OP Malhotra Class 12 Maths Solutions Chapter 5 Determinants Exercise 5 (A)

Question 1. Evaluate the following determinants:
(i) \( \left|\begin{array}{ll} 2 & 5 \\ 4 & 1 \end{array}\right| \)
(ii) \( \left|\begin{array}{ll} 3 & 5 \\ 1 & 2 \end{array}\right| \)
(iii) \( \left|\begin{array}{rr} a & b \\ -b & a \end{array}\right| \)
(iv) \( \left|\begin{array}{rr} x+2 & 2 x+5 \\ 3 x-1 & x-3 \end{array}\right| \)
(v) \( \left|\begin{array}{rr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right| \)
(vi) \( \left|\begin{array}{cc} y-x & -x^2+x y-y^2 \\ x+y & x^2+x y+y^2 \end{array}\right| \)
Answer:
(i) To find the value of this determinant, we multiply the numbers diagonally. First, multiply 2 by 1, which gives 2. Then, multiply 5 by 4, which gives 20. Subtract the second result from the first result. So, \(2 - 20\) equals \(-18\). Determinants help us find solutions for systems of linear equations.
\( \left|\begin{array}{ll} 2 & 5 \\ 4 & 1 \end{array}\right| = 2 \times 1 - 5 \times 4 = 2 - 20 = -18 \)
(ii) For this determinant, we multiply 3 by 2 to get 6. Then, we multiply 5 by 1 to get 5. When we subtract 5 from 6, the result is 1. This simple calculation gives us the determinant's value.
\( \left|\begin{array}{ll} 3 & 5 \\ 1 & 2 \end{array}\right| = 3 \times 2 - 1 \times 5 = 6 - 5 = 1 \)
(iii) Here, we have letters instead of numbers. We multiply 'a' by 'a' to get \(a^2\). Next, we multiply 'b' by '\(-b\)' to get \(-b^2\). When we subtract \(-b^2\) from \(a^2\), the two minus signs make a plus, so the answer is \(a^2 + b^2\). This formula is often used in complex number calculations.
\( \left|\begin{array}{rr} a & b \\ -b & a \end{array}\right| = a \times a - b \times (-b) = a^2 + b^2 \)
(iv) To solve this, we multiply diagonally. First, multiply \((x + 2)\) by \((x - 3)\). Then, multiply \((2x + 5)\) by \((3x - 1)\). Subtract the second product from the first. After expanding and simplifying all the terms, the final answer comes out to be \(-5x^2 - 14x - 1\). This method helps simplify complex algebraic expressions.
\( \left|\begin{array}{rr} x+2 & 2 x+5 \\ 3 x-1 & x-3 \end{array}\right| = (x + 2)(x - 3) - (2x + 5)(3x - 1) \)
\( = (x^2 - 3x + 2x - 6) - (6x^2 - 2x + 15x - 5) \)
\( = (x^2 - x - 6) - (6x^2 + 13x - 5) \)
\( = x^2 - x - 6 - 6x^2 - 13x + 5 \)
\( = -5x^2 - 14x - 1 \)
(v) Here, we deal with trigonometric functions. We multiply \(\cos \theta\) by \(\cos \theta\) to get \(\cos^2 \theta\). Then, we multiply \(-\sin \theta\) by \(\sin \theta\) to get \(-\sin^2 \theta\). When we subtract \((-\sin^2 \theta)\), it becomes \(+\sin^2 \theta\). So, we have \(\cos^2 \theta + \sin^2 \theta\), which is a basic trigonometric identity always equal to 1. This identity is fundamental in trigonometry.
\( \left|\begin{array}{rr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right| = \cos \theta \times \cos \theta - (-\sin \theta) \times \sin \theta \)
\( = \cos^2 \theta + \sin^2 \theta \)
\( = 1 \)
(vi) To evaluate this determinant, we multiply the diagonal elements. First, multiply \((y - x)\) by \((x^2 + xy + y^2)\), which simplifies to \((y^3 - x^3)\). Next, multiply \((x + y)\) by \((-x^2 + xy - y^2)\), which simplifies to \(-(x^3 + y^3)\). Now, subtract the second product from the first. So, \((y^3 - x^3) - (-(x^3 + y^3))\) becomes \((y^3 - x^3) + (x^3 + y^3)\). The \(-x^3\) and \(+x^3\) cancel out, leaving us with \(y^3 + y^3 = 2y^3\). Determinants involving algebraic expressions are common in higher mathematics.
\( \left|\begin{array}{cc} y-x & -x^2+x y-y^2 \\ x+y & x^2+x y+y^2 \end{array}\right| \)
\( = (y-x)(x^2+xy+y^2) - (x+y)(-x^2+xy-y^2) \)
\( = (y^3-x^3) + (x+y)(x^2-xy+y^2) \)
\( = y^3-x^3 + x^3+y^3 \)
\( = 2y^3 \)
In simple words: For each part, multiply the numbers or expressions along the main diagonal, then multiply along the other diagonal. Subtract the second result from the first. Simplify your answer.

🎯 Exam Tip: Remember the basic formula for a 2x2 determinant: \( \left|\begin{array}{ll} a & b \\ c & d \end{array}\right| = ad - bc \). Be careful with signs, especially when terms are negative or when subtracting an expression.

Question 2. Prove that \( \left|\begin{array}{rr} \sin 10^{\circ} & -\cos 10^{\circ} \\ \sin 80^{\circ} & \cos 80^{\circ} \end{array}\right| = 1 \)
Answer: To prove this, we calculate the determinant. We multiply \(\sin 10^\circ\) by \(\cos 80^\circ\). Then we multiply \(-\cos 10^\circ\) by \(\sin 80^\circ\). When we subtract the second product from the first, we get \(\sin 10^\circ \cos 80^\circ - (-\cos 10^\circ \sin 80^\circ)\), which simplifies to \(\sin 10^\circ \cos 80^\circ + \cos 10^\circ \sin 80^\circ\). This is the formula for \(\sin(A+B)\), which here means \(\sin(10^\circ + 80^\circ)\). This simplifies to \(\sin 90^\circ\), and we know \(\sin 90^\circ\) is equal to 1. This confirms the proof. Understanding trigonometric identities is key here.
\( \left|\begin{array}{rr} \sin 10^{\circ} & -\cos 10^{\circ} \\ \sin 80^{\circ} & \cos 80^{\circ} \end{array}\right| \)
\( = \sin 10^\circ \cos 80^\circ - (-\cos 10^\circ \sin 80^\circ) \)
\( = \sin 10^\circ \cos 80^\circ + \cos 10^\circ \sin 80^\circ \)
\( = \sin(10^\circ + 80^\circ) \)
\( = \sin 90^\circ \)
\( = 1 \)
In simple words: Calculate the determinant by multiplying diagonally and subtracting. This gives a sum that is the formula for \(\sin(10^\circ + 80^\circ)\). This is \(\sin 90^\circ\), which is 1.

🎯 Exam Tip: Recognize trigonometric identities like \(\sin(A+B) = \sin A \cos B + \cos A \sin B\) immediately, as they simplify calculations significantly. Also, know standard trigonometric values like \(\sin 90^\circ = 1\).

Question 3. If \( \left|\begin{array}{rr} 3 & m \\ 4 & 5 \end{array}\right| = 3 \), find the value of m.
Answer: We are given a determinant and its value. First, we calculate the determinant: multiply 3 by 5 to get 15, and multiply 'm' by 4 to get 4m. Subtract 4m from 15. The problem tells us this result is equal to 3. So, \(15 - 4m = 3\). To solve for 'm', we subtract 15 from both sides, getting \(-4m = -12\). Then, divide by -4, which gives \(m = 3\). Solving equations involving determinants is a basic skill in linear algebra.
Given \( \left|\begin{array}{rr} 3 & m \\ 4 & 5 \end{array}\right| = 3 \)
\( \implies 3 \times 5 - m \times 4 = 3 \)
\( \implies 15 - 4m = 3 \)
\( \implies -4m = 3 - 15 \)
\( \implies -4m = -12 \)
\( \implies m = \frac{-12}{-4} \)
\( \implies m = 3 \)
In simple words: Calculate the determinant which is \(15 - 4m\). Set it equal to 3. Then, solve the simple equation for 'm'. You will find that 'm' is 3.

🎯 Exam Tip: Always set up the determinant equation correctly first, then use standard algebraic methods to solve for the unknown variable. Pay attention to signs when moving terms across the equals sign.

Question 4. (i) Find the value of x, if \( \left|\begin{array}{rr} x-1 & x-2 \\ x & x-3 \end{array}\right| = 0 \)
(ii) \( \left|\begin{array}{ll} 3 x & 7 \\ -2 & 4 \end{array}\right|=\left|\begin{array}{ll} 8 & 7 \\ 6 & 4 \end{array}\right| \)
Answer:
(i) To find 'x', we first expand the determinant. Multiply \((x - 1)\) by \((x - 3)\), then subtract the product of 'x' and \((x - 2)\). Set the whole expression equal to 0. Expand the terms to get \(x^2 - 4x + 3 - (x^2 - 2x) = 0\). The \(x^2\) terms cancel out, leaving \(-2x + 3 = 0\). By moving 3 to the other side and dividing by -2, we find that \(x = \frac{3}{2}\). This shows how determinants can be used to set up and solve equations.
Given \( \left|\begin{array}{rr} x-1 & x-2 \\ x & x-3 \end{array}\right| = 0 \)
\( \implies (x - 1)(x - 3) - x(x - 2) = 0 \)
\( \implies (x^2 - 3x - x + 3) - (x^2 - 2x) = 0 \)
\( \implies x^2 - 4x + 3 - x^2 + 2x = 0 \)
\( \implies -2x + 3 = 0 \)
\( \implies -2x = -3 \)
\( \implies x = \frac{3}{2} \)
(ii) Here, we have two determinants set equal to each other. First, calculate the left determinant: multiply \(3x\) by 4 to get \(12x\), then subtract the product of 7 and -2, which is -14. So, the left side is \(12x - (-14) = 12x + 14\). Next, calculate the right determinant: multiply 8 by 4 to get 32, then subtract the product of 7 and 6, which is 42. So, the right side is \(32 - 42 = -10\). Now set the two results equal: \(12x + 14 = -10\). Subtract 14 from both sides to get \(12x = -24\), then divide by 12 to find \(x = -2\). This demonstrates solving for a variable by equating two determinant values.
Given \( \left|\begin{array}{ll} 3 x & 7 \\ -2 & 4 \end{array}\right|=\left|\begin{array}{ll} 8 & 7 \\ 6 & 4 \end{array}\right| \)
\( \implies (3x)(4) - (7)(-2) = (8)(4) - (7)(6) \)
\( \implies 12x - (-14) = 32 - 42 \)
\( \implies 12x + 14 = -10 \)
\( \implies 12x = -10 - 14 \)
\( \implies 12x = -24 \)
\( \implies x = \frac{-24}{12} \)
\( \implies x = -2 \)
In simple words: For (i), expand the determinant and set the quadratic equation to zero, then solve for 'x'. For (ii), calculate both determinants separately, set them equal, and solve the linear equation for 'x'.

🎯 Exam Tip: When dealing with algebraic expressions in determinants, be careful with signs during expansion and simplification. For equations with two determinants, calculate each side fully before equating them.

Question 5. Determine the value of k for which \( \left|\begin{array}{rr} k & k \\ 4 & 2 k \end{array}\right| = 0 \)
Answer: We need to find the values of 'k' that make this determinant equal to zero. First, calculate the determinant: multiply 'k' by \(2k\) to get \(2k^2\), then subtract the product of 'k' and 4, which is \(4k\). So, we have the equation \(2k^2 - 4k = 0\). To solve this, factor out \(2k\), which gives \(2k(k - 2) = 0\). For this product to be zero, either \(2k = 0\) or \((k - 2) = 0\). This means \(k = 0\) or \(k = 2\). These are the two values for 'k'. Finding roots of a quadratic equation often involves factoring.
Given \( \left|\begin{array}{rr} k & k \\ 4 & 2 k \end{array}\right| = 0 \)
\( \implies k \times (2k) - k \times 4 = 0 \)
\( \implies 2k^2 - 4k = 0 \)
\( \implies 2k(k - 2) = 0 \)
This equation holds true if either of the factors is zero.
\( \implies 2k = 0 \) or \( k - 2 = 0 \)
\( \implies k = 0 \) or \( k = 2 \)
In simple words: Calculate the determinant. Set the resulting expression to zero. Factor the equation to find two possible values for 'k'. The values are 0 and 2.

🎯 Exam Tip: When a determinant equals zero, it usually means there are specific conditions for the variables. Factoring is a key method for solving quadratic equations that arise from determinants.