Numerical elements Question 2 Detailed Solution

\( P=\begin{bmatrix} 1 & \alpha & 3\ 1 & 3 & 3\ 2 & 4 & 4 \end{bmatrix} \)

For 3X3 matrix,

\( |Adj A|=|A|^{2} \)

\( |Adj A|=16 \)

\( 1(12-12)-\alpha(4-6)+3(4-6)=16 \)

\( 2\alpha-6=16 \)

\( 2\alpha=22 \)

\( \alpha=11 \)

Numerical elements Question 3:

If \(\rm p=\begin{bmatrix}1&\alpha&3\\\ 1&3&3\\\ 2&4&4\end{bmatrix}\) is the adjoint of the 3 × 3 matrix A and det A = 4, then a is equal to 

  1. 4
  2. 11
  3. 5
  4. 0

Answer (Detailed Solution Below)

Option 2 : 11

Numerical elements Question 3 Detailed Solution

Calculation

|p| = 2α - 6 = (det A)² = 16

⇒ α = 11

Hence option 2 is correct

Numerical elements Question 4:

Let αβ ≠ 0 and A = \(\rm \begin{bmatrix}\beta&\alpha&3\\ \alpha&\alpha&\beta\\\ -\beta &\alpha&2\alpha\end{bmatrix}\ If \ B=\rm \begin{bmatrix}3\alpha&-9&3\alpha\\\ -\alpha&7&-2\alpha\\\ -2\alpha&5&-2\beta\end{bmatrix}\)  is the matrix of cofactors of the elements of A, then det(AB) is equal to :

  1. 343
  2. 125
  3. 64
  4. 216

Answer (Detailed Solution Below)

Option 4 : 216

Numerical elements Question 4 Detailed Solution

Concept:

The co-factor of Aij = (– 1)i+jMij, where Mij is the minor obtained by removing the ith row and jth column.

Calculation:

Given, A = \(\rm \begin{bmatrix}\beta&\alpha&3\\ \alpha&\alpha&\beta\\\ -\beta &\alpha&2\alpha\end{bmatrix}\) and B = \(\rm\begin{bmatrix}3\alpha&-9&3\alpha\\\ -\alpha&7&-2\alpha\\\ -2\alpha&5&-2\beta\end{bmatrix}\)

Now, equating co-factor fo A21  = (2α2 – 3α) = α

α = 2, 0

⇒ α = 2 [∵ α ≠ 0] 

Now, 2α2αβ = 3α 

⇒ 8 – 2β = 6

⇒ β = 1

∴ |AB| = |A cof (A)| = |A|3 

∴ |A| = \(\left|\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 2 & 1 \\ -1 & 2 & 4 \end{array}\right|\) = 6 - 2(9) + 3(6) = 6

⇒ |A|3 63 = 216

∴ The value of det(AB) is equal to 216.

The correct answer is Option 4.

Numerical elements Question 5:

Find the value of the determinant: \(\begin{vmatrix}5^2&5^3&5^4\\5^3&5^4&5^5\\5^5&5^6&5^7\end{vmatrix}\)

  1. 0
  2. 53
  3. 55
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : 0

Numerical elements Question 5 Detailed Solution

Concept:

Determinants:

  • A linear combination of the rows/columns does not affect the value of the determinant.
  • If two rows/columns of a given matrix are interchanged, then the value of the determinant gets multiplied by - 1.
  • If a row/column of a given matrix is multiplied by a scalar k, then the value of the determinant is also multiplied by k.

 

Calculation:

Let D = \(\begin{vmatrix}5^2&5^3&5^4\\5^3&5^4&5^5\\5^5&5^6&5^7\end{vmatrix}\).

Dividing R1 by scalar 52 and R2 by 53, we get:

⇒ D = \((5^2)(5^3)\begin{vmatrix}1&5&5^2\\1&5&5^2\\5^5&5^6&5^7\end{vmatrix}\)

Using R1 → R1 - R2, we get:

⇒ D = \((5^2)(5^3)\begin{vmatrix}0&0&0\\1&5&5^2\\5^5&5^6&5^7\end{vmatrix}\)

Expanding along R1, we get:

⇒ D = 0.

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