Ch 4  ·  Q–
0%
Chapter 4 Exercise 4.1 Solutions

Determinants

Step-by-Step Solutions to the NCERT Class 12 Mathematics Chapter 4 Exercise 4.1

Class 12 Mathematics Exercise 4.1 NCERT Solutions Determinants Class 12 Mathematics Chapter 4 CBSE Board Exam JEE Main CUET Determinants of Order 2 Determinants of Order 3 Evaluation of Determinants Cofactor Expansion Minors and Cofactors Properties of Determinants Step-by-Step Solutions
8 Questions
20–25 min Ideal time
Q1 Now at
Q1
NUMERIC3 marks
Evaluate the determinant \[\begin{vmatrix}2 & 4\\-5 & -1\end{vmatrix}\]
📘 Concept & Theory
Concept/Theory

A determinant of order \(2\times2\) has the general form

\[\begin{vmatrix}a & b\\c & d\end{vmatrix}\]

Its value is calculated using the formula

\[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc\]

Thus, for a \(2\times2\) determinant, multiply the entries of the principal diagonal, multiply the entries of the other diagonal, and subtract the second product from the first.

🗺️ Solution Roadmap
Step-by-step Plan
  1. Identify the four entries of the determinant with \(a\), \(b\), \(c\), and \(d\).

  2. Apply the formula \(ad-bc\).

  3. Substitute the values carefully, paying particular attention to the negative signs.

  4. Evaluate each product separately.

  5. Perform the final subtraction and state the value of the determinant.

✏️ Solution
Complete Solution
Step-by-step Solution  ·  4 steps
  1. Given
    \[\begin{vmatrix}2 & 4\\-5 & -1\end{vmatrix}\]
  2. Compare this determinant with the standard form
    \[\begin{vmatrix}a & b\\c & d\end{vmatrix}\]
  3. Therefore,
    \[a=2,\quad b=4,\quad c=-5,\quad d=-1\]
  4. For a determinant of order \(2\times2\),
    \[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc\]
  5. Substituting the given values, we get
    \[\begin{aligned}\Delta&=(2)(-1)-(4)(-5)\\&=-2-(-20)\\&=18\end{aligned}\]
🎯 Exam Significance
Exam Significance

This question tests one of the most fundamental skills in the chapter: evaluating a determinant of order \(2\times2\). Although the calculation is short, correct handling of negative signs is essential. Similar questions can appear as direct one-mark or short-answer questions in board examinations and can also form the first step of longer determinant problems.

Students should remember the formula \(ad-bc\) rather than relying only on visual diagonal multiplication. Writing the substitution explicitly also reduces sign errors and makes the solution clear and logically complete.

Significance for Competitive Entrance Examinations

In competitive examinations, determinants are frequently used as intermediate tools in questions involving matrices, equations of lines and planes, areas, coordinate geometry, inverse matrices, and systems of linear equations. Even when the determinant calculation itself is elementary, speed and accuracy are important because a sign error can invalidate an otherwise correct solution.

For entrance-exam preparation, students should develop the habit of recognizing the \(2\times2\) determinant immediately and applying \(ad-bc\) without unnecessary expansion.

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  6 points
  1. For a \(2\times2\) determinant, use the formula \(ad-bc\).

  2. For this problem, \(a=2\), \(b=4\), \(c=-5\), and \(d=-1\).

  3. Negative signs must be handled carefully during substitution.

  4. Subtracting a negative quantity changes the operation into addition.

  5. The determinant evaluates to \(18\).

  6. The formula \(ad-bc\) is a basic building block for more advanced determinant problems.

↑ Top
1 / 8  ·  13%
Q2 →
Q2
NUMERIC3 marks
Evaluate the following determinants:
(i)

\[\begin{vmatrix}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{vmatrix}\]

(ii)

\[\begin{vmatrix}x^{2}-x+1 & x-1\\x+1 & x+1\end{vmatrix}\]

📘 Concept & Theory
Concept/Theory

Both parts involve the evaluation of a determinant of order \(2\times2\). For a determinant

\[\begin{vmatrix}a & b\\c & d\end{vmatrix}\]

the standard formula is

\[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc\]

In part (i), after applying this formula, the expression reduces to the fundamental trigonometric identity

\[\sin^2\theta+\cos^2\theta=1\]

In part (ii), the determinant produces algebraic products that must be expanded and simplified carefully. In particular, the products involving \(x+1\) should be expanded systematically to avoid sign errors.

🗺️ Solution Roadmap
Step-by-step Plan
  1. Identify the four entries of each \(2\times2\) determinant.

  2. Apply the formula \(ad-bc\).

  3. For part (i), simplify the resulting trigonometric expression using \(\sin^2\theta+\cos^2\theta=1\).

  4. For part (ii), expand each algebraic product separately.

  5. Combine like terms and obtain the simplified polynomial.

  6. State the final value of each determinant clearly.

✏️ Solution
Complete Solution
Step-by-step Solution  ·  8 steps
  1. (i)
  2. Given
    \[\Delta=\begin{vmatrix}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{vmatrix}\]
  3. Compare this with the standard form
    \[\begin{vmatrix}a & b\\c & d\end{vmatrix}\]
  4. Therefore,
    \[a=\cos\theta,\quad b=-\sin\theta,\quad c=\sin\theta,\quad d=\cos\theta\]
  5. Using the formula
    \[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc\]
  6. we obtain
    \[\begin{aligned}\Delta&=(\cos\theta)(\cos\theta)-(-\sin\theta)(\sin\theta)\\ &=\cos^2\theta + \sin^2\theta)\\&=1\end{aligned}\]
  7. (ii)
  8. Given
    \[\Delta=\begin{vmatrix}x^2-x+1 & x-1\\x+1 & x+1\end{vmatrix}\]
  9. Comparing with
    \[\begin{vmatrix}a & b\\c & d\end{vmatrix}\]
  10. we have
    \[a=x^2-x+1,\quad b=x-1,\quad c=x+1,\quad d=x+1\]
  11. Using
    \[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc\]
  12. we obtain
    \[\begin{aligned} \Delta &=(x^2-x+1)(x+1)-(x-1)(x+1)\\ &=x^2(x+1)-x(x+1)+1(x+1)-(x-1)(x+1)\\ &=x^2(x+1)-x(x+1)+1(x+1)-(x^2-1)\\ &\qquad\quad\scriptsize \color{orange}\text{(Using the identity } (a-b)(a+b)=a^2-b^2)\\ &=x^3+1-(x^2-1)\\ &=x^3+1-x^2+1\\ &=x^3-x^2+2 \end{aligned}\]
💡 Answer
Final Answer

(i)

\[\boxed{1}\]

(ii)

\[\boxed{x^3-x^2+2}\]

🎯 Exam Significance
Exam Significance

This question combines the basic evaluation formula for a \(2\times2\) determinant with two important simplification techniques. Part (i) checks whether the student can correctly handle negative entries and recognize the identity \(\sin^2\theta+\cos^2\theta=1\). Part (ii) tests determinant evaluation together with polynomial expansion and simplification.

Such questions are particularly useful for building accuracy in the early part of the Determinants chapter. In board examinations, showing the intermediate substitution and simplification steps makes the method transparent and helps avoid losing marks because of an unexplained calculation.

Significance for Competitive Entrance Examinations

For competitive examinations, the main advantage is speed with accuracy. A \(2\times2\) determinant should be recognized immediately as \(ad-bc\), after which the remaining work is usually algebraic or trigonometric simplification.

Part (i) reinforces the ability to identify standard identities quickly, while part (ii) develops algebraic manipulation skills that frequently appear inside more complex determinant and matrix problems. The identity

\[ (x-1)(x+1)=x^2-1 \]
is especially useful for reducing calculation time.

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  8 points
  1. For every \(2\times2\) determinant, use \(ad-bc\).

  2. Negative entries require careful sign handling.

  3. The identity

    \[ \sin^2\theta+\cos^2\theta=1 \]
    immediately simplifies the determinant in part (i).

  4. The product

    \[ (x^2-x+1)(x+1) \]
    simplifies to \(x^3+1\).

  5. The difference of squares identity gives

    \[ (x-1)(x+1)=x^2-1. \]

  6. The final result for part (i) is \(1\).

  7. The final result for part (ii) is \(x^3-x^2+2\).

  8. Writing each algebraic and sign-changing step explicitly is useful for both board-exam presentation and error-free competitive-exam preparation.

← Q1
2 / 8  ·  25%
Q3 →
Q3
NUMERIC3 marks
if \[A=\begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}\] then show that \[\left|2A\right|=4\left|A\right|\]
📘 Concept & Theory
Concept/Theory

This question illustrates an important property of determinants involving multiplication of a matrix by a scalar.

If \(A\) is a square matrix of order \(n\) and \(k\) is a scalar, then

\[ |kA|=k^n|A| \]

Here, \(A\) is a matrix of order \(2\times2\). Therefore, \(n=2\). Hence,

\[ |kA|=k^2|A| \]

Taking \(k=2\), we obtain

\[ |2A|=2^2|A|=4|A| \]

The question asks us to verify this property directly by evaluating both sides.

🗺️ Solution Roadmap
Step-by-step Plan
  1. Write the given matrix \(A\).

  2. Multiply every element of \(A\) by \(2\) to obtain \(2A\).

  3. Evaluate the determinant \(|2A|\) using the \(2\times2\) determinant formula \(ad-bc\).

  4. Evaluate the determinant \(|A|\) using the same formula.

  5. Multiply \(|A|\) by \(4\).

  6. Compare \(|2A|\) with \(4|A|\).

  7. If both values are equal, the required result is proved.

✏️ Solution
Complete Solution
Step-by-step Solution  ·  15 steps
  1. Given
    \[A=\begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}\]
  2. Step 1: Find \(2A\)
  3. Multiplying every element of \(A\) by \(2\), we get
    \[2A=2\begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}\]
  4. Therefore,
    \[2A=\begin{bmatrix}2(1) & 2(2)\\2(4) & 2(2)\end{bmatrix}\]
  5. Hence,
    \[2A=\begin{bmatrix}2 & 4\\8 & 4\end{bmatrix}\]
  6. Step 2: Evaluate \(|2A|\)
  7. We have
    \[|2A|=\begin{vmatrix}2 & 4\\8 & 4\end{vmatrix}\]
  8. For a \(2\times2\) determinant,
    \[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc\]
  9. Therefore,
    \[\begin{aligned}|2A|&=(2)(4)-(4)(8)\\&=8-32\\&=-24\end{aligned}\]
  10. Step 3: Evaluate \(|A|\)
  11. From the given matrix,
    \[|A|=\begin{vmatrix}1 & 2\\4 & 2\end{vmatrix}\]
  12. Using the formula \(ad-bc\),
    \[\begin{aligned} |A|&=(1)(2)-(2)(4)\\ &=2-8\\ &=-6 \end{aligned}\]
  13. Step 4: Evaluate \(4|A|\)
  14. |A|=-6
  15. Therefore,
    \[4|A|=-24\]
  16. Step 5: Compare Both Sides
  17. \[|2A|=-24\]
    and
    \[4|A|=-24\]
  18. Hence,
    \[\boxed{|2A|=4|A|}\]
  19. Hence Proved
  20. Alternative Verification Using the Determinant Property
  21. The matrix \(A\) is of order \(2\times2\). Therefore, for any scalar \(k\),
    \[|kA|=k^2|A|\]
  22. Putting \(k=2\),
    \[|2A|=2^2|A|\]
  23. Thus,
    \[\boxed{|2A|=4|A|}\]
    This gives a direct theoretical verification of the result, while the preceding calculation verifies it numerically for the given matrix.
🎯 Exam Significance
Exam Significance

This question introduces an important determinant property: when every element of a determinant of order \(n\) is multiplied by a scalar \(k\), the determinant is multiplied by \(k^n\). Since the given determinant is of order \(2\), multiplying the matrix by \(2\) multiplies its determinant by \(2^2=4\).

For board examinations, students should be able to both apply this property and verify it through direct calculation. Questions of this type help establish the determinant properties that are used extensively in the subsequent exercises.

Significance for Competitive Entrance Examinations

This property is particularly useful for reducing calculation time in competitive examinations. Instead of evaluating a transformed determinant from scratch, students can often determine its value immediately from the scalar-multiplication rule.

The general result

\[ |kA|=k^n|A| \]
should be memorised with the order \(n\) of the determinant in mind. A common mistake is to write \(k|A|\) irrespective of the order of the matrix. For a \(2\times2\) determinant the factor is \(k^2\), whereas for a \(3\times3\) determinant it is \(k^3\).

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  7 points
  1. For a square matrix \(A\) of order \(n\),

    \[ |kA|=k^n|A| \]

  2. For a \(2\times2\) matrix,

    \[ |kA|=k^2|A| \]

  3. For \(k=2\),

    \[ |2A|=4|A| \]

  4. For the given matrix,

    \[ |A|=-6 \]

  5. The determinant of \(2A\) is

    \[ |2A|=-24 \]

  6. Also,

    \[ 4|A|=4(-6)=-24 \]

  7. Therefore,

    \[ \boxed{|2A|=4|A|} \]

← Q2
3 / 8  ·  38%
Q4 →
Q4
NUMERIC3 marks
if \[A=\begin{bmatrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{bmatrix}\] then show that \[\left|3A\right|=27\left|A\right|\]
📘 Concept & Theory
Concept/Theory

This question illustrates the scalar multiplication property of determinants. If \(A\) is a square matrix of order \(n\) and \(k\) is a scalar, then

\[ |kA|=k^n|A| \]

Here, \(A\) is a matrix of order \(3\times3\). Therefore, \(n=3\). Hence,

\[ |kA|=k^3|A| \]

Taking \(k=3\), we immediately obtain

\[ |3A|=3^3|A|=27|A| \]

The result can also be verified directly by calculating both \(|3A|\) and \(27|A|\). Since the question asks us to show the result, the direct verification is useful for understanding the property and for strengthening determinant-calculation skills.

🗺️ Solution Roadmap
Step-by-step Plan
  1. Write the given matrix \(A\).

  2. Multiply every entry of \(A\) by \(3\) to obtain \(3A\).

  3. Evaluate \(|3A|\) by expanding along the first row or another convenient row.

  4. Evaluate \(|A|\) using the same determinant-expansion principle.

  5. Calculate \(27|A|\).

  6. Compare the values of \(|3A|\) and \(27|A|\).

  7. Conclude that \(|3A|=27|A|\).

✏️ Solution
Complete Solution
Step-by-step Solution  ·  24 steps
  1. Step 1: Write the given matrix
  2. Given
    \[A=\begin{bmatrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{bmatrix}\]
  3. Step 2: Find \(3A\)
  4. Multiplying every element of \(A\) by \(3\),
    \[3A=3\begin{bmatrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{bmatrix}\]
  5. Therefore,
    \[3A=\begin{bmatrix}3(1) & 3(0) & 3(1)\\3(0) & 3(1) & 3(2)\\3(0) & 3(0) & 3(4)\end{bmatrix}\]
  6. Hence,
    \[3A=\begin{bmatrix}3 & 0 & 3\\0 & 3 & 6\\0 & 0 & 12\end{bmatrix}\]
  7. Step 3: Evaluate \(|3A|\)
  8. We have
    \[|3A|=\begin{vmatrix}3 & 0 & 3\\0 & 3 & 6\\0 & 0 & 12\end{vmatrix}\]
  9. Expand the determinant along the first row. For a \(3\times3\) determinant,
    \[\begin{vmatrix}a & b & c\\d & e & f\\g & h & i\end{vmatrix} =a\begin{vmatrix}e & f\\h & i\end{vmatrix}- b\begin{vmatrix}d & f\\g & i\end{vmatrix}+ c\begin{vmatrix}d & e\\g & h\end{vmatrix}\]
  10. Thus,
    \[|3A|=3\begin{vmatrix}3 & 6\\0 & 12\end{vmatrix}- 0\begin{vmatrix}0 & 6\\0 & 12\end{vmatrix}+ 3\begin{vmatrix}0 & 3\\0 & 0\end{vmatrix}\]
    The second and third terms are zero because their corresponding determinants are multiplied by zero or contain a zero row.
  11. Therefore,
    \[|3A|=3\begin{vmatrix}3 & 6\\0 & 12\end{vmatrix}\]
  12. Using the \(2\times2\) determinant formula,
    \[\begin{vmatrix}3 & 6\\0 & 12\end{vmatrix}=(3)(12)-(6)(0)\]
  13. Hence,
    \[=36\]
  14. Therefore,
    \[|3A|=3(36)\]
  15. Thus,
    \[\boxed{|3A|=108}\]
  16. Step 4: Evaluate \(|A|\)
  17. Now consider
    \[|A|=\begin{vmatrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{vmatrix}\]
  18. Expanding along the first row,
    \[|A|=1\begin{vmatrix}1 & 2\\0 & 4\end{vmatrix}-0\begin{vmatrix}0 & 2\\0 & 4\end{vmatrix}+1\begin{vmatrix}0 & 1\\0 & 0\end{vmatrix}\]
    The second term is zero because it is multiplied by \(0\), and the third \(2\times2\) determinant is zero because its first column consists entirely of zeros.
  19. Therefore,
    \[|A|=\begin{vmatrix}1 & 2\\0 & 4\end{vmatrix}\]
  20. Using the \(2\times2\) determinant formula,
    \[|A|=(1)(4)-(2)(0)\]
  21. Hence,
    \[\boxed{|A|=4}\]
  22. Step 5: Calculate \(27|A|\)
  23. Since
    \[|A|=4\]
  24. therfore,
    \[27|A|=27(4)=108\]
  25. Step 6: Compare Both Sides
  26. \[|3A|=108\]
    and
    \[27|A|=108\]
  27. Therefore,
    \[\boxed{|3A|=27|A|}\]
  28. Hence Proved
  29. Direct Verification Using the Determinant Property
  30. The matrix \(A\) is of order \(3\times3\). Therefore, the scalar multiplication property gives
    \[|kA|=k^3|A|\]
  31. Putting \(k=3\),
    \[|3A|=3^3|A|\]
  32. Since
    \[3^3=27\]
  33. we obtain
    \[\boxed{|3A|=27|A|}\]
    This explains why the factor is \(27\), rather than \(3\). A scalar \(3\) occurs in each of the three rows or columns of the determinant, producing the factor \(3\times3\times3=27\).
🎯 Exam Significance
Exam Significance

This problem is important because it develops the scalar multiplication property of determinants of order \(3\). Students must understand that when every element of a \(3\times3\) determinant is multiplied by \(3\), the determinant is multiplied by \(3^3=27\), not merely by \(3\).

The direct calculation also reinforces determinant expansion and the evaluation of \(2\times2\) minors. In a board examination, writing the determinant expansion carefully and showing the intermediate calculations provides a complete and logically verifiable solution.

Significance for Competitive Entrance Examinations

For competitive examinations, the determinant property provides a major shortcut. Once the order of the determinant is identified, there is usually no need to calculate the transformed determinant entry by entry.

For a \(3\times3\) determinant,

\[ |kA|=k^3|A|. \]
Therefore, the moment a question contains \(|3A|\), a student should recognize the factor \(3^3=27\). This saves calculation time and reduces the possibility of arithmetic errors.

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  8 points
  1. For a square matrix \(A\) of order \(n\),

    \[ |kA|=k^n|A| \]

  2. For a \(3\times3\) matrix,

    \[ |kA|=k^3|A| \]

  3. For \(k=3\),

    \[ |3A|=3^3|A|=27|A| \]

  4. For the given matrix,

    \[ |A|=4 \]

  5. Therefore,

    \[ 27|A|=27(4)=108 \]

  6. Direct calculation also gives

    \[ |3A|=108 \]

  7. Hence,

    \[ \boxed{|3A|=27|A|} \]

  8. The exponent in the scalar multiplication property is the order of the determinant.

← Q3
4 / 8  ·  50%
Q5 →
Q5
NUMERIC3 marks
Evaluate the following determinants:
(i)
\[\begin{vmatrix}3 & -1 & -2\\0 & 0 & -1\\3 & -5 & 0\end{vmatrix}\]
(ii)
\[\begin{vmatrix}3 & -4 & 5\\1 & 1 & -2\\2 & 3 & 1\end{vmatrix}\]
(iii)
\[\begin{vmatrix}0 & 1 & 2\\-1 & 0 & -3\\-2 & 3 & 0\end{vmatrix}\]
(iv)
\[\begin{vmatrix}2 & -1 & -2\\0 & 2 & -1\\3 & -5 & 0\end{vmatrix}\]
📘 Concept & Theory
Concept/Theory

A determinant of order \(3\times3\) can be evaluated by expanding along any row or column. The most efficient choice is generally the row or column containing the greatest number of zeros.

For a determinant

\[\Delta=\begin{vmatrix}a & b & c\\d & e & f\\g & h & i\end{vmatrix},\]

expansion along the first row gives

\[\Delta=a\begin{vmatrix}e & f\\h & i\end{vmatrix}-b\begin{vmatrix}d & f\\g & i\end{vmatrix}+c\begin{vmatrix}d & e\\g & h\end{vmatrix}\]

The signs of the cofactors follow the pattern

\[\begin{matrix}+ & - & +\\- & + & -\\+ & - & +\end{matrix}\]

Therefore, when expanding along the second row, the signs are \(-,+,-\), and when expanding along the third row, they are \(+,-,+\).

In these four problems, selecting a suitable row and applying the \(3\times3\) expansion formula reduces the calculation to determinants of order \(2\times2\).

🗺️ Solution Roadmap
Step-by-step Plan
  1. Inspect each determinant and identify a convenient row or column for expansion.

  2. Apply the appropriate cofactor signs carefully.

  3. Reduce the \(3\times3\) determinant to \(2\times2\) determinants.

  4. Evaluate each \(2\times2\) determinant using

    \[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc\]

  5. Simplify the resulting arithmetic step by step.

  6. Check all negative signs before stating the final answer.

✏️ Solution
(i) \(\qquad\begin{vmatrix}3 & -1 & -2\\0 & 0 & -1\\3 & -5 & 0\end{vmatrix}\)
Step-by-step Solution  ·  4 steps
  1. Given
    \[\begin{vmatrix}3 & -1 & -2\\0 & 0 & -1\\3 & -5 & 0\end{vmatrix}\]
  2. The second row contains two zeros, so expansion along the second row is the most efficient method.
  3. The second row is
    \[\begin{matrix}0 & 0 & -1\end{matrix}\]
  4. Expanding along the second row, the first two terms vanish and only the third term remains. Since the cofactor sign at position \((2,3)\) is positive,
    \[\Delta=0\begin{vmatrix}-1 & -2\\-5 & 0\end{vmatrix}+0\begin{vmatrix}3 & -2\\3 & 0\end{vmatrix}-(-1)\begin{vmatrix}3 & -1\\3 & -5\end{vmatrix}\]
  5. Since \(-(-1)=1\),
    \[\Delta=\begin{vmatrix}3 & -1\\3 & -5\end{vmatrix}\]
  6. Using the \(2\times2\) determinant formula,
    \[\begin{aligned}\Delta &=(3)(-5)-(-1)(3)\\&=-15+3\\&=-12\end{aligned}\]
✏️ Solution
(ii) \(\qquad \begin{vmatrix}3 & -4 & 5\\1 & 1 & -2\\2 & 3 & 1\end{vmatrix}\)
Step-by-step Solution  ·  5 steps
  1. Given
    \[\begin{vmatrix}3 & -4 & 5\\1 & 1 & -2\\2 & 3 & 1\end{vmatrix}\]
  2. There are no zeros in this determinant, so we can expand along the first row.
  3. Using the cofactor signs \(+,-,+\),
    \[\Delta=3\begin{vmatrix}1 & -2\\3 & 1\end{vmatrix}-(-4)\begin{vmatrix}1 & -2\\2 & 1\end{vmatrix}+5\begin{vmatrix}1 & 1\\2 & 3\end{vmatrix}\]
  4. Evaluate the first \(2\times2\) determinant:
    \[\begin{aligned}\begin{vmatrix}1 & -2\\3 & 1\end{vmatrix}&=(1)(1)-(-2)(3)\\&=1+6\\&=7\end{aligned}\]
  5. Evaluate the second \(2\times2\) determinant:
    \[\begin{aligned}\begin{vmatrix}1 & -2\\2 & 1\end{vmatrix}&=(1)(1)-(-2)(2)\\&=1+4\\&=5\end{aligned}\]
  6. Evaluate the third \(2\times2\) determinant:
    \[\begin{aligned}\begin{vmatrix}1 & 1\\2 & 3\end{vmatrix}&=(1)(3)-(1)(2)\\&=3-2\\&=1\end{aligned}\]
  7. Substituting these values,
    \[\begin{aligned}\Delta &=3(7)-(-4)(5)+5(1)\\&=21+20+5\\&=46\end{aligned}\]
✏️ Solution
(iii) \(\qquad \begin{vmatrix}0 & 1 & 2\\-1 & 0 & -3\\-2 & 3 & 0\end{vmatrix}\)
Step-by-step Solution  ·  6 steps
  1. Given
    \[\begin{vmatrix}0 & 1 & 2\\-1 & 0 & -3\\-2 & 3 & 0\end{vmatrix}\]
  2. We expand along the first row. The cofactor signs are \(+,-,+\).
    \[\Delta=0\begin{vmatrix}0 & -3\\3 & 0\end{vmatrix}-1\begin{vmatrix}-1 & -3\\-2 & 0\end{vmatrix}+2\begin{vmatrix}-1 & 0\\-2 & 3\end{vmatrix}\]
  3. The first term is zero:
    \[0\begin{vmatrix}0 & -3\\3 & 0\end{vmatrix}=0\]
  4. Now evaluate the second \(2\times2\) determinant:
    \[\begin{aligned}\begin{vmatrix}-1 & -3\\-2 & 0\end{vmatrix}&=(-1)(0)-(-3)(-2)\\&=0-6\\&=-6\end{aligned}\]
  5. Now evaluate the third \(2\times2\) determinant:
    \[\begin{aligned}\begin{vmatrix}-1 & 0\\-2 & 3\end{vmatrix}&=(-1)(3)-(0)(-2)\\&=-3-0\\&=-3\end{aligned}\]
  6. Hence, the third term is
    \[2(-3)=-6\]
  7. Combining all three terms,
    \[\Delta=0+6-6=0\]
✏️ Solution
(iv) \(\qquad \begin{vmatrix}2 & -1 & -2\\0 & 2 & -1\\3 & -5 & 0\end{vmatrix}\)
Step-by-step Solution  ·  6 steps
  1. Given
    \[\begin{vmatrix}2 & -1 & -2\\0 & 2 & -1\\3 & -5 & 0\end{vmatrix}\]
  2. We expand along the first row using the cofactor signs \(+,-,+\).
    \[\Delta=2\begin{vmatrix}2 & -1\\-5 & 0\end{vmatrix}-(-1)\begin{vmatrix}0 & -1\\3 & 0\end{vmatrix}+(-2)\begin{vmatrix}0 & 2\\3 & -5\end{vmatrix}\]
  3. Since \(-(-1)=1\) and the third coefficient is \(-2\), this becomes
    \[\Delta=2\begin{vmatrix}2 & -1\\-5 & 0\end{vmatrix}+\begin{vmatrix}0 & -1\\3 & 0\end{vmatrix}-2\begin{vmatrix}0 & 2\\3 & -5\end{vmatrix}\]
  4. Evaluate the first \(2\times2\) determinant:
    \[\begin{aligned}\begin{vmatrix}2 & -1\\-5 & 0\end{vmatrix}&=(2)(0)-(-1)(-5)\\&=0-5\\&=-5\end{aligned}\]
  5. Evaluate the second \(2\times2\) determinant:
    \[\begin{aligned}\begin{vmatrix}0 & -1\\3 & 0\end{vmatrix}&=(0)(0)-(-1)(3)\\&=0+3\\&=3\end{aligned}\]
  6. Evaluate the third \(2\times2\) determinant:
    \[\begin{aligned}\begin{vmatrix}0 & 2\\3 & -5\end{vmatrix}&=(0)(-5)-(2)(3)\\&=0-6\\&=-6\end{aligned}\]
  7. Substituting these values,
    \[\begin{aligned}\Delta &=2(-5)+3-2(-6)\\&=-10+3+12\\&=5\end{aligned}\]
💡 Answer
Final Answer

\[ \boxed{\text{(i)}\ -12} \]

\[ \boxed{\text{(ii)}\ 46} \]

\[ \boxed{\text{(iii)}\ 0} \]

\[ \boxed{\text{(iv)}\ 5} \]

🎯 Exam Significance
Exam Significance

This question is important for mastering the expansion of a determinant of order \(3\). It tests whether students can select an efficient row or column, apply the correct cofactor signs, reduce the problem to \(2\times2\) determinants, and simplify the resulting expressions accurately.

Part (i) is particularly useful because the second row contains two zeros, making expansion along that row highly efficient. Part (ii) develops systematic first-row expansion, while parts (iii) and (iv) reinforce careful handling of negative entries and signs.

In board examinations, students should not skip the intermediate minors when the question asks for an evaluation. Writing the expansion and subsequent \(2\times2\) calculations clearly helps demonstrate the complete method and reduces sign-related errors.

Significance for Competitive Entrance Examinations

For competitive examinations, the key skill is choosing the most economical route. A row or column containing zeros should generally be preferred because every zero eliminates an entire cofactor term.

Students should also develop immediate recognition of the cofactor sign pattern

\[\begin{matrix}+ & - & +\\- & + & -\\+ & - & +\end{matrix}\]
This is essential for avoiding sign errors under time pressure.

Part (iii) also provides a useful structural observation. The determinant is skew-symmetric:

\[ A^{T}=-A. \]
Since a skew-symmetric matrix of odd order has determinant zero, the answer can be recognized immediately once this property is known. However, the direct expansion remains valuable for learning the fundamental method.

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  8 points
  1. A \(3\times3\) determinant can be expanded along any row or column.

  2. Choose a row or column containing zeros whenever possible.

  3. The cofactor sign pattern is

    \[\begin{matrix}+ & - & +\\- & + & -\\+ & - & +\end{matrix}\]

  4. Every \(3\times3\) determinant evaluation ultimately reduces to \(2\times2\) determinants.

  5. For a \(2\times2\) determinant,

    \[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc\]

  6. For the determinant exactly given in part (i), the correct answer is

    \[ \boxed{-12} \]

  7. The answers are

    \[ \boxed{-12,\ 46,\ 0,\ 5} \]

  8. Sign accuracy is as important as arithmetic accuracy when evaluating determinants.

← Q4
5 / 8  ·  63%
Q6 →
Q6
NUMERIC3 marks
If \[A=\begin{bmatrix}1 & 1 & -2\\2 & 1 & -3\\5 & 4 & -9\end{bmatrix}\] find \(|A|\).
📘 Concept & Theory
Concept/Theory

To evaluate a determinant of order \(3\times3\), we can expand it along any row or column. For a determinant

\[\begin{vmatrix}a & b & c\\d & e & f\\g & h & i\end{vmatrix}\]

expansion along the first row gives

\[\begin{vmatrix}a & b & c\\d & e & f\\g & h & i\end{vmatrix}=a\begin{vmatrix}e & f\\h & i\end{vmatrix}-b\begin{vmatrix}d & f\\g & i\end{vmatrix}+ c\begin{vmatrix}d & e\\g & h\end{vmatrix}\]

The cofactor signs along the first row are therefore

\[+,-,+\]

Each resulting \(2\times2\) determinant is evaluated using

\[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc\]

🗺️ Solution Roadmap
Step-by-step Plan
  1. Write the determinant corresponding to the given matrix \(A\).

  2. Expand along the first row.

  3. Apply the cofactor sign pattern \(+,-,+\).

  4. Evaluate each resulting \(2\times2\) determinant separately.

  5. Multiply each minor by its corresponding first-row element.

  6. Combine the three terms carefully, paying particular attention to the negative entries.

  7. State the final value of \(|A|\).

✏️ Solution
Complete Solution
Step-by-step Solution  ·  14 steps
  1. Given
    \[A=\begin{bmatrix}1 & 1 & -2\\2 & 1 & -3\\5 & 4 & -9\end{bmatrix}\]
  2. Therefore,
    \[|A|=\begin{vmatrix}1 & 1 & -2\\2 & 1 & -3\\5 & 4 & -9\end{vmatrix}\]
  3. Step 1: Expand along the first row
  4. The first row is
    \[\begin{matrix}1 & 1 & -2\end{matrix}\]
  5. Using the cofactor signs \(+,-,+\),
    \[|A|=1\begin{vmatrix}1 & -3\\4 & -9\end{vmatrix}-1\begin{vmatrix}2 & -3\\5 & -9\end{vmatrix}+(-2)\begin{vmatrix}2 & 1\\5 & 4\end{vmatrix}\]
    Notice that the third term is written as \(+(-2)\) because the third entry of the first row is \(-2\), while the cofactor sign in the third position is positive.
  6. Step 2: Evaluate the first \(2\times2\) determinant
  7. Consider
    \[\begin{vmatrix}1 & -3\\4 & -9\end{vmatrix}\]
  8. Using \(ad-bc\),
    \[\begin{aligned}\begin{vmatrix}1 & -3\\4 & -9\end{vmatrix}&=(1)(-9)-(-3)(4)\\&=-9+12\\&=3\end{aligned}\]
  9. Step 3: Evaluate the second \(2\times2\) determinant
  10. Consider
    \[\begin{vmatrix}2 & -3\\5 & -9\end{vmatrix}\]
  11. Using \(ad-bc\),
    \[\begin{aligned}\begin{vmatrix}2 & -3\\5 & -9\end{vmatrix}&=(2)(-9)-(-3)(5)\\&=-18+15\\&=-3\end{aligned}\]
  12. Remember that the second term in the cofactor expansion carries a negative sign. Therefore, its contribution is
    \[-1(-3)=3\]
  13. Step 4: Evaluate the third \(2\times2\) determinant
  14. Consider
    \[\begin{vmatrix}2 & 1\\5 & 4\end{vmatrix}\]
  15. Using \(ad-bc\),
    \[\begin{aligned}\begin{vmatrix}2 & 1\\5 & 4\end{vmatrix}&=(2)(4)-(1)(5)\\&=8-5\\&=3\end{aligned}\]
  16. The corresponding element in the first row is \(-2\), and its cofactor sign is positive. Therefore, the third contribution is
    \[(-2)(3)=-6\]
  17. Step 5: Combine all three contributions
  18. Thus,
    \[|A|=3+3-6\]
  19. Therefore,
    \[|A|=6-6\]
  20. Hence,
    \[\boxed{|A|=0}\]
🎯 Exam Significance
Exam Significance

This problem is a direct application of cofactor expansion for a determinant of order \(3\). It is particularly useful for strengthening the ability to reduce a \(3\times3\) determinant to three \(2\times2\) determinants and then simplify them accurately.

The problem also demonstrates why every sign in a cofactor expansion matters. The second term has a negative cofactor sign, while the third matrix element itself is negative. Keeping these two facts separate prevents common sign mistakes.

Significance for Competitive Entrance Examinations

For competitive examinations, determinant evaluation should become systematic and fast. Students should immediately recognize the cofactor pattern \(+,-,+\) when expanding along the first row and avoid combining too many arithmetic operations into a single step.

This problem also illustrates the importance of checking structural shortcuts before using them. A row-dependence argument is valid only when the proposed linear relation is actually verified. When no convenient relation is evident, direct cofactor expansion provides a dependable method.

Common Error Alert
  • The cofactor signs along the first row are \(+,-,+\).
  • The third entry of the first row is \(-2\), so the third contribution is \(+(-2)\) times its minor.
  • For
    \[\begin{vmatrix}1 & -3\\4 & -9\end{vmatrix},\]
    the value is \(3\), not \(-21\) or another value obtained by mishandling the negative product.
  • For
    \[\begin{vmatrix}2 & -3\\5 & -9\end{vmatrix},\]
    the value is \(-3\).
  • The second cofactor term therefore becomes
    \[-1(-3)=3\]
  • For
    \[\begin{vmatrix}2 & 1\\5 & 4\end{vmatrix},\]
    the value is \(3\), making the third contribution \(-6\).
  • The final calculation is
    \[3+3-6=0\]
🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  7 points
  1. A \(3\times3\) determinant can be evaluated by cofactor expansion along any row or column.

  2. For first-row expansion, the cofactor signs are \(+,-,+\).

  3. Every resulting \(2\times2\) determinant is evaluated using

    \[ ad-bc. \]

  4. For this determinant, the three cofactor contributions are \(3\), \(3\), and \(-6\).

  5. Therefore,

    \[ |A|=3+3-6=0. \]

  6. The final answer is

    \[ \boxed{|A|=0}. \]

  7. Never assume row dependence without verifying the proposed linear relation exactly.

← Q5
6 / 8  ·  75%
Q7 →
Q7
NUMERIC3 marks
Find the values of \(x\), if:
(i) \(\begin{vmatrix}2 & 4\\5 & 1\end{vmatrix}=\begin{vmatrix}2x & 4\\6 & x\end{vmatrix}\)

(ii) \(\begin{vmatrix}2 & 3\\4 & 5\end{vmatrix}=\begin{vmatrix}x & 3\\2x & 5\end{vmatrix}\)
📘 Concept & Theory
Concept/Theory

The determinant of a matrix of order \(2\times2\) is evaluated using the formula

\[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc\]

Therefore, when two \(2\times2\) determinants are equal, we can evaluate both determinants using this formula and then equate their resulting algebraic expressions.

If the resulting equation is quadratic, both possible roots must be considered. If the resulting equation is linear, it will normally give a single value of \(x\).

🗺️ Solution Roadmap
Step-by-step Plan
  1. Evaluate the determinant on the left-hand side using \(ad-bc\).

  2. Evaluate the determinant on the right-hand side using the same formula.

  3. Equate the two resulting algebraic expressions.

  4. Simplify the equation carefully.

  5. Solve the resulting equation for \(x\).

  6. For a quadratic equation, retain all valid roots.

  7. Verify the obtained value or values in the original determinant equation whenever appropriate.

✏️ Solution
(i) \(\quad\begin{vmatrix}2 & 4\\5 & 1\end{vmatrix}=\begin{vmatrix}2x & 4\\6 & x\end{vmatrix}\)
Step-by-step Solution  ·  4 steps
  1. Given — \(\quad\begin{vmatrix}2 & 4\\5 & 1\end{vmatrix}=\begin{vmatrix}2x & 4\\6 & x\end{vmatrix}\)
  2. Step 1: Evaluate the determinant on the left-hand side
  3. Using
    \[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc,\]
  4. we get
    \[\begin{aligned}\begin{vmatrix}2 & 4\\5 & 1\end{vmatrix}&=(2)(1)-(4)(5)\\&=2-20\\&=-18\end{aligned}\]
  5. Step 2: Evaluate the determinant on the right-hand side
  6. \[\begin{aligned}\begin{vmatrix}2x & 4\\6 & x\end{vmatrix}&=(2x)(x)-(4)(6)\\&=2x^2-24\end{aligned}\]
  7. Step 3: Equate the two determinants
  8. \[\begin{aligned} -18&=2x^2-24\\ \Rightarrow 2x^2&=24-18\\\ x^2&=\dfrac63\\ x&=\pm\sqrt{3} \end{aligned}\]
✏️ Solution
(ii) \(\quad\begin{vmatrix}2 & 3\\4 & 5\end{vmatrix}=\begin{vmatrix}x & 3\\2x & 5\end{vmatrix}\)
Step-by-step Solution  ·  4 steps
  1. Given
    \[\begin{vmatrix}2 & 3\\4 & 5\end{vmatrix}=\begin{vmatrix}x & 3\\2x & 5\end{vmatrix}\]
  2. Step 1: Evaluate the determinant on the left-hand side
  3. Using the \(2\times2\) determinant formula,
    \[\begin{aligned}\begin{vmatrix}2 & 3\\4 & 5\end{vmatrix}&=(2)(5)-(3)(4)\\&=10-12\\&=-2\end{aligned}\]
  4. Step 2: Evaluate the determinant on the right-hand side
  5. \[\begin{aligned}\begin{vmatrix}x & 3\\2x & 5\end{vmatrix}&=(x)(5)-(3)(2x)\\&=5x-6x\\&=-x\end{aligned}\]
  6. Step 3: Equate the two determinants
  7. Since the determinants are equal,
    \[-2=-x\]
  8. Step 4: Solve for \(x\)
  9. Multiplying both sides by \(-1\),
    \[2=x\]
💡 Answer
Final Answer
\[\boxed{\text{(i)}\quad x=\pm\sqrt{3}}\]
\[\boxed{\text{(ii)}\quad x=2}\]
🎯 Exam Significance
Exam Significance

This question tests one of the fundamental applications of determinants: converting an equality between determinants into an algebraic equation in an unknown variable.

For board examinations, students should clearly show the evaluation of both determinants before solving the resulting equation. In part (i), the equation is quadratic, so both roots must be considered. Writing only \(x=\sqrt{3}\) would be incomplete because \(x=-\sqrt{3}\) also satisfies \(x^2=3\).

Part (ii) reinforces the handling of algebraic entries inside a determinant and demonstrates how a determinant equation can reduce directly to a linear equation.

Significance for Competitive Entrance Examinations

These problems are useful for developing speed and accuracy in determinant-based algebraic manipulation. In an entrance examination, the essential process is immediate:

\[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc\]

Students should simplify each determinant before equating the expressions. This avoids unnecessary expansion and reduces the possibility of sign errors.

Part (i) is especially important because the equation involves \(x^2\). Whenever an equation gives \(x^2=k\), both square-root possibilities should be considered unless additional restrictions on \(x\) are given.

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  7 points
  1. The determinant of a \(2\times2\) matrix is

    \[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc\]

  2. When two determinants are equal, evaluate each one and then equate their values.

  3. Algebraic entries such as \(x\) or \(2x\) are handled in exactly the same way as numerical entries.

  4. A quadratic equation obtained from a determinant may have two valid values of \(x\).

  5. For part (i),

    \[\boxed{x=\pm\sqrt{3}}\]

  6. For part (ii),

    \[\boxed{x=2}\]

  7. Verification by substitution is a useful way to detect sign or algebraic errors.

← Q6
7 / 8  ·  88%
Q8 →
Q8
NUMERIC3 marks
If \[\begin{vmatrix}x & 2\\18 & x\end{vmatrix}=\begin{vmatrix}6 & 2\\18 & 6\end{vmatrix},\]

then \(x\) is equal to:

  1. \(6\)
  2. \(\pm6\)
  3. \(-6\)
  4. \(0\)
📘 Concept & Theory
Concept/Theory

For a determinant of order \(2\times2\),

\[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc\]

Therefore, the determinant equation can be converted into an ordinary algebraic equation by evaluating both determinants separately.

An important point in this question is that the resulting equation contains \(x^2\). If

\[ x^2=a, \]
where \(a>0\), then
\[ x=\pm\sqrt{a}. \]
Thus, both positive and negative roots must be considered unless the question imposes an additional restriction on \(x\).

🗺️ Solution Roadmap
Step-by-step Plan
  1. Evaluate the determinant containing \(x\).

  2. Evaluate the numerical determinant on the right-hand side.

  3. Equate the two determinant values.

  4. Solve the resulting equation for \(x\).

  5. Compare the obtained values with the given options.

✏️ Solution
Complete Solution
Step-by-step Solution  ·  6 steps
  1. Given
    \[\begin{vmatrix}x & 2\\18 & x\end{vmatrix}=\begin{vmatrix}6 & 2\\18 & 6\end{vmatrix},\]
  2. Step 1: Evaluate the determinant on the left-hand side
  3. Using
    \[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc,\]
  4. we obtain
    \[\begin{aligned}\begin{vmatrix}x & 2\\18 & x\end{vmatrix}&=(x)(x)-(2)(18)\\&=x^2-36\end{aligned}\]
  5. Step 2: Evaluate the determinant on the right-hand side
  6. \[\begin{aligned}\begin{vmatrix}6 & 2\\18 & 6\end{vmatrix}&=(6)(6)-(2)(18)\\&=36-36\\&=0\end{aligned}\]
  7. Step 3: Equate the two values
  8. Since the determinants are equal,
    \[\begin{aligned}x^2-36&=0\\x^2&=36\\x&=\pm\sqrt{36}\\x&=\pm6\end{aligned}\]
  9. Step 5: Match with the given options
  10. The obtained values are
    \[x=6\quad\text{or}\quad x=-6\]
  11. Hence, the correct option is
    \[\boxed{\text{(B) }\pm6}\]
🎯 Exam Significance
Exam Significance

This problem reinforces the fundamental \(2\times2\) determinant formula and its application to equations involving an unknown variable. It also tests whether students correctly handle the two roots generated by a square equation.

In a board examination, students should explicitly write

\[ x^2=36 \]
followed by
\[ x=\pm6. \]
Writing only \(x=6\) would omit one valid solution.

Significance for Competitive Entrance Examinations

This is a typical quick-evaluation problem. The determinant on the right immediately evaluates to zero:

\[ 6(6)-2(18)=0. \]
Therefore, the left determinant must also be zero:
\[ x^2-36=0. \]

Recognising the difference of two squares gives an even faster route:

\[ x^2-36=(x-6)(x+6)=0. \]
Hence,
\[ x=6\quad\text{or}\quad x=-6. \]

Such factorisation-based recognition can save valuable time in entrance examinations while also reducing arithmetic work.

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  7 points
  1. For a \(2\times2\) determinant,

    \[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc\]

  2. Evaluate both determinants before solving the resulting equation.

  3. The numerical determinant on the right is

    \[36-36=0\]

  4. Consequently,

    \[x^2-36=0\]

  5. Using factorisation,

    \[(x-6)(x+6)=0\]

  6. Hence,

    \[\boxed{x=\pm6}\]

  7. The correct multiple-choice answer is

    \[\boxed{\text{(B) }\pm6}\]

← Q7
8 / 8  ·  100%
↑ Back to top
🎓

Chapter Complete!

All 8 solutions for Determinants covered.

↑ Review from the top
📚
ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
Sharing this chapter
NCERT Class 12 Determinants Exercise 4.1 Q8 Solution
NCERT Class 12 Determinants Exercise 4.1 Q8 Solution — Complete Notes & Solutions · academia-aeternum.com
NCERT Class 12 Mathematics Chapter 4 Determinants Exercise 4.1 Question 8 focuses on finding the value of an unknown variable by using the basic properties of determinants. In this multiple-choice question, two determinants of order 2 are equated, requiring students to evaluate each determinant using the formula ad − bc and then solve the resulting algebraic equation. The problem provides an excellent opportunity to strengthen accuracy in determinant calculations, especially when negative signs…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
Share on
academia-aeternum.com/class-12/mathematics/determinants/exercises/exercise-4.1/ Copy link
💡
Exam tip: Sharing chapter notes with your study group creates a reinforcement loop. Teaching a concept is the fastest path to mastering it.

Recent posts

    Determinants — Learning Resources

    📄 Detailed Notes
    🧠 Practice MCQs
    ✔️ True / False
    🎯 Advance MCQs

    Frequently Asked Questions

    The correct answer is x = ±6, which corresponds to option (B).

    For a determinant |a b; c d|, use the formula ad - bc.

    Its determinant is x² - 36.

    Its value is 6 × 6 - 2 × 18 = 36 - 36 = 0.

    Factorise as (x - 6)(x + 6) = 0, giving x = 6 or x = -6.

    Because x² = 36 has two real solutions, x = 6 and x = -6.

    Option (B), x = ±6, is correct.

    The question tests evaluation of 2×2 determinants and solving the resulting algebraic equation.

    No. x = -6 also satisfies the determinant equation, so the complete answer is x = ±6.

    Yes. The correct option is (B) ±6; “16” is a typographical error.

    Get in Touch

    Let's Connect

    Questions, feedback, or suggestions?
    We'd love to hear from you.