GSEB Class 12 Maths Solutions Chapter 4 Determinants Exercise 4.1

Note: When a matrix is multiplied by k, each element of the matrix is multiplied by k, whereas in the case of a determinant, only elements of a row or a column are multiplied by k.

Question 1. \( \left|\begin{array}{cc} 2 & 4 \\ -5 & -1 \end{array}\right| \)
Answer: We have the determinant:
\[ \left|\begin{array}{cc} 2 & 4 \\ -5 & -1 \end{array}\right| = 2 \times (-1) - 4 \times (-5) \]
\[ = -2 - (-20) \]
\[ = -2 + 20 \]
\[ = 18 \]
In simple words: To calculate this determinant, you multiply the numbers on the main diagonal (top-left by bottom-right), then subtract the product of the numbers on the other diagonal (top-right by bottom-left).

Exam Tip: Remember the formula for a \( 2 \times 2 \) determinant: \( \left|\begin{array}{cc} a & b \\ c & d \end{array}\right| = ad - bc \). This is a fundamental concept for solving matrix problems.

Question 2.
(i) \( \left|\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right| \)
(ii) \( \left|\begin{array}{cc} x^{2}-x+1 & x-1 \\ x+1 & x+1 \end{array}\right| \)
Answer:
(i) To evaluate this determinant:
\[ \left|\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right| = (\cos \theta \times \cos \theta) - (-\sin \theta \times \sin \theta) \]
\[ = \cos^2\theta - (-\sin^2\theta) \]
\[ = \cos^2\theta + \sin^2\theta \]
\[ = 1 \]
(ii) To evaluate the second determinant:
\[ \left|\begin{array}{cc} x^{2}-x+1 & x-1 \\ x+1 & x+1 \end{array}\right| = (x^2-x+1)(x+1) - (x-1)(x+1) \]
First, let's expand the products:
\( (x^2-x+1)(x+1) = x^3 + x^2 - x^2 - x + x + 1 = x^3 + 1 \)
\( (x-1)(x+1) = x^2 - 1 \)
Now, substitute these back into the determinant expression:
\[ = (x^3+1) - (x^2-1) \]
\[ = x^3+1-x^2+1 \]
\[ = x^3-x^2+2 \]
In simple words: For part (i), we used a trigonometric identity. For part (ii), we used algebraic multiplication and subtraction to simplify the expression.

Exam Tip: Always remember basic trigonometric identities like \( \cos^2\theta + \sin^2\theta = 1 \) and algebraic identities like \( (a-b)(a+b) = a^2-b^2 \) as they frequently simplify determinant calculations.

Question 3. If \( A = \left[\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right] \), then show that \( |2A| = 4|A| \).
Answer: First, let's calculate \( 2A \):
\[ 2A = 2 \left[\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right] = \left[\begin{array}{ll} 2 \times 1 & 2 \times 2 \\ 2 \times 4 & 2 \times 2 \end{array}\right] = \left[\begin{array}{ll} 2 & 4 \\ 8 & 4 \end{array}\right] \]
Now, find the determinant of \( 2A \):
\[ |2A| = \left|\begin{array}{cc} 2 & 4 \\ 8 & 4 \end{array}\right| = (2 \times 4) - (4 \times 8) = 8 - 32 = -24 \]
Next, find the determinant of \( A \):
\[ |A| = \left|\begin{array}{cc} 1 & 2 \\ 4 & 2 \end{array}\right| = (1 \times 2) - (2 \times 4) = 2 - 8 = -6 \]
Now, calculate \( 4|A| \):
\[ 4|A| = 4 \times (-6) = -24 \]
Since \( |2A| = -24 \) and \( 4|A| = -24 \), we have shown that \( |2A| = 4|A| \).
In simple words: We first multiplied the matrix A by 2, then found its determinant. Next, we found the determinant of A and multiplied it by 4. Both results were the same, proving the relationship.

Exam Tip: Remember that for a scalar k and an \( n \times n \) matrix A, \( |kA| = k^n |A| \). In this case, \( n=2 \), so \( |2A| = 2^2 |A| = 4|A| \). This property is useful for quickly solving such problems.

Question 4. If \( A = \left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right] \), then show that \( |3A| = 27|A| \).
Answer: First, let's calculate \( 3A \):
\[ 3A = 3 \left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right] = \left[\begin{array}{lll} 3 \times 1 & 3 \times 0 & 3 \times 1 \\ 3 \times 0 & 3 \times 1 & 3 \times 2 \\ 3 \times 0 & 3 \times 0 & 3 \times 4 \end{array}\right] = \left[\begin{array}{lll} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{array}\right] \]
Now, find the determinant of \( 3A \). Since it's an upper triangular matrix, its determinant is the product of its diagonal elements:
\[ |3A| = 3 \times 3 \times 12 = 9 \times 12 = 108 \]
Alternatively, we can take out common factors from rows/columns. Take out 3 from R₁, R₂, and R₃ each:
\[ |3A| = 3 \times 3 \times 3 \left|\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right] = 27 \left|\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right] \]
Now, find the determinant of \( A \). Since A is also an upper triangular matrix, its determinant is the product of its diagonal elements:
\[ |A| = \left|\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right| = 1 \times 1 \times 4 = 4 \]
Now, calculate \( 27|A| \):
\[ 27|A| = 27 \times 4 = 108 \]
Since \( |3A| = 108 \) and \( 27|A| = 108 \), we have shown that \( |3A| = 27|A| \).
In simple words: We multiplied matrix A by 3 and then found its determinant. We also found the determinant of A and multiplied it by 27. Both calculations gave the same result, confirming the property.

Exam Tip: For a scalar k and an \( n \times n \) matrix A, the property \( |kA| = k^n |A| \) holds. Here, A is a \( 3 \times 3 \) matrix, so \( n=3 \), and thus \( |3A| = 3^3 |A| = 27|A| \). Recognizing this property can save calculation time.

Question 5. Evaluate the determinants:
(i) \( \left|\begin{array}{ccc} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array}\right| \)
(ii) \( \left|\begin{array}{ccc} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array}\right| \)
(iii) \( \left|\begin{array}{ccc} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{array}\right| \)
(iv) \( \left|\begin{array}{ccc} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array}\right| \)
Answer:
(i) To evaluate \( \left|\begin{array}{ccc} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array}\right| \), let's expand along the first row:
\[ = 3 \left|\begin{array}{cc} 0 & -1 \\ -5 & 0 \end{array}\right| - (-1) \left|\begin{array}{cc} 0 & -1 \\ 3 & 0 \end{array}\right| + (-2) \left|\begin{array}{cc} 0 & 0 \\ 3 & -5 \end{array}\right| \]
\[ = 3(0 \times 0 - (-1) \times (-5)) + 1(0 \times 0 - (-1) \times 3) - 2(0 \times (-5) - 0 \times 3) \]
\[ = 3(0 - 5) + 1(0 - (-3)) - 2(0 - 0) \]
\[ = 3(-5) + 1(3) - 2(0) \]
\[ = -15 + 3 - 0 \]
\[ = -12 \]
Alternatively, expanding with elements of the second row, which contains zeros:
\[ = -0 \left|\begin{array}{cc} -1 & -2 \\ -5 & 0 \end{array}\right| + 0 \left|\begin{array}{cc} 3 & -2 \\ 3 & 0 \end{array}\right| - (-1) \left|\begin{array}{cc} 3 & -1 \\ 3 & -5 \end{array}\right| \]
\[ = 0 + 0 + 1((3 \times (-5)) - (-1 \times 3)) \]
\[ = 1(-15 - (-3)) \]
\[ = 1(-15 + 3) \]
\[ = -12 \]
(ii) To evaluate \( \left|\begin{array}{ccc} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array}\right| \), let's expand along the first row:
\[ = 3 \left|\begin{array}{cc} 1 & -2 \\ 3 & 1 \end{array}\right| - (-4) \left|\begin{array}{cc} 1 & -2 \\ 2 & 1 \end{array}\right| + 5 \left|\begin{array}{cc} 1 & 1 \\ 2 & 3 \end{array}\right| \]
\[ = 3(1 \times 1 - (-2) \times 3) + 4(1 \times 1 - (-2) \times 2) + 5(1 \times 3 - 1 \times 2) \]
\[ = 3(1 + 6) + 4(1 + 4) + 5(3 - 2) \]
\[ = 3(7) + 4(5) + 5(1) \]
\[ = 21 + 20 + 5 \]
\[ = 46 \]
(iii) To evaluate \( \left|\begin{array}{ccc} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{array}\right| \), let's expand along the first row:
\[ = 0 \left|\begin{array}{cc} 0 & -3 \\ 3 & 0 \end{array}\right| - 1 \left|\begin{array}{cc} -1 & -3 \\ -2 & 0 \end{array}\right| + 2 \left|\begin{array}{cc} -1 & 0 \\ -2 & 3 \end{array}\right| \]
\[ = 0 - 1((-1 \times 0) - (-3) \times (-2)) + 2((-1 \times 3) - (0 \times (-2))) \]
\[ = 0 - 1(0 - 6) + 2(-3 - 0) \]
\[ = -1(-6) + 2(-3) \]
\[ = 6 - 6 \]
\[ = 0 \]
(iv) To evaluate \( \left|\begin{array}{ccc} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array}\right| \), let's expand along the first column because of the zero:
\[ = 2 \left|\begin{array}{cc} 2 & -1 \\ -5 & 0 \end{array}\right| - 0 \left|\begin{array}{cc} -1 & -2 \\ -5 & 0 \end{array}\right| + 3 \left|\begin{array}{cc} -1 & -2 \\ 2 & -1 \end{array}\right| \]
\[ = 2((2 \times 0) - (-1) \times (-5)) - 0 + 3((-1 \times (-1)) - (-2) \times 2) \]
\[ = 2(0 - 5) - 0 + 3(1 - (-4)) \]
\[ = 2(-5) + 3(1 + 4) \]
\[ = -10 + 3(5) \]
\[ = -10 + 15 \]
\[ = 5 \]
In simple words: For each part, we calculated the determinant by expanding along a row or column, which helps break down the bigger problem into smaller \( 2 \times 2 \) determinants that are easier to solve. We carefully followed the sign convention for each term.

Exam Tip: When evaluating a \( 3 \times 3 \) determinant, choose to expand along the row or column that contains the most zeros. This minimizes the number of \( 2 \times 2 \) determinants you need to calculate, reducing potential errors.

Question 6. If \( A = \left|\begin{array}{ccc} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{array}\right| \), find \( |A| \).
Answer: To find \( |A| \), we can expand along the first row:
\[ |A| = 1 \left|\begin{array}{cc} 1 & -3 \\ 4 & -9 \end{array}\right| - 1 \left|\begin{array}{cc} 2 & -3 \\ 5 & -9 \end{array}\right| + (-2) \left|\begin{array}{cc} 2 & 1 \\ 5 & 4 \end{array}\right| \]
\[ = 1((1 \times (-9)) - (-3 \times 4)) - 1((2 \times (-9)) - (-3 \times 5)) - 2((2 \times 4) - (1 \times 5)) \]
\[ = 1(-9 - (-12)) - 1(-18 - (-15)) - 2(8 - 5) \]
\[ = 1(-9 + 12) - 1(-18 + 15) - 2(3) \]
\[ = 1(3) - 1(-3) - 6 \]
\[ = 3 + 3 - 6 \]
\[ = 0 \]
In simple words: We calculated the determinant of matrix A by expanding it based on the first row. We did the cross-multiplication for each smaller determinant and then summed them up to get the final answer.

Exam Tip: Determinants that evaluate to zero often indicate that the matrix is singular or that its rows/columns are linearly dependent. Always double-check your arithmetic, especially with negative numbers.

Question 7. Find the value of x, if
(i) \( \left|\begin{array}{cc} 2 & 4 \\ 5 & 1 \end{array}\right| = \left|\begin{array}{cc} 2x & 4 \\ 6 & x \end{array}\right| \)
(ii) \( \left|\begin{array}{cc} 2 & 3 \\ 4 & 5 \end{array}\right| = \left|\begin{array}{cc} x & 3 \\ 2x & 5 \end{array}\right| \)
Answer:
(i) First, evaluate the left-hand side determinant:
\[ \left|\begin{array}{cc} 2 & 4 \\ 5 & 1 \end{array}\right| = (2 \times 1) - (4 \times 5) = 2 - 20 = -18 \]
Next, evaluate the right-hand side determinant:
\[ \left|\begin{array}{cc} 2x & 4 \\ 6 & x \end{array}\right| = (2x \times x) - (4 \times 6) = 2x^2 - 24 \]
Now, set the two determinants equal to each other:
\( -18 = 2x^2 - 24 \)
Add 24 to both sides:
\( -18 + 24 = 2x^2 \)
\( 6 = 2x^2 \)
Divide by 2:
\( x^2 = 3 \)
Take the square root of both sides:
\( x = \pm\sqrt{3} \)
(ii) First, evaluate the left-hand side determinant:
\[ \left|\begin{array}{cc} 2 & 3 \\ 4 & 5 \end{array}\right| = (2 \times 5) - (3 \times 4) = 10 - 12 = -2 \]
Next, evaluate the right-hand side determinant:
\[ \left|\begin{array}{cc} x & 3 \\ 2x & 5 \end{array}\right| = (x \times 5) - (3 \times 2x) = 5x - 6x = -x \]
Now, set the two determinants equal to each other:
\( -2 = -x \)
Multiply both sides by -1:
\( x = 2 \)
In simple words: For both parts, we calculated the value of each determinant separately. Then, we set them equal to form an equation and solved for x using basic algebra.

Exam Tip: Be careful with signs when calculating determinants, especially when dealing with negative numbers. When solving for x, remember to consider both positive and negative roots for \( x^2 \) equations.

Question 8. If \( \left|\begin{array}{cc} x & 2 \\ 18 & x \end{array}\right| = \left|\begin{array}{cc} 6 & 2 \\ 18 & 6 \end{array}\right| \), then x is equal to
(a) 6
(b) ± 6
(c) - 6
(d) 6, 6
Answer: (b) ± 6
In simple words: We calculated the value of each determinant, set them equal, and solved the resulting equation for x, finding both positive and negative solutions.

Exam Tip: When solving equations involving \( x^2 \), always remember that there are usually two solutions (a positive and a negative root), unless x is zero. In MCQ questions, check all options that fit the solution.