PYQ Core Concepts

Cheatsheet Content

### Chapter 1: Relations and Functions #### Important Equations & Concepts * **Relation $(R)$:** A subset of $A \times B$. * **Domain:** Set of all first elements of the ordered pairs in R. * **Range:** Set of all second elements of the ordered pairs in R. * **Co-domain:** The entire set B. * **Types of Relations:** * **Reflexive:** $(a, a) \in R$ for every $a \in A$. * **Symmetric:** If $(a, b) \in R$, then $(b, a) \in R$. * **Transitive:** If $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$. * **Equivalence Relation:** A relation that is reflexive, symmetric, and transitive. * **Function $(f: A \to B)$:** A special type of relation where every element of set A has one and only one image in set B. * **Domain:** Set A. * **Co-domain:** Set B. * **Range:** Set of all images of elements of A under f. * **Types of Functions:** * **One-one (Injective):** Different elements of A have different images in B. * If $f(x_1) = f(x_2) \implies x_1 = x_2$. * **Onto (Surjective):** Every element in B is the image of some element in A. * For every $y \in B$, there exists $x \in A$ such that $f(x) = y$. (Range = Co-domain) * **One-one and Onto (Bijective):** A function that is both injective and surjective. * **Composition of Functions $(f \circ g)$:** If $f: A \to B$ and $g: B \to C$, then $g \circ f: A \to C$ is defined as $(g \circ f)(x) = g(f(x))$. * $(f \circ g)(x) = f(g(x))$. (Note: Order matters) * **Invertible Function:** A function $f: X \to Y$ is invertible if there exists a function $g: Y \to X$ such that $g \circ f = I_X$ and $f \circ g = I_Y$. * A function is invertible if and only if it is bijective. * The inverse function is denoted by $f^{-1}$. * **Binary Operations $(*)$:** A binary operation on a set A is a function $*: A \times A \to A$. * **Commutative:** $a * b = b * a$ for all $a, b \in A$. * **Associative:** $(a * b) * c = a * (b * c)$ for all $a, b, c \in A$. * **Identity Element $(e)$:** $a * e = e * a = a$ for all $a \in A$. * **Inverse Element $(a^{-1})$:** For each $a \in A$, there exists $a^{-1} \in A$ such that $a * a^{-1} = a^{-1} * a = e$. * **Continuity of a Function:** A function $f(x)$ is continuous at a point $x=c$ if: 1. $f(c)$ is defined. 2. $\lim_{x \to c^-} f(x)$ exists. 3. $\lim_{x \to c^+} f(x)$ exists. 4. $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$. * For piecewise functions, check continuity at the 'joining' points. #### Sure Questions ##### Question 1 1. **(a)** A function f defined on the set of natural numbers N as $f(n) = 2n$. Then f is: (A) one-one and onto (B) one-one but not onto (C) not one-one and not onto (D) onto but not one-one **(1 Mark)** **(b)** If $R = \{(1,4), (2,5), (3,6)\}$ is a relation from $A = \{1,2,3\}$ to $B = \{4,5,6,7\}$. Check whether R is a bijection or not. Justify your answer. **(2 Marks) (March 2025 Model)** **Answer:** **(a) (B) one-one but not onto** **Steps:** * **For one-one:** Let $f(n_1) = f(n_2)$. Then $2n_1 = 2n_2 \implies n_1 = n_2$. So, f is one-one. * **For onto:** Let $y = 2n$. If $y=3$ (an odd natural number), then $n = 3/2$, which is not a natural number. So, f is not onto. **(b) R is one-one but not onto.** **Justification:** * **One-one:** Each element in A has a unique image in B. $f(1)=4$, $f(2)=5$, $f(3)=6$. * **Onto:** The element $7 \in B$ does not have a pre-image in A. Therefore, the function is not onto. Since it is not onto, it cannot be a bijection. ##### Question 2 1. **(a)** What is the minimum number of ordered pairs to form a reflexive relation on $A = \{1,2,3\}$? **(1 Mark)** **(b)** Show that the relation $R = \{(a, b): |a – b| \text{ is even} \}$ defined on $A = \{1,2,3,4,5\}$ is an equivalence relation. **(3 Marks) (March 2025 Model)** **Answer:** **(a) 3** **Explanation:** For a set A with n elements, a reflexive relation must contain at least n ordered pairs of the form (a,a). For $A=\{1,2,3\}$, these are $(1,1), (2,2), (3,3)$. **(b) Steps:** To prove R is an equivalence relation, we need to show it is reflexive, symmetric, and transitive. * **Reflexive:** For any $a \in A$, $|a - a| = 0$, which is an even number. Thus, $(a,a) \in R$ for all $a \in A$. Therefore, R is reflexive. * **Symmetric:** Let $(a,b) \in R$. This means $|a - b|$ is even. Since $|a - b| = |b - a|$, it follows that $|b - a|$ is also even. Thus, $(b,a) \in R$. Therefore, R is symmetric. * **Transitive:** Let $(a,b) \in R$ and $(b,c) \in R$. This means $|a - b|$ is even and $|b - c|$ is even. If $|a - b|$ is even, then $a - b = 2k_1$ for some integer $k_1$. If $|b - c|$ is even, then $b - c = 2k_2$ for some integer $k_2$. Adding these two equations: $(a - b) + (b - c) = 2k_1 + 2k_2$ $\implies a - c = 2(k_1 + k_2)$. This implies that $a - c$ is an even number, and so $|a - c|$ is even. Thus, $(a,c) \in R$. Therefore, R is transitive. Since R is reflexive, symmetric, and transitive, it is an equivalence relation. ##### Question 3 1. If $f: R \to R$ be defined by $f(x) = \begin{cases} kx + 1 & \text{if } x \le \pi \\ \cos x & \text{if } x > \pi \end{cases}$ **(a)** For what value of k, f is continuous? **(1 Mark)** **(b)** Find $f(2)$. **(2 Marks) (March 2025 Model)** **Answer:** **(a) Steps for continuity:** For f to be continuous at $x = \pi$, the left-hand limit, right-hand limit, and the function value at $x=\pi$ must be equal. * LHL at $x = \pi$: $\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (kx + 1) = k\pi + 1$ * RHL at $x = \pi$: $\lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} (\cos x) = \cos \pi = -1$ * Function value at $x = \pi$: $f(\pi) = k\pi + 1$ For continuity: $k\pi + 1 = -1$ $k\pi = -2$ $k = -\frac{2}{\pi}$ **(b) Steps for $f(2)$:** Since $2 #### Tricky Questions ##### Question 4 1. Let $f: R \to R$ be given by $f(x) = x^3$. **(i)** Show that the function f is injective. **(2 Marks)** **(ii)** Let A and B be two sets and $f: A \times B \to B \times A$ defined as $f(a, b) = (b, a)$. Show that f is one-one and onto. **(2 Marks) (Model 2025)** **Answer:** **(i) Steps for injectivity:** To show f is injective (one-one), assume $f(x_1) = f(x_2)$ for $x_1, x_2 \in R$. $x_1^3 = x_2^3$ Taking the cube root of both sides (which is a unique operation for real numbers): $x_1 = x_2$ Since $f(x_1) = f(x_2) \implies x_1 = x_2$, the function f is injective. **(ii) Steps for one-one and onto:** * **One-one:** Let $(a_1, b_1) \in A \times B$ and $(a_2, b_2) \in A \times B$. Assume $f(a_1, b_1) = f(a_2, b_2)$. By definition of f, this means $(b_1, a_1) = (b_2, a_2)$. For two ordered pairs to be equal, their corresponding components must be equal: $b_1 = b_2$ and $a_1 = a_2$. Therefore, $(a_1, b_1) = (a_2, b_2)$. Since $f(a_1, b_1) = f(a_2, b_2) \implies (a_1, b_1) = (a_2, b_2)$, the function f is one-one. * **Onto:** Let $(b, a)$ be an arbitrary element in the codomain $B \times A$. We need to find an element $(x, y) \in A \times B$ such that $f(x, y) = (b, a)$. By definition of f, $f(x, y) = (y, x)$. So, we need $(y, x) = (b, a)$. This implies $y = b$ and $x = a$. Since $a \in A$ and $b \in B$, the element $(a, b) \in A \times B$ is a pre-image of $(b, a)$. Thus, for every element in the codomain, there exists a pre-image in the domain. Therefore, the function f is onto. Since f is both one-one and onto, it is a bijection. ##### Question 5 1. Consider the relation R in the set $A = \{1,2,3 ... 13,14\}$ defined as $R = \{(x, y) : 3x – y = 0\}$. Examine whether the relation R is reflexive, symmetric and transitive. **(3 Marks)** **Answer:** **Steps:** First, list the elements of R: $3x - y = 0 \implies y = 3x$. * For $x=1, y=3 \implies (1,3) \in R$ * For $x=2, y=6 \implies (2,6) \in R$ * For $x=3, y=9 \implies (3,9) \in R$ * For $x=4, y=12 \implies (4,12) \in R$ * For $x=5, y=15$, but $15 \notin A$, so we stop. So, $R = \{(1,3), (2,6), (3,9), (4,12)\}$. * **Reflexive:** A relation R on set A is reflexive if $(a,a) \in R$ for all $a \in A$. Here, $(1,1) \notin R$ (since $3(1)-1 = 2 \ne 0$). Therefore, R is not reflexive. * **Symmetric:** A relation R on set A is symmetric if whenever $(a,b) \in R$, then $(b,a) \in R$. Here, $(1,3) \in R$. But $(3,1) \notin R$ (since $3(3)-1 = 8 \ne 0$). Therefore, R is not symmetric. * **Transitive:** A relation R on set A is transitive if whenever $(a,b) \in R$ and $(b,c) \in R$, then $(a,c) \in R$. Here, $(1,3) \in R$. For $(3,c)$, we have $(3,9) \in R$. Now, we must check if $(1,9) \in R$. $3(1)-9 = -6 \ne 0$. So $(1,9) \notin R$. Therefore, R is not transitive. --- ### Chapter 2: Inverse Trigonometric Functions #### Important Equations & Concepts * **Definition:** Inverse trigonometric functions are inverse functions of the trigonometric functions. They return an angle for a given trigonometric ratio. * **Principal Value Branches:** | Function | Domain | Range (Principal Value Branch) | | :------------------ | :----------- | :----------------------------- | | $\sin^{-1} x$ | $[-1, 1]$ | $[-\pi/2, \pi/2]$ | | $\cos^{-1} x$ | $[-1, 1]$ | $[0, \pi]$ | | $\tan^{-1} x$ | $(-\infty, \infty)$ | $(-\pi/2, \pi/2)$ | | $\csc^{-1} x$ | $(-\infty, -1] \cup [1, \infty)$ | $[-\pi/2, \pi/2] - \{0\}$ | | $\sec^{-1} x$ | $(-\infty, -1] \cup [1, \infty)$ | $[0, \pi] - \{\pi/2\}$ | | $\cot^{-1} x$ | $(-\infty, \infty)$ | $(0, \pi)$ | * **Properties of Inverse Trigonometric Functions:** 1. **Identity Properties:** * $\sin^{-1}(\sin x) = x$ for $x \in [-\pi/2, \pi/2]$ * $\cos^{-1}(\cos x) = x$ for $x \in [0, \pi]$ * $\tan^{-1}(\tan x) = x$ for $x \in (-\pi/2, \pi/2)$ * And similarly for their reciprocals. 2. **Reciprocal Identities:** * $\sin^{-1}(1/x) = \csc^{-1} x$, for $x \ge 1$ or $x \le -1$ * $\cos^{-1}(1/x) = \sec^{-1} x$, for $x \ge 1$ or $x \le -1$ * $\tan^{-1}(1/x) = \cot^{-1} x$, for $x > 0$ 3. **Negative Argument Identities:** * $\sin^{-1}(-x) = -\sin^{-1} x$, for $x \in [-1, 1]$ * $\tan^{-1}(-x) = -\tan^{-1} x$, for $x \in R$ * $\csc^{-1}(-x) = -\csc^{-1} x$, for $|x| \ge 1$ * $\cos^{-1}(-x) = \pi - \cos^{-1} x$, for $x \in [-1, 1]$ * $\sec^{-1}(-x) = \pi - \sec^{-1} x$, for $|x| \ge 1$ * $\cot^{-1}(-x) = \pi - \cot^{-1} x$, for $x \in R$ 4. **Complementary Angle Identities:** * $\sin^{-1} x + \cos^{-1} x = \pi/2$, for $x \in [-1, 1]$ * $\tan^{-1} x + \cot^{-1} x = \pi/2$, for $x \in R$ * $\csc^{-1} x + \sec^{-1} x = \pi/2$, for $|x| \ge 1$ 5. **Addition/Subtraction Formulas:** * $\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, if $xy -1$ * $2\tan^{-1} x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$, if $|x| #### Sure Questions ##### Question 1 1. **(a)** The graph below depicts: (A) $y = \sin^{-1} x$ (B) $y = \cos^{-1} x$ (C) $y = \csc^{-1} x$ (D) $y = \cot^{-1} x$ **(1 Mark)** **(b)** Write the simplest form of $\tan^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} \right)$; $x \ne 0$. **(2 Marks) (March 2025 Model)** **Answer:** **(a) (A) $y = \sin^{-1} x$** **Explanation:** The graph shows the domain $[-1,1]$ and range $[-\pi/2, \pi/2]$, which is characteristic of the principal value branch of $\sin^{-1} x$. **(b) Steps for simplest form:** Let $x = \tan \theta$. Then $\theta = \tan^{-1} x$. Substitute $x = \tan \theta$ into the expression: $\tan^{-1} \left( \frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta} \right)$ Using the identity $1+\tan^2 \theta = \sec^2 \theta$: $= \tan^{-1} \left( \frac{\sqrt{\sec^2 \theta}-1}{\tan \theta} \right)$ Since the range of $\tan^{-1} x$ is $(-\pi/2, \pi/2)$, we can assume $\theta$ is in this range, so $\sec \theta > 0$. $= \tan^{-1} \left( \frac{\sec \theta-1}{\tan \theta} \right)$ Convert to $\sin \theta$ and $\cos \theta$: $= \tan^{-1} \left( \frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}} \right)$ $= \tan^{-1} \left( \frac{1-\cos \theta}{\sin \theta} \right)$ Using half-angle identities: $1-\cos \theta = 2\sin^2(\theta/2)$ and $\sin \theta = 2\sin(\theta/2)\cos(\theta/2)$: $= \tan^{-1} \left( \frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} \right)$ $= \tan^{-1} \left( \frac{\sin(\theta/2)}{\cos(\theta/2)} \right)$ $= \tan^{-1} (\tan(\theta/2))$ $= \theta/2$ Substitute back $\theta = \tan^{-1} x$: $= \frac{1}{2} \tan^{-1} x$ ##### Question 2 1. **(i)** The principal value of $\sin^{-1}\left(\frac{1}{2}\right)$ is: (A) $\frac{\pi}{4}$ (B) $\frac{\pi}{3}$ (C) $\pi$ (D) $\frac{\pi}{6}$ **(1 Mark)** **(ii)** Prove that $2\sin^{-1}(x) = \sin^{-1}(2x\sqrt{1-x^2})$, for $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$. **(2 Marks) (Feb 2025 Model)** **Answer:** **(i) (D) $\frac{\pi}{6}$** **Explanation:** We know that $\sin(\pi/6) = 1/2$, and $\pi/6$ lies in the principal value branch of $\sin^{-1} x$, which is $[-\pi/2, \pi/2]$. **(ii) Steps for proof:** Let $\sin^{-1} x = \theta$. Then $x = \sin \theta$. The given condition $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$ implies $-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}$. Consider the RHS: $\sin^{-1}(2x\sqrt{1-x^2})$ Substitute $x = \sin \theta$: $= \sin^{-1}(2\sin \theta \sqrt{1-\sin^2 \theta})$ Using the identity $1-\sin^2 \theta = \cos^2 \theta$: $= \sin^{-1}(2\sin \theta \sqrt{\cos^2 \theta})$ Since $-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}$, $\cos \theta \ge 0$, so $\sqrt{\cos^2 \theta} = \cos \theta$: $= \sin^{-1}(2\sin \theta \cos \theta)$ Using the double angle identity $2\sin \theta \cos \theta = \sin(2\theta)$: $= \sin^{-1}(\sin(2\theta))$ Given $-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}$, multiplying by 2 gives $-\frac{\pi}{2} \le 2\theta \le \frac{\pi}{2}$. In this interval, $\sin^{-1}(\sin y) = y$. So, $\sin^{-1}(\sin(2\theta)) = 2\theta$. Substitute back $\theta = \sin^{-1} x$: $= 2\sin^{-1} x$ Thus, LHS = RHS. ##### Question 3 1. **(i)** The principal value range of $f(x) = \tan^{-1} x$ is: (a) $[-\frac{\pi}{2}, \frac{\pi}{2}]$ (b) $[0, \pi]$ (c) $(-\frac{\pi}{2}, \frac{\pi}{2})$ (d) $(0, \pi)$ **(1 Mark)** **(ii)** For the principal value range, evaluate $\cot^{-1}(-1) + \csc^{-1}(-\sqrt{2}) + \sec^{-1}(2)$. **(2 Marks) (Dec 2024 Model)** **Answer:** **(i) (c) $(-\frac{\pi}{2}, \frac{\pi}{2})$** **Explanation:** The principal value branch of $\tan^{-1} x$ is the open interval $(-\pi/2, \pi/2)$. **(ii) Steps for evaluation:** Let $A = \cot^{-1}(-1)$. We know $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$. So, $A = \pi - \cot^{-1}(1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$. Let $B = \csc^{-1}(-\sqrt{2})$. We know $\csc^{-1}(-x) = -\csc^{-1}(x)$. So, $B = -\csc^{-1}(\sqrt{2})$. Since $\csc(\pi/4) = \sqrt{2}$, $B = -\frac{\pi}{4}$. Let $C = \sec^{-1}(2)$. Since $\sec(\pi/3) = 2$, $C = \frac{\pi}{3}$. Now, add the values: $A + B + C = \frac{3\pi}{4} - \frac{\pi}{4} + \frac{\pi}{3}$ $= \frac{2\pi}{4} + \frac{\pi}{3}$ $= \frac{\pi}{2} + \frac{\pi}{3}$ Find a common denominator: $= \frac{3\pi}{6} + \frac{2\pi}{6} = \frac{5\pi}{6}$ #### Tricky Questions ##### Question 4 1. If $\sin^{-1} x = y$, then which of the following is true? (a) $0 \le y \le \pi$ (b) $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$ (c) $0 --- ### Chapter 3: Matrices #### Important Equations & Concepts * **Matrix:** A rectangular array of numbers or functions. * Order of a matrix: $m \times n$ (m rows, n columns). * **Types of Matrices:** * **Row Matrix:** Has only one row. * **Column Matrix:** Has only one column. * **Square Matrix:** Number of rows = number of columns ($m=n$). * **Diagonal Matrix:** A square matrix where all non-diagonal elements are zero. * **Scalar Matrix:** A diagonal matrix where all diagonal elements are equal. * **Identity Matrix $(I)$:** A diagonal matrix where all diagonal elements are 1. ($I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, $I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$) * **Zero Matrix $(O)$:** All elements are zero. * **Matrix Operations:** * **Addition/Subtraction:** Only possible for matrices of the same order. Performed element-wise. * **Scalar Multiplication:** Multiply each element of the matrix by the scalar. $k[a_{ij}] = [ka_{ij}]$. * **Matrix Multiplication:** If $A$ is $m \times n$ and $B$ is $n \times p$, then $AB$ is $m \times p$. * $(AB)_{ij} = \sum_{k=1}^n A_{ik}B_{kj}$. * Matrix multiplication is generally NOT commutative ($AB \ne BA$). * Matrix multiplication IS associative $(AB)C = A(BC)$. * Distributive property: $A(B+C) = AB+AC$. * **Transpose of a Matrix $(A^T)$:** Obtained by interchanging rows and columns. * $(A^T)_{ij} = A_{ji}$. * Properties: $(A^T)^T = A$; $(A+B)^T = A^T+B^T$; $(kA)^T = kA^T$; $(AB)^T = B^T A^T$. * **Special Matrices based on Transpose:** * **Symmetric Matrix:** $A^T = A$. (Elements $a_{ij} = a_{ji}$) * **Skew-Symmetric Matrix:** $A^T = -A$. (Elements $a_{ij} = -a_{ji}$, diagonal elements are 0) * Any square matrix A can be expressed as the sum of a symmetric and a skew-symmetric matrix: $A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$. * **Elementary Operations (Row/Column Operations):** 1. Interchange of any two rows (or columns): $R_i \leftrightarrow R_j$ or $C_i \leftrightarrow C_j$. 2. Multiplication of elements of any row (or column) by a non-zero scalar: $R_i \to kR_i$ or $C_i \to kC_i$. 3. Addition to the elements of any row (or column) the corresponding elements of another row (or column) multiplied by a non-zero scalar: $R_i \to R_i + kR_j$ or $C_i \to C_i + kC_j$. * **Invertible Matrix:** A square matrix A is invertible if there exists a square matrix B of the same order such that $AB = BA = I$. * B is called the inverse of A, denoted as $A^{-1}$. * $(AB)^{-1} = B^{-1}A^{-1}$. * $(A^T)^{-1} = (A^{-1})^T$. * A matrix is invertible if and only if its determinant is non-zero ($\det(A) \ne 0$). Such a matrix is called **non-singular**. * If $\det(A) = 0$, the matrix is **singular** and its inverse does not exist. #### Sure Questions ##### Question 1 1. **(i)** If A and B are symmetric matrices of the same order, then AB - BA is: (a) Symmetric matrix (b) Skew symmetric matrix (c) Identity matrix (d) Zero matrix **(1 Mark)** **(ii)** For what value of x, $\begin{bmatrix} 1 & 2 & 0 \\ 1 & x & 2 \\ 1 & 1 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$? **(2 Marks) (Dec 2025 Model)** **Answer:** **(i) (b) Skew symmetric matrix** **Steps:** Given A and B are symmetric matrices, so $A^T = A$ and $B^T = B$. Let $C = AB - BA$. We need to find $C^T$: $C^T = (AB - BA)^T$ Using the property $(X-Y)^T = X^T - Y^T$: $C^T = (AB)^T - (BA)^T$ Using the property $(XY)^T = Y^T X^T$: $C^T = B^T A^T - A^T B^T$ Substitute $A^T = A$ and $B^T = B$: $C^T = BA - AB$ $C^T = -(AB - BA)$ $C^T = -C$ Since $C^T = -C$, AB - BA is a skew symmetric matrix. **(ii) Steps for finding x:** Perform the matrix multiplication: $\begin{bmatrix} 1 & 2 & 0 \\ 1 & x & 2 \\ 1 & 1 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} (1)(0) + (2)(1) + (0)(2) \\ (1)(0) + (x)(1) + (2)(2) \\ (1)(0) + (1)(1) + (2)(2) \end{bmatrix} = \begin{bmatrix} 0+2+0 \\ 0+x+4 \\ 0+1+4 \end{bmatrix} = \begin{bmatrix} 2 \\ x+4 \\ 5 \end{bmatrix}$ We are given that this product is equal to $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$. So, $\begin{bmatrix} 2 \\ x+4 \\ 5 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$. This implies: $2 = 0$ (which is a contradiction) $x+4 = 0 \implies x = -4$ $5 = 0$ (another contradiction) **Conclusion:** If the question strictly means the entire resulting vector must be zero, then no such $x$ exists. However, in exam settings, such questions often imply finding the value of $x$ that makes the relevant component zero. If we assume the question implies finding $x$ such that the second component is zero, then $x=-4$. Given the context of "sure questions", it's likely testing the multiplication and solving for $x$ from the $x+4=0$ equation. **Assuming the intent was to find x for $x+4=0$, then $x=-4$.** ##### Question 2 1. Express the matrix $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix. **(2 Marks) (March 2025 Model)** **Answer:** **Steps:** Let A be written as $A = P + Q$, where P is a symmetric matrix and Q is a skew-symmetric matrix. The formulas are: $P = \frac{1}{2}(A + A^T)$ and $Q = \frac{1}{2}(A - A^T)$. First, find the transpose of A, $A^T$: $A^T = \begin{bmatrix} 3 & -1 \\ 1 & 2 \end{bmatrix}$ * **Calculate P (Symmetric part):** $A + A^T = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} + \begin{bmatrix} 3 & -1 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 3+3 & 1-1 \\ -1+1 & 2+2 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 4 \end{bmatrix}$ $P = \frac{1}{2} \begin{bmatrix} 6 & 0 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}$ To verify P is symmetric, check $P^T = P$: $\begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}^T = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}$. So, P is symmetric. * **Calculate Q (Skew-symmetric part):** $A - A^T = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & -1 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 3-3 & 1-(-1) \\ -1-1 & 2-2 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$ $Q = \frac{1}{2} \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ To verify Q is skew-symmetric, check $Q^T = -Q$: $\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}^T = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = - \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$. So, Q is skew-symmetric. * **Sum of P and Q:** $P + Q = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 3+0 & 0+1 \\ 0-1 & 2+0 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ This is equal to matrix A. Therefore, $A = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$. ##### Question 3 1. If $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, find k so that $A^2 = KA - 2I$. **(3 Marks) (2022 Model)** **Answer:** **Steps:** First, calculate $A^2$: $A^2 = A \cdot A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} (3)(3)+(-2)(4) & (3)(-2)+(-2)(-2) \\ (4)(3)+(-2)(4) & (4)(-2)+(-2)(-2) \end{bmatrix}$ $A^2 = \begin{bmatrix} 9-8 & -6+4 \\ 12-8 & -8+4 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}$ Next, calculate $KA$: $KA = k \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix}$ Next, calculate $2I$: $2I = 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ Now, substitute these into the given equation $A^2 = KA - 2I$: $\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ $\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k-2 & -2k-0 \\ 4k-0 & -2k-2 \end{bmatrix}$ $\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix}$ Equate the corresponding elements: * From the top-left element: $1 = 3k - 2 \implies 3k = 3 \implies k = 1$. * From the top-right element: $-2 = -2k \implies k = 1$. * From the bottom-left element: $4 = 4k \implies k = 1$. * From the bottom-right element: $-4 = -2k - 2 \implies -2k = -2 \implies k = 1$. All equations yield $k=1$. Therefore, the value of k is 1. #### Tricky Questions ##### Question 4 1. If $A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$, then show that $A^2 - 5A + 7I = 0$. **(2 Marks) (March 2013 Model)** **Answer:** **Steps:** First, calculate $A^2$: $A^2 = A \cdot A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} (1)(1)+(2)(2) & (1)(2)+(2)(1) \\ (2)(1)+(1)(2) & (2)(2)+(1)(1) \end{bmatrix}$ $A^2 = \begin{bmatrix} 1+4 & 2+2 \\ 2+2 & 4+1 \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}$ Next, calculate $5A$: $5A = 5 \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 10 \\ 10 & 5 \end{bmatrix}$ Next, calculate $7I$: $7I = 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$ Now, substitute these into the expression $A^2 - 5A + 7I$: $A^2 - 5A + 7I = \begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix} - \begin{bmatrix} 5 & 10 \\ 10 & 5 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$ $= \begin{bmatrix} 5-5+7 & 4-10+0 \\ 4-10+0 & 5-5+7 \end{bmatrix}$ $= \begin{bmatrix} 7 & -6 \\ -6 & 7 \end{bmatrix}$ **Conclusion:** The result $\begin{bmatrix} 7 & -6 \\ -6 & 7 \end{bmatrix}$ is not the zero matrix. Therefore, the statement $A^2 - 5A + 7I = 0$ is **false** for the given matrix A. This suggests a potential typo in the question's coefficients, as such "show that" questions typically lead to a true statement. A common related identity is the Cayley-Hamilton theorem, which for this matrix A, would yield $A^2 - 2A - 3I = 0$. ##### Question 5 1. If $A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$, then prove that $A^2 = \begin{bmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{bmatrix}$. **(2 Marks)** **Answer:** **Steps:** Calculate $A^2 = A \cdot A$: $A^2 = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$ Perform matrix multiplication: $= \begin{bmatrix} (\cos \theta)(\cos \theta) + (\sin \theta)(-\sin \theta) & (\cos \theta)(\sin \theta) + (\sin \theta)(\cos \theta) \\ (-\sin \theta)(\cos \theta) + (\cos \theta)(-\sin \theta) & (-\sin \theta)(\sin \theta) + (\cos \theta)(\cos \theta) \end{bmatrix}$ $= \begin{bmatrix} \cos^2 \theta - \sin^2 \theta & \sin \theta \cos \theta + \sin \theta \cos \theta \\ -\sin \theta \cos \theta - \sin \theta \cos \theta & -\sin^2 \theta + \cos^2 \theta \end{bmatrix}$ Simplify the terms using trigonometric identities: * $\cos^2 \theta - \sin^2 \theta = \cos 2\theta$ * $\sin \theta \cos \theta + \sin \theta \cos \theta = 2\sin \theta \cos \theta = \sin 2\theta$ * $-\sin \theta \cos \theta - \sin \theta \cos \theta = -2\sin \theta \cos \theta = -\sin 2\theta$ * $-\sin^2 \theta + \cos^2 \theta = \cos^2 \theta - \sin^2 \theta = \cos 2\theta$ Substitute these identities back into the matrix: $A^2 = \begin{bmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{bmatrix}$ This matches the required matrix. --- ### Chapter 4: Determinants #### Important Equations & Concepts * **Determinant:** A scalar value that can be computed from the elements of a square matrix. * For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\det(A) = |A| = ad - bc$. * For $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$, $\det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$. * Determinant can be expanded along any row or column. * **Properties of Determinants:** 1. If rows/columns are interchanged, determinant changes sign. 2. If any two rows/columns are identical, determinant is 0. 3. If elements of a row/column are multiplied by a scalar $k$, determinant is multiplied by $k$. 4. $\det(A^T) = \det(A)$. 5. $\det(AB) = \det(A)\det(B)$. 6. If elements of any row/column are expressed as sum of two terms, determinant can be expressed as sum of two determinants. 7. If a scalar multiple of one row/column is added to another row/column, determinant remains unchanged. * **Minors $(M_{ij})$:** The determinant of the submatrix obtained by deleting the $i$-th row and $j$-th column. * **Cofactors $(C_{ij})$:** $C_{ij} = (-1)^{i+j} M_{ij}$. * **Adjoint of a Matrix $(\text{adj}(A))$:** The transpose of the cofactor matrix. $\text{adj}(A) = [C_{ij}]^T$. * For $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$. * **Inverse of a Matrix $(A^{-1})$:** $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$, provided $\det(A) \ne 0$. * If $\det(A) = 0$, $A^{-1}$ does not exist (A is a singular matrix). * **System of Linear Equations (Matrix Method):** For a system $AX = B$: * If $\det(A) \ne 0$ (non-singular), then $X = A^{-1}B$ (unique solution, consistent). * If $\det(A) = 0$ (singular): * If $(\text{adj}(A))B \ne O$ (zero matrix), then no solution (inconsistent). * If $(\text{adj}(A))B = O$, then infinitely many solutions (consistent). * **Area of a Triangle:** The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is: Area $= \frac{1}{2} \left| \det \begin{bmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{bmatrix} \right|$. * If the area is 0, the three points are collinear. #### Sure Questions ##### Question 1 1. Consider the system of equations: $2x + 3y + 3z = 5$ $x - 2y + z = -4$ $3x - y - 2z = 3$ **(i)** Write its matrix form. **(1 Mark)** **(ii)** Prove that the system is consistent. **(1 Mark)** **(iii)** Solve the system using matrix method. **(4 Marks) (March 2025 Model)** **Answer:** **(i) Matrix form:** The system of equations can be written as $AX = B$, where: $A = \begin{bmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix}$. **(ii) Proving consistency:** A system of linear equations is consistent if $\det(A) \ne 0$. Calculate the determinant of A: $\det(A) = 2 \begin{vmatrix} -2 & 1 \\ -1 & -2 \end{vmatrix} - 3 \begin{vmatrix} 1 & 1 \\ 3 & -2 \end{vmatrix} + 3 \begin{vmatrix} 1 & -2 \\ 3 & -1 \end{vmatrix}$ $= 2((-2)(-2) - (1)(-1)) - 3((1)(-2) - (1)(3)) + 3((1)(-1) - (-2)(3))$ $= 2(4 - (-1)) - 3(-2 - 3) + 3(-1 - (-6))$ $= 2(5) - 3(-5) + 3(5)$ $= 10 + 15 + 15 = 40$ Since $\det(A) = 40 \ne 0$, the matrix A is non-singular. Therefore, the system of equations is consistent and has a unique solution. **(iii) Solving using matrix method:** To solve $AX=B$, we find $X = A^{-1}B$. First, find $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$. We already have $\det(A) = 40$. Now, find the cofactors of A: * $C_{11} = \begin{vmatrix} -2 & 1 \\ -1 & -2 \end{vmatrix} = 4 - (-1) = 5$ * $C_{12} = -\begin{vmatrix} 1 & 1 \\ 3 & -2 \end{vmatrix} = -(-2 - 3) = 5$ * $C_{13} = \begin{vmatrix} 1 & -2 \\ 3 & -1 \end{vmatrix} = -1 - (-6) = 5$ * $C_{21} = -\begin{vmatrix} 3 & 3 \\ -1 & -2 \end{vmatrix} = -(-6 - (-3)) = -(-3) = 3$ * $C_{22} = \begin{vmatrix} 2 & 3 \\ 3 & -2 \end{vmatrix} = -4 - 9 = -13$ * $C_{23} = -\begin{vmatrix} 2 & 3 \\ 3 & -1 \end{vmatrix} = -(-2 - 9) = 11$ * $C_{31} = \begin{vmatrix} 3 & 3 \\ -2 & 1 \end{vmatrix} = 3 - (-6) = 9$ * $C_{32} = -\begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} = -(2 - 3) = 1$ * $C_{33} = \begin{vmatrix} 2 & 3 \\ 1 & -2 \end{vmatrix} = -4 - 3 = -7$ The cofactor matrix $C = \begin{bmatrix} 5 & 5 & 5 \\ 3 & -13 & 11 \\ 9 & 1 & -7 \end{bmatrix}$. The adjoint of A, $\text{adj}(A) = C^T$: $\text{adj}(A) = \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}$. Now, find $A^{-1}$: $A^{-1} = \frac{1}{40} \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}$. Finally, calculate $X = A^{-1}B$: $X = \frac{1}{40} \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix} \begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix}$ $X = \frac{1}{40} \begin{bmatrix} (5)(5)+(3)(-4)+(9)(3) \\ (5)(5)+(-13)(-4)+(1)(3) \\ (5)(5)+(11)(-4)+(-7)(3) \end{bmatrix}$ $X = \frac{1}{40} \begin{bmatrix} 25 - 12 + 27 \\ 25 + 52 + 3 \\ 25 - 44 - 21 \end{bmatrix}$ $X = \frac{1}{40} \begin{bmatrix} 40 \\ 80 \\ -40 \end{bmatrix}$ $X = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$ So, $x=1, y=2, z=-1$. ##### Question 2 1. Consider the following system of equation: $3x - 2y + 3z = 8$ $2x + y - z = 1$ $4x - 3y + 2z = 4$ **(i)** Write the matrix form $AX = B$. **(1 Mark)** **(ii)** Find $A^{-1}$. **(2 Marks)** **(iii)** Solve the system by matrix method. **(3 Marks) (Feb 2025 Model)** **Answer:** **(i) Matrix form:** $A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, $B = \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix}$. **(ii) Finding $A^{-1}$:** First, find $\det(A)$: $\det(A) = 3 \begin{vmatrix} 1 & -1 \\ -3 & 2 \end{vmatrix} - (-2) \begin{vmatrix} 2 & -1 \\ 4 & 2 \end{vmatrix} + 3 \begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix}$ $= 3(2 - 3) + 2(4 - (-4)) + 3(-6 - 4)$ $= 3(-1) + 2(8) + 3(-10)$ $= -3 + 16 - 30 = 13 - 30 = -17$ Since $\det(A) = -17 \ne 0$, $A^{-1}$ exists. Cofactors of A: * $C_{11} = \begin{vmatrix} 1 & -1 \\ -3 & 2 \end{vmatrix} = 2 - 3 = -1$ * $C_{12} = -\begin{vmatrix} 2 & -1 \\ 4 & 2 \end{vmatrix} = -(4 - (-4)) = -8$ * $C_{13} = \begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix} = -6 - 4 = -10$ * $C_{21} = -\begin{vmatrix} -2 & 3 \\ -3 & 2 \end{vmatrix} = -(-4 - (-9)) = -(5) = -5$ * $C_{22} = \begin{vmatrix} 3 & 3 \\ 4 & 2 \end{vmatrix} = 6 - 12 = -6$ * $C_{23} = -\begin{vmatrix} 3 & -2 \\ 4 & -3 \end{vmatrix} = -(-9 - (-8)) = -(-1) = 1$ * $C_{31} = \begin{vmatrix} -2 & 3 \\ 1 & -1 \end{vmatrix} = 2 - 3 = -1$ * $C_{32} = -\begin{vmatrix} 3 & 3 \\ 2 & -1 \end{vmatrix} = -(-3 - 6) = -(-9) = 9$ * $C_{33} = \begin{vmatrix} 3 & -2 \\ 2 & 1 \end{vmatrix} = 3 - (-4) = 7$ Cofactor matrix $C = \begin{bmatrix} -1 & -8 & -10 \\ -5 & -6 & 1 \\ -1 & 9 & 7 \end{bmatrix}$. $\text{adj}(A) = C^T = \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix}$. $A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = -\frac{1}{17} \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix}$. **(iii) Solving using matrix method:** $X = A^{-1}B = -\frac{1}{17} \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix} \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix}$ $X = -\frac{1}{17} \begin{bmatrix} (-1)(8)+(-5)(1)+(-1)(4) \\ (-8)(8)+(-6)(1)+(9)(4) \\ (-10)(8)+(1)(1)+(7)(4) \end{bmatrix}$ $X = -\frac{1}{17} \begin{bmatrix} -8 - 5 - 4 \\ -64 - 6 + 36 \\ -80 + 1 + 28 \end{bmatrix}$ $X = -\frac{1}{17} \begin{bmatrix} -17 \\ -34 \\ -51 \end{bmatrix}$ $X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ So, $x=1, y=2, z=3$. ##### Question 3 1. If $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & -1 \\ 1 & 1 & -1 \end{bmatrix}$. **(i)** Find the inverse of A. **(3 Marks)** **(ii)** Express the system of equations: $x + y + z = 1$ $x - y - z = 1$ $x + y - z = -1$ in $AX = B$ form. **(1 Mark)** **(iii)** Find the solution of the above system of equations. **(2 Marks) (IMP 2024 Model)** **Answer:** **(i) Finding $A^{-1}$:** First, find $\det(A)$: $\det(A) = 1 \begin{vmatrix} -1 & -1 \\ 1 & -1 \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \\ 1 & -1 \end{vmatrix} + 1 \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix}$ $= 1(1 - (-1)) - 1(-1 - (-1)) + 1(1 - (-1))$ $= 1(2) - 1(0) + 1(2)$ $= 2 - 0 + 2 = 4$ Since $\det(A) = 4 \ne 0$, $A^{-1}$ exists. Cofactors of A: * $C_{11} = \begin{vmatrix} -1 & -1 \\ 1 & -1 \end{vmatrix} = 1 - (-1) = 2$ * $C_{12} = -\begin{vmatrix} 1 & -1 \\ 1 & -1 \end{vmatrix} = -(-1 - (-1)) = 0$ * $C_{13} = \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} = 1 - (-1) = 2$ * $C_{21} = -\begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -(-1 - 1) = 2$ * $C_{22} = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -1 - 1 = -2$ * $C_{23} = -\begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = -(1 - 1) = 0$ * $C_{31} = \begin{vmatrix} 1 & 1 \\ -1 & -1 \end{vmatrix} = -1 - (-1) = 0$ * $C_{32} = -\begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -(-1 - 1) = 2$ * $C_{33} = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -1 - 1 = -2$ Cofactor matrix $C = \begin{bmatrix} 2 & 0 & 2 \\ 2 & -2 & 0 \\ 0 & 2 & -2 \end{bmatrix}$. $\text{adj}(A) = C^T = \begin{bmatrix} 2 & 2 & 0 \\ 0 & -2 & 2 \\ 2 & 0 & -2 \end{bmatrix}$. $A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{4} \begin{bmatrix} 2 & 2 & 0 \\ 0 & -2 & 2 \\ 2 & 0 & -2 \end{bmatrix} = \begin{bmatrix} 1/2 & 1/2 & 0 \\ 0 & -1/2 & 1/2 \\ 1/2 & 0 & -1/2 \end{bmatrix}$. **(ii) Matrix form $AX = B$:** $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & -1 \\ 1 & 1 & -1 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, $B = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}$. **(iii) Solving the system:** $X = A^{-1}B = \frac{1}{4} \begin{bmatrix} 2 & 2 & 0 \\ 0 & -2 & 2 \\ 2 & 0 & -2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}$ $X = \frac{1}{4} \begin{bmatrix} (2)(1)+(2)(1)+(0)(-1) \\ (0)(1)+(-2)(1)+(2)(-1) \\ (2)(1)+(0)(1)+(-2)(-1) \end{bmatrix}$ $X = \frac{1}{4} \begin{bmatrix} 2+2+0 \\ 0-2-2 \\ 2+0+2 \end{bmatrix}$ $X = \frac{1}{4} \begin{bmatrix} 4 \\ -4 \\ 4 \end{bmatrix}$ $X = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$ So, $x=1, y=-1, z=1$. #### Tricky Questions ##### Question 4 1. If $A = \begin{bmatrix} 2 & 3 \\ 5 & 4 \end{bmatrix}$, find $A^2 - 7A - 2I = 0$. **(3 Marks) (March 2014 Model)** **Answer:** **Steps:** First, calculate $A^2$: $A^2 = A \cdot A = \begin{bmatrix} 2 & 3 \\ 5 & 4 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 5 & 4 \end{bmatrix} = \begin{bmatrix} (2)(2)+(3)(5) & (2)(3)+(3)(4) \\ (5)(2)+(4)(5) & (5)(3)+(4)(4) \end{bmatrix}$ $A^2 = \begin{bmatrix} 4+15 & 6+12 \\ 10+20 & 15+16 \end{bmatrix} = \begin{bmatrix} 19 & 18 \\ 30 & 31 \end{bmatrix}$ Next, calculate $7A$: $7A = 7 \begin{bmatrix} 2 & 3 \\ 5 & 4 \end{bmatrix} = \begin{bmatrix} 14 & 21 \\ 35 & 28 \end{bmatrix}$ Next, calculate $2I$: $2I = 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ Now, substitute these into the expression $A^2 - 7A - 2I$: $A^2 - 7A - 2I = \begin{bmatrix} 19 & 18 \\ 30 & 31 \end{bmatrix} - \begin{bmatrix} 14 & 21 \\ 35 & 28 \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ $= \begin{bmatrix} 19-14-2 & 18-21-0 \\ 30-35-0 & 31-28-2 \end{bmatrix}$ $= \begin{bmatrix} 3 & -3 \\ -5 & 1 \end{bmatrix}$ **Conclusion:** The result $\begin{bmatrix} 3 & -3 \\ -5 & 1 \end{bmatrix}$ is not the zero matrix. Therefore, the statement $A^2 - 7A - 2I = 0$ is **false** for the given matrix A. This indicates a potential typo in the question's coefficients. (As discussed in Chapter 3, Question 4, the Cayley-Hamilton theorem for this matrix gives $A^2 - 6A - 7I = 0$.) ##### Question 5 1. If $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$, then show that $A^2 - 4A + I = 0$. **(3 Marks) (March 2013 Model)** **Answer:** **Steps:** First, calculate $A^2$: $A^2 = A \cdot A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} (2)(2)+(3)(1) & (2)(3)+(3)(2) \\ (1)(2)+(2)(1) & (1)(3)+(2)(2) \end{bmatrix}$ $A^2 = \begin{bmatrix} 4+3 & 6+6 \\ 2+2 & 3+4 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix}$ Next, calculate $4A$: $4A = 4 \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 12 \\ 4 & 8 \end{bmatrix}$ Next, calculate $I$: $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ Now, substitute these into the expression $A^2 - 4A + I$: $A^2 - 4A + I = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ 4 & 8 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ $= \begin{bmatrix} 7-8+1 & 12-12+0 \\ 4-4+0 & 7-8+1 \end{bmatrix}$ $= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ This is the zero matrix. Therefore, $A^2 - 4A + I = 0$ is proven. --- ### Chapter 5: Continuity and Differentiability #### Important Equations & Concepts * **Continuity at a point $c$:** $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$. If a function is continuous at every point in its domain, it is a continuous function. * **Differentiability at a point $c$:** A function $f(x)$ is differentiable at $x=c$ if the derivative $f'(c)$ exists. $f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$. This limit exists if and only if the left-hand derivative (LHD) and right-hand derivative (RHD) are equal. * LHD = $\lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h}$ * RHD = $\lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h}$ * If a function is differentiable at $x=c$, then it is continuous at $x=c$. The converse is not always true (e.g., $|x|$ at $x=0$). * **Derivatives of Standard Functions:** * $\frac{d}{dx}(c) = 0$ * $\frac{d}{dx}(x^n) = nx^{n-1}$ * $\frac{d}{dx}(e^x) = e^x$ * $\frac{d}{dx}(a^x) = a^x \ln a$ * $\frac{d}{dx}(\ln x) = \frac{1}{x}$ * $\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$ * $\frac{d}{dx}(\sin x) = \cos x$ * $\frac{d}{dx}(\cos x) = -\sin x$ * $\frac{d}{dx}(\tan x) = \sec^2 x$ * $\frac{d}{dx}(\cot x) = -\csc^2 x$ * $\frac{d}{dx}(\sec x) = \sec x \tan x$ * $\frac{d}{dx}(\csc x) = -\csc x \cot x$ * $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$ * $\frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1-x^2}}$ * $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$ * $\frac{d}{dx}(\cot^{-1} x) = -\frac{1}{1+x^2}$ * $\frac{d}{dx}(\sec^{-1} x) = \frac{1}{|x|\sqrt{x^2-1}}$ * $\frac{d}{dx}(\csc^{-1} x) = -\frac{1}{|x|\sqrt{x^2-1}}$ * **Rules of Differentiation:** * **Sum/Difference Rule:** $\frac{d}{dx}(u \pm v) = \frac{du}{dx} \pm \frac{dv}{dx}$ * **Product Rule:** $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$ * **Quotient Rule:** $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$ * **Chain Rule:** $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ (If $y=f(u)$ and $u=g(x)$) * **Logarithmic Differentiation:** Used for functions of the form $f(x)^{g(x)}$ or complex products/quotients. 1. Take natural logarithm on both sides: $\ln y = g(x) \ln f(x)$. 2. Differentiate implicitly with respect to x. 3. Solve for $\frac{dy}{dx}$. * **Implicit Differentiation:** Used when y cannot be easily expressed as an explicit function of x. 1. Differentiate both sides of the equation with respect to x. 2. Remember that y is a function of x, so use the chain rule for terms involving y (e.g., $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$). 3. Collect all terms with $\frac{dy}{dx}$ on one side and the rest on the other. 4. Factor out $\frac{dy}{dx}$ and solve. * **Parametric Differentiation:** If $x = f(t)$ and $y = g(t)$, then: * $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, provided $\frac{dx}{dt} \ne 0$. * $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$. #### Sure Questions ##### Question 1 1. Find $\frac{d}{dx} (2^x) = $ **(1 Mark)** **(ii)** If $\sin x + \cos y = xy$, find $\frac{dy}{dx}$. **(2 Marks) (Model 2025)** **Answer:** **(i) Steps for differentiation:** Using the formula $\frac{d}{dx}(a^x) = a^x \ln a$: $\frac{d}{dx}(2^x) = 2^x \ln 2$ **(ii) Steps for implicit differentiation:** Given $\sin x + \cos y = xy$. Differentiate both sides with respect to x: $\frac{d}{dx}(\sin x) + \frac{d}{dx}(\cos y) = \frac{d}{dx}(xy)$ $\cos x + (-\sin y) \frac{dy}{dx} = (1)y + x \frac{dy}{dx}$ (using chain rule for $\cos y$ and product rule for $xy$) $\cos x - y = x \frac{dy}{dx} + \sin y \frac{dy}{dx}$ Factor out $\frac{dy}{dx}$ on the right side: $\cos x - y = (x + \sin y) \frac{dy}{dx}$ Isolate $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{\cos x - y}{x + \sin y}$ ##### Question 2 1. **(i)** Prove that the function $f(x) = x - |x|$ is continuous at $x = 0$ in R. **(2 Marks)** **(ii)** If $x = at^2$, $y = 2at$, then find $\frac{d^2y}{dx^2}$. **(2 Marks) (March 2025 Model)** **Answer:** **(i) Steps for proving continuity at x=0:** The function is $f(x) = x - |x|$. We need to check LHL, RHL, and $f(0)$. * Case 1: $x ##### Question 3 1. If $y = x^{\sin x}$, find $\frac{dy}{dx}$. **(2 Marks) (Model 2025)** **Answer:** **Steps for logarithmic differentiation:** Given $y = x^{\sin x}$. This is a function of the form $f(x)^{g(x)}$, so we use logarithmic differentiation. Take the natural logarithm of both sides: $\ln y = \ln (x^{\sin x})$ Using the logarithm property $\ln(a^b) = b \ln a$: $\ln y = \sin x \ln x$ Now, differentiate both sides with respect to x: $\frac{d}{dx}(\ln y) = \frac{d}{dx}(\sin x \ln x)$ Using the chain rule for $\ln y$ and the product rule for $\sin x \ln x$: $\frac{1}{y} \frac{dy}{dx} = (\cos x)(\ln x) + (\sin x)\left(\frac{1}{x}\right)$ $\frac{1}{y} \frac{dy}{dx} = \cos x \ln x + \frac{\sin x}{x}$ Isolate $\frac{dy}{dx}$: $\frac{dy}{dx} = y \left( \cos x \ln x + \frac{\sin x}{x} \right)$ Substitute back $y = x^{\sin x}$: $\frac{dy}{dx} = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right)$ #### Tricky Questions ##### Question 4 1. If $x = a(\cos \theta + \theta \sin \theta)$ and $y = a(\sin \theta - \theta \cos \theta)$, find $\frac{dy}{dx}$. **(3 Marks)** **Answer:** **Steps for parametric differentiation:** Given $x = a(\cos \theta + \theta \sin \theta)$ and $y = a(\sin \theta - \theta \cos \theta)$. First, find $\frac{dx}{d\theta}$: $\frac{dx}{d\theta} = a \frac{d}{d\theta}(\cos \theta + \theta \sin \theta)$ $= a \left( -\sin \theta + (1)\sin \theta + \theta(\cos \theta) \right)$ (using product rule for $\theta \sin \theta$) $= a \left( -\sin \theta + \sin \theta + \theta \cos \theta \right)$ $= a(\theta \cos \theta)$ Next, find $\frac{dy}{d\theta}$: $\frac{dy}{d\theta} = a \frac{d}{d\theta}(\sin \theta - \theta \cos \theta)$ $= a \left( \cos \theta - ((1)\cos \theta + \theta(-\sin \theta)) \right)$ (using product rule for $\theta \cos \theta$) $= a \left( \cos \theta - (\cos \theta - \theta \sin \theta) \right)$ $= a \left( \cos \theta - \cos \theta + \theta \sin \theta \right)$ $= a(\theta \sin \theta)$ Now, find $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$: $\frac{dy}{dx} = \frac{a(\theta \sin \theta)}{a(\theta \cos \theta)}$ Assuming $\theta \ne 0$ and $\cos \theta \ne 0$: $\frac{dy}{dx} = \frac{\sin \theta}{\cos \theta} = \tan \theta$. ##### Question 5 1. If $y = ( \sin^{-1} x )^2$, then show that $(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = 2$. **(2 Marks) (March 2024 Model)** **Answer:** **Steps:** Given $y = ( \sin^{-1} x )^2$. First, find $\frac{dy}{dx}$: $\frac{dy}{dx} = 2(\sin^{-1} x) \cdot \frac{d}{dx}(\sin^{-1} x)$ (using chain rule) $\frac{dy}{dx} = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}}$ Rearrange to simplify further differentiation: $\sqrt{1-x^2} \frac{dy}{dx} = 2\sin^{-1} x$ Square both sides: $(1-x^2) \left(\frac{dy}{dx}\right)^2 = (2\sin^{-1} x)^2 = 4(\sin^{-1} x)^2$ Since $y = (\sin^{-1} x)^2$, we have: $(1-x^2) \left(\frac{dy}{dx}\right)^2 = 4y$ Now, differentiate both sides with respect to x. Use product rule on the left side: $\frac{d}{dx}((1-x^2)) \left(\frac{dy}{dx}\right)^2 + (1-x^2) \frac{d}{dx}\left(\left(\frac{dy}{dx}\right)^2\right) = \frac{d}{dx}(4y)$ $(-2x) \left(\frac{dy}{dx}\right)^2 + (1-x^2) \left(2 \frac{dy}{dx} \frac{d^2y}{dx^2}\right) = 4 \frac{dy}{dx}$ Divide the entire equation by $2 \frac{dy}{dx}$ (assuming $\frac{dy}{dx} \ne 0$, which is true unless $x=0$): $-x \frac{dy}{dx} + (1-x^2) \frac{d^2y}{dx^2} = 2$ Rearrange the terms to match the required form: $(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = 2$. This completes the proof.