Maths 2 Unit Semester Exam Papers PDF

# Find the Rank of the Matrix ## Example 8 Find the rank of the Matrix, by reducing it to the normal form [JNTU (H) June 2010 (Set No. 3)] ### Solution : Let A = $$\begin{bmatrix} 2 & 1& 3 & 5 \\ 4 & 2 & 1 & 3 \\ 8 & 4& 7 & 13 \\ 8 & 4 & -3 & -1 \end{bmatrix}$$ $$\begin{bmatrix} 2 & 1& 3 & 5 \...

# Find the Rank of the Matrix ## Example 8 Find the rank of the Matrix, by reducing it to the normal form [JNTU (H) June 2010 (Set No. 3)] ### Solution : Let A = $$\begin{bmatrix} 2 & 1& 3 & 5 \\ 4 & 2 & 1 & 3 \\ 8 & 4& 7 & 13 \\ 8 & 4 & -3 & -1 \end{bmatrix}$$ $$\begin{bmatrix} 2 & 1& 3 & 5 \\ 0 & 0 & -5 & -7 \\ 0 & 0 & -5 & -7\\ 0 & 0 & -15 & -21\\ \end{bmatrix}$$ (Applying R₂ - 2R₁, R₃ - 4R₁, R₄ - 4R₁) $$\begin{bmatrix} 2 & 1& 3 & 5 \\ 0 & 0 & -5 & -7 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix}$$ (Applying R₄ - 3R₂, R₃ - R₂) $$\begin{bmatrix} 10 & 5 & 0 & 4 \\ 0 & 0 & -5 & -7\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ (Applying 5R₁ + 3R₂) $$\begin{bmatrix} 1 & 0 & 0 & 4 \\ 0 & 0 & 1 & -7\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ (Applying $C₁ \implies \frac{C₁}{10}, C₂ \implies \frac{C₂} {5}, C₃ \implies \frac{C₃} {5}$) $$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & -7\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ (Applying C₂ - C₁, C₄ - 4C₁) $$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ (Applying C₄ +7C₃) $$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ (Applying C₂ → C₃) This is of the form $$\begin{bmatrix} I_2 & O \\ O & O \end{bmatrix} $$ which is in normal form. Rank of A = 2 # Find the characteristic roots of the matrix ## Example 1 Find the characteristic roots of the matrix A = $$\begin{bmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix}$$ [JNTU 2008, (H) June 2009 (Set No. 4)] ### Solution: Given matrix is A = $$\begin{bmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix}$$ The characteristic equation of A is |A - λI| = 0 $ \begin{bmatrix} 2-λ & 2 & 1 \\ 1 & 3-λ & 1 \\ 1 & 2 & 2 - λ \end{bmatrix} $ = 0 i.e., $(2-λ)[(3-λ)(2-λ)-2]-2[2-λ-1]+1[2-3+λ]=0$ ⇒ $(2-λ)(λ² - 5λ + 4) - 2 (1 -λ) + (λ - 1) = 0$ ⇒ $(2-λ)(λ² - 5λ + 4) - (λ - 1) = 0$ ⇒ $2λ^2 − 10λ + 8 - λ^3 + 5λ^2 - 4λ -λ + 1 = 0$ ⇒ $-λ^3 + 7λ^2 − 15λ + 9 = 0$ ⇒ $λ^3 - 7λ^2 + 15λ - 9 = 0$ ⇒ $(λ-1)(λ^2 - 6λ + 9) = 0 $ ⇒ $(λ-1)(λ-3)^2 = 0$ :. λ = 1, 1, 3 Hence the characteristic roots of A are 1 , 1, 3. # Reduce Quadratic Form to the Normal Form ## Example 1 Reduce the quadratic form 3x² + 2y² + 3z² - 2xy - 2yz to the normal form by orthogonal transformation. [JNTU (H) Dec. 2011 (Set No. 1)] ### Solution: The matrix A of the quadratic form is $$\begin{bmatrix} 3 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 3 \end{bmatrix}$$ Now we shall diagonalize the matrix A by orthogonal transformation. For this, we require the eigen values and eigen vectors. So, we need to find the eigen values and eigenvectors of A. Characteristic equation of A is |A -λI| = 0 $$\begin{bmatrix} 3-λ & -1 & 0 \\ -1 & 2-λ & -1 \\ 0 & -1 & 3-λ \end{bmatrix}$$ = 0 i.e., $(3-λ)[(2-λ)(3-λ)-1] + 1[-1(3-λ)] = 0$ ⇒ $(3-λ)(λ^2 - 5λ + 5) - (3-λ) = 0$ ⇒ $(3-λ)(λ^2 - 5λ + 4) = 0$ ⇒ $(3-λ)(λ-1)(λ-4) = 0$ :. λ = 3, λ = 1, λ = 4, which are all different. **When λ = 3, we have** -y = 0 -x -y -z = 0 -y = 0 :. y = 0, x = -z. The corresponding eigen vector is X₁ = $$\begin{bmatrix} 1\\ 0\\ -1 \end{bmatrix}$$ **When λ = 1, the eigen vector is given by** 2x - y = 0 -x + y - z = 0 -y + 2z = 0 Let z= k so that y = 2k and 2x= 2k ⇒ x= k Then, $$\begin{bmatrix} x\\ y\\ z \end{bmatrix}$$ = $$\begin{bmatrix} k\\ 2k\\ k \end{bmatrix}$$ = k $$\begin{bmatrix} 1\\ 2\\ 1 \end{bmatrix}$$ Thus the corresponding eigen vector is X₂ = $$\begin{bmatrix} 1\\ 2\\ 1 \end{bmatrix}$$ **Similarly for λ = 4, we have the eigen vector is:** X₃ = $$\begin{bmatrix} 1\\ -1\\ 1 \end{bmatrix}$$ We observe that these three vectors are mutually orthogonal. We normalize these vectors and obtain e₁ = $$\begin{bmatrix} 1/\sqrt{2}\\ 0\\ -1/\sqrt{2} \end{bmatrix}$$ , e₂ = $$\begin{bmatrix} 1/\sqrt{6}\\ 2/\sqrt{6}\\ 1/\sqrt{6} \end{bmatrix}$$ , e₃ = $$\begin{bmatrix} 1/\sqrt{3}\\ -1/\sqrt{3}\\ 1/\sqrt{3} \end{bmatrix}$$ Let P = The modal matrix in normalized form = [e₁ e₂ e₃] = $$\begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{6} & 1/\sqrt{3}\\ 0 & 2/\sqrt{6} & - 1/\sqrt{3}\\ -1/\sqrt{2} & 1/\sqrt{6} & 1/\sqrt{3} \end{bmatrix}$$ **Diagonalization:** Since P is orthogonal matrix P⁻¹ = Pᵀ so PᵀAP = D where D is the diagonal matrix. :. D = PᵀAP = $$\begin{bmatrix} 1/\sqrt{2} & 0 & - 1/\sqrt{2}\\ 1/\sqrt{6} & 2/\sqrt{6} & 1/\sqrt{6}\\ 1/\sqrt{3} & - 1/\sqrt{3} & 1/\sqrt{3} \end{bmatrix}$$ $$\begin{bmatrix} 3 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 3 \end{bmatrix}$$ $$\begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{6} & 1/\sqrt{3}\\ 0 & 2/\sqrt{6} & - 1/\sqrt{3}\\ -1/\sqrt{2} & 1/\sqrt{6} & 1/\sqrt{3} \end{bmatrix}$$ = $$\begin{bmatrix} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{bmatrix}$$ = diag[3, 1, 4] The quadratic form will be reduced to the normal form. **Hence the transformed Quadratic form is:** XᵀAX = YᵀDY = $$\begin{bmatrix} y₁& y₂ & y₃ \end{bmatrix}$$ $$\begin{bmatrix} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{bmatrix}$$ $$\begin{bmatrix} y₁\\ y₂\\ y₃ \end{bmatrix}$$ = 3y₁² + y₂² + 4y₃² by the orthogonal transformation X = PY. i.e., x = (1/√2)y₁ + (1/√6)y₂ + (1/√3)y₃ y = (2/√6)y₂ - (1/√3)y₃ z = (-1/√2)y₁ + (1/√6)y₂ + (1/√3)y₃ # Find the nature of the quadratic form ## Example 2 Find the nature of the quadratic form 2x² + 2y² + 2z² + 2yz. [JNTU (H) Jan. 2012 (Set No. 3), Dec, 2019] ### Solution: Given quadratic form is 2x² + 2y² + 2z² + 2yz Matrix of the quadratic form is A = $$\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 2 \end{bmatrix}$$ Characteristic equation of A is $$\begin{vmatrix} 2-λ & 0 & 0 \\ 0 & 2-λ & 1 \\ 0 & 1 & 2-λ \end{vmatrix}$$ = 0 [Expand by R₁] ⇒ (2-λ)[(2-λ)² -1] = 0 ⇒ (2-λ)(λ² - 4λ + 3) = 0 ⇒ (2-λ)(λ - 3)(λ - 1) = 0 :. The roots of the characteristic equation are 1, 2, 3. All the roots are positive. The Q. F. is + ve definite. # Find the Eigen values and Eigen vectors of the matrix ## Example 10 Find the Eigen values and the corresponding Eigen vectors of the matrix $$\begin{bmatrix} 2 & 2 & 0 \\ 2 & 5 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$ [JNTU Sep. 2008, (H) Dec. 2011 (Set No. 2)] ### Solution: Let A = $$\begin{bmatrix} 2 & 2 & 0 \\ 2 & 5 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$ The characteristic equation of 'A' is |A - λI| = 0. i.e., $$\begin{vmatrix} 2-λ & 2 & 0 \\ 2 & 5-λ & 0 \\ 0 & 0 & 3-λ \end{vmatrix}$$ = 0 ⇒ (2-λ)[(5-λ)(3-λ)]-2[2(3-λ)]=0 ⇒ (2-λ)(15-5λ-3λ+λ²)-4(3-λ)=0 ⇒ (2-λ)(λ²-8λ+15)-4(3-λ)=0 ⇒ (2-λ)(λ²-8λ+15)-4(3-λ)=0 ⇒ (2-λ)(λ²-8λ+15)-4(3-λ)=0 ⇒ 2λ² - 16λ + 30 - λ³ + 8λ² - 15λ -12 + 4λ = 0 ⇒ -λ³ + 10λ² - 27λ + 18 = 0 ⇒ λ³ - 10λ² + 27λ - 18 = 0 We observe that λ = 1 is a root. Dividing with (λ -1), we get (λ -1)(λ² - 9λ + 18) = 0 i.e., (λ -1)(λ² - 6λ - 3λ + 18) = 0 i.e., (λ -1)[λ(λ - 6)-3(λ -6)]=0 or (λ -1)(λ - 3)(λ - 6) = 0 ∴ λ = 1, 3, 6 Thus, the Eigen values of 'A' are 1, 3, 6. **Case 1:** The Eigen vector corresponding to λ = 1 is given by $$\begin{bmatrix} 1 & 2 & 0 \\ 2 & 4 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$ $$\begin{bmatrix} x₁ \\ x₂ \\ x₃ \end{bmatrix}$$ = $$\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ [From (1)] ⇒ x₁ + 2x₂ = 0 2x₁ + 4x₂ = 0 and 2x₃ = 0 ⇒ x₃ = 0 Let x₂ = k. Then (2)x₁+ 2k = 0⇒ x₁ = -2k :. The Eigen vector corresponding to λ = 1 is X₁ = $$\begin{bmatrix} -2\\ 1\\ 0 \end{bmatrix}$$ **Case 2:** The Eigen vector corresponding to λ = 3 is given by $$\begin{bmatrix} -1 & 2 & 0 \\ 2 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ $$\begin{bmatrix} x₁ \\ x₂ \\ x₃ \end{bmatrix}$$ = $$\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ [From (1)] ⇒-x₁ + 2x₂ = 0 ⇒ 2x₂ = x₁ and 2x₁ + 2x₂ = 0 ⇒ x₁ + x₂ = 0 From (3) and (4), we get x₂ = 0 ⇒ x₂ = 0 From (3), we get x₁ = 0 Let x₃ = k. The Eigen vector corresponding to λ = 3 is X₂ = $$\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$$ **Case 3:** The Eigen vector corresponding to λ = 6 is given by $$\begin{bmatrix} -4 & 2 &0 \\ 2 & -1 & 0 \\ 0 & 0 & -3 \end{bmatrix}$$ $$\begin{bmatrix} x₁ \\ x₂ \\ x₃ \end{bmatrix}$$ = $$\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ [From (1)] ⇒-4x₁+2x₂ = 0 2x₁-x₂=0. 2x₁ = x₂ and -3x₃ = 0 Let x₁ = k. Then x₂ = 2x₁ = 2k :. The eigen vector corresponding to λ = 6 is X₃ = $$\begin{bmatrix} 1\\ 2\\ 0 \end{bmatrix}$$ Hence, the Eigen vectors corresponding to λ = 1, 3, 6 are $$\begin{bmatrix} -2\\ 1\\ 0 \end{bmatrix}$$ # Determine the modal matrix P and verify P⁻¹AP is a diagonal matrix. ## Example 2 Determine the modal matrix P of A = $$\begin{bmatrix} -2 & 2 & -3 \\ 2 & -1 & -2 \\ -1 & -2 & -λ \end{bmatrix}$$ and verify P⁻¹AP is a diagonal matrix. ### Solution: The characteristic equation of A is $$\begin{vmatrix} -2-λ & 2 & -3 \\ 2 & -1-λ & -2 \\ -1 & -2 & -λ \end{vmatrix}$$= 0 ⇒ (λ - 5)(λ + 3)² = 0 Thus, the eigen values of A are λ = 5, λ = -3, and λ = -3. **When λ = 5, we have** $$\begin{bmatrix} -7 & 2 & -3\\ 2 & -6 & -2\\ -1 & -2 & -5 \end{bmatrix}$$ $$\begin{bmatrix} x\\ y\\ z \end{bmatrix}$$ = $$\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$$ The corresponding eigen vector is X₁ = $$\begin{bmatrix} 1\\ 2\\ 1 \end{bmatrix}$$ Similarly, for the eigen value λ = -3, we can have two linearly independent vectors X₂ = $$\begin{bmatrix} 2\\ -1\\ 0 \end{bmatrix}$$ and X₃ = $$\begin{bmatrix} 3\\ 0\\ 1 \end{bmatrix}$$ Observe that X₁, X₂ and X₃ are not orthogonal. But they are linearly independent. Consider P = (X₁ X₂ X₃) :. P = $$\begin{bmatrix} 1 & 2 & 3 \\ 2 & -1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$$ = Modal matrix of A. Now det P = 1(-1) - 2(2) + 3(0 -1) = -1 - 4 - 3 = -8 :. P⁻¹ = 1/det P(adj P) = 1/-8 $$\begin{bmatrix} -1 & -2 & 3 \\ 2 & 4 & 6 \\ -1 & -2 & -5 \end{bmatrix}$$ P⁻¹A = (1/-8) $$\begin{bmatrix} -1 & -2 & 3 \\ 2 & 4 & 6 \\ -1 & -2 & -5 \end{bmatrix}$$ $$\begin{bmatrix} -2 & 2 & -3 \\ 2 & -1 & -2 \\ -1 & -2 & -λ \end{bmatrix}$$ = (1/-8) $$\begin{bmatrix} -1 & -2 & 3 \\ 2 & 4 & 6 \\ -1 & -2 & -5 \end{bmatrix}$$ $$\begin{bmatrix} -2 & 2 & -3 \\ 2 & -1 & -2 \\ -1 & -2 & -λ \end{bmatrix}$$ = (1/8) $$\begin{bmatrix} -5 & -10 & 15 \\ -6 & -12 & -18 \\ 3 & 6 & 15 \end{bmatrix}$$ and P⁻¹AP = (1/8) $$\begin{bmatrix} -5 & -10 & 15 \\ -6 & -12 & -18 \\ 3 & 6 & 15 \end{bmatrix}$$ $$\begin{bmatrix} -2 & 2 & -3 \\ 2 & -1 & -2 \\ -1 & -2 & -λ \end{bmatrix}$$ = (1/8) $$\begin{bmatrix} -40 & 0 & 0 \\ 0 & 24 & 0 \\ 0 & 0 & 24 \end{bmatrix}$$ = $$\begin{bmatrix} 5 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -3 \end{bmatrix}$$ = diag(5, -3, -3) Hence, P⁻¹AP is a diagonal matrix. # Find nature of the quadratic form, index and signature. ## Example 6 Find nature of the quadratic form, index and signature of 10x² + 2y² + 5z² -4xy -10xz +6yz. [JNTU Aril 2007 (Set No. 4), Sep. 2008 (Set No.1), (H) Dec. 2018] Find the transformation that will transform 10x² + 2y² + 5z² + 6yz − 10zx − 4xy into a sum of squares and find its reduced form. ### Solution: The given quadratic form is 10x² + 2y² + 5z² − 4xy − 10xz + 6yz. Its matrix is given by A = $$\begin{bmatrix} 10 & -2 & -5 \\ -2 & 2 & 3 \\ -5 & 3 & 5 \end{bmatrix}$$ We write A = I₃ AI₃ i.e. $$\begin{bmatrix} 10 & -2 & -5 \\ -2 & 2 & 3 \\ -5 & 3 & 5 \end{bmatrix}$$ = $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$\begin{bmatrix} 10 & -2 & -5 \\ -2 & 2 & 3 \\ -5 & 3 & 5 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ = $$\begin{bmatrix} 10 & -2 & -5 \\ -2 & 2 & 3 \\ -5 & 3 & 5 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ ⇒ $$\begin{bmatrix} 10 & -2 & -5 \\ 0 & 8 & 10 \\ 0 & 4 & 5 \end{bmatrix}$$ = $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$\begin{bmatrix} 10 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 4 & 1 \end{bmatrix}$$ = $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ (Applying R₂ → 5R₂ + R₁; R₃ → 2R₃+ R₁) ⇒ $$\begin{bmatrix} 10 & 0 & 0 \\ 0 & 40 & 20 \\ 0 & 20 & 10 \end{bmatrix}$$ = $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 5 & 1 \end{bmatrix}$$ $$\begin{bmatrix} 10 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 4 & 2 \end{bmatrix}$$ = $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 5 & 1 \end{bmatrix}$$ (Applying C₂ → 5C₂ + C₁; C₃ → 2C₃ + C₁) ⇒ $$\begin{bmatrix} 10 & 0 & 0 \\ 0 & 40 & 20 \\ 0 & 100 & 20 \end{bmatrix}$$ = $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 5 & 0 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 4 & 1 \end{bmatrix}$$ = $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 5 & 0 \end{bmatrix}$$ (Applying R₃ → 2R₃ - R₂) ⇒ $$\begin{bmatrix} 10 & 0 & 0 \\ 0 & 40 & 20 \\ 0 & 0 & 20 \end{bmatrix}$$ = $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 5 & -5 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 4 & 1 \end{bmatrix}$$ = $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 5 & -5 \end{bmatrix}$$ (Applying C₃ → 2C₃ - C₂) Thus, the given quadratic form is reduced into normal form. i.e., D = PAP where D = $$\begin{bmatrix} 10 & 0 & 0 \\ 0 & 40 & 0 \\ 0 & 0 & 20 \end{bmatrix}$$ and P = $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 5 & -5 \end{bmatrix}$$ The linear transformation is X = PY i.e., $$\begin{bmatrix} x\\ y\\ z \end{bmatrix}$$ = $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 5 & -5 \end{bmatrix}$$ $$\begin{bmatrix} y₁\\ y₂\\ y₃ \end{bmatrix}$$ i.e. x = y₁ + y₂ + y₃; y = 5y₂ - 5y₃; z = 4y₃ The given quadratic form is reduced to XᵀAX = (PY)ᵀA(PY) = Yᵀ(PᵀAP)Y = YᵀDY = $$\begin{bmatrix} y₁& y₂& y₃ \end{bmatrix}$$ $$\begin{bmatrix} 10 & 0 & 0 \\ 0 & 40 & 0 \\ 0 & 0 & 20 \end{bmatrix}$$ $$\begin{bmatrix} y₁ \\ y₂\\ y₃ \end{bmatrix}$$ = 10y₁² + 40y₂² Nature of the quadratic form is positive semi-definite (Here r = 2, n = 3, s = 2) Index, s = no. of positive terms in normal form = 2 Signature = 2s - r = 2(2) - 2 = 2. # Using Cayley-Hamilton theorem, find A⁸, if A = $$\begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}$$ ## Example 10 Using Cayley-Hamilton theorem, find (i) A⁸, if A = $$\begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}$$ (ii) A³, where A = $$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$ [JNTU July-2003] ### Solution: (i) Given A = $$\begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}$$ Characteristic equation of A is |A - λI| = 0 i.e. $$\begin{vmatrix} 1-λ & 2 \\ 2 & -1-λ \end{vmatrix}$$ = 0 ⇒ λ² - 5 = 0 By Cayley-Hamilton theorem, A satisfies its characteristic equation. So, we must have A² = 5I. :. A⁸ = 5A⁶ = 5(A²) (A²) (A²) = 5(5I) (5I) (5I) = 625I. This is left as an exercise to the reader. # Verify Cayley-Hamilton theorem for the matrix A = $$\begin{bmatrix} 1 & 2 & -1 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$$ ## Example 11 (i) If A = $$\begin{bmatrix} 1 & 2 & -1 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$$ verify Cayley-Hamilton theorem. Find A⁴ and A⁻¹ using Cayley-Hamilton theorem. [JNTU Sep. 2006 (Set No. 4)] (ii) Verify Cayley-Hamilton theorem for the matrix A = $$\begin{bmatrix} 1 & 2 & -1 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$$ and hence find A⁻¹ ### Solution: Given A = $$\begin{bmatrix} 1 & 2 & -1 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$$ Characteristic equation of A is given by |A - λI| =0 $$\begin{vmatrix} 1-λ & 2 & -1 \\ 2 & 1-λ & -2 \\ 2 & -2 & 1-λ \end{vmatrix}$$ = 0 ⇒ $$\begin{vmatrix} 3-λ & 0 & λ \\ 2 & 1-λ & -2\\ 2 & -2 & 1-λ \end{vmatrix}$$ = 0 ⇒ (3-λ)[(1-λ)²-4]-[-4-2(1-λ)]=0 ⇒ (3-λ)(λ²-2λ-3)-[-4-2+2λ]=0 ⇒ (3-λ)(λ²-2λ-3)+6-2λ = 0 ⇒ 3λ²-6λ-9-λ³+2λ²+3λ+6-2λ=0 ⇒ -λ³+5λ²-5λ-3=0 ⇒ λ³-5λ²+5λ+3=0 i.e., A³ - 5A² + 5A + 3I = O (2) Now A² = $$\begin{bmatrix} 1 & 2 & -1 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 2 & -1 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$$ = $$\begin{bmatrix} 3 & 6 & -6 \\ 0 & 9 & -6 \\ 0 & 0 & 3 \end{bmatrix}$$ and A³ = $$\begin{bmatrix} 1 & 2 & -1 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$$ $$\begin{bmatrix} 3 & 6 & -6 \\ 0 & 9 & -6 \\ 0 & 0 & 3 \end{bmatrix}$$ = $$\begin{bmatrix} 3 & 24 & -21 \\ 6 & 21 &