Previous Year Questions- Linear Algebra - 2 | Engineering Mathematics for Electrical Engineering - Electrical Engineering (EE) PDF Download

Q26: A system matrix is given as follows
The absolute value of the ratio of the maximum eigenvalue to the minimum eigenvalue is _______.      (SET-1 (2014))
(a) 1
(b) 2
(c) 3
(d) 4
Ans: (c)
Sol: Characteristic equation ∣A−λI∣= 0
The absolute values of λ are ∣𝜆∣ = 1, 2, 3
 Ratio of maximum and minimum eigen value is = 3 : 1 = 3/1 = 3

Q27: Given a system of equations
x + 2y + 2z = b₁
5x + y + 3z = b₂
What of the following is true regarding its solutions      (SET-1(2014))
(a) The system has a unique solution for any given b₁ and b₂
(b) The system will have infinitely many solutions for any given b₁ and b₂
(c) Whether or not a solution exists depends on the given b₁ and b_{2
(d) The systems would have no solution for any values of b1 and b2}
Ans: (b)

Q28: A Matrix has eigenvalues -1 and -2. The corresponding eigenvectors are  respectively. The matrix is     (2013)
(a)
(b)
(c)
(d)
Ans: d
Sol: AX = λX
From equation (i) and (iii), a = 0 and b = 1
From equation (ii) and (iv), c = -2 and d = -3

Q29: The equation  has      (2013)
(a) no solution
(b) only one solution
(c) non-zero unique solution
(d) multiple solutions
Ans: (d)
Sol: i.e. $𝑥1$x₁ and x₂ are having infinite number of solutions. 
⇒ Multiple solutions are these.

Q30: Given that  the value of A³ is      (2012 )
(a) 15A + 12I
(b) 19A + 30I
(c) 17A + 15I
(d) 17A + 21I
Ans: (b) 
Sol: Characteristic equation of A is
(by Cayley Hamilton theorem)
⇒ A² = −5A − 6I
Multiplying by A on both sides, we have,
A³= −5A² − 6A  
⇒ A³ =−5(−5A−6I)−6A = 19A + 30I

Q31: The matrix  is decomposed into a product of a lower triangular matrix [L] and an upper triangular matrix [U]. The properly decomposed [L] and [U] matrices respectively are     (2011)
(a)
(b)
(c)
(d)
Ans: (d)
Sol: Let us try Dolittle's decomposition by putting 𝑙₁₁ = 1 and 𝑙₂₂ = 1
So one possible breakdown is
But this is not any of the choises given.
So let us do Crout's decomposition, by putting $𝑢11=1$u₁₁ = 1 and u₂₂ = 1

Q32:  For the set of equations, $𝑥1+2𝑥2+𝑥3+4𝑥4=2$x₁ + 2x₂ + x₃ + 4x₄ = 2 and 3𝑥₁ + 6𝑥₂ + 3𝑥₃ + 12𝑥₄ = 6. The following statement is true.      (2010)
(a) Only the trivial solution 𝑥₁ = 𝑥₂ = 𝑥₃ = 𝑥₄ = 0 exists
(b) There are no solutions
(c) A unique non-trivial solution exists
(d) Multiple non-trivial solutions exist
Ans: (d)
Sol: The augmental matrix is
Performing gauss-elimination on this we get
rank(A) = rank(A∣B) = 1
So, system is consistent.
Since, system's rank = 1 is less than the number of variables, only infinite (multiple) non-trival solution exists.

Q33: An eigenvector of  is       (2010)
(a)
(b)
(c)
(d)
Ans: (b)
Sol: Given,

P is triangular. So eigen values are the diagonal elements themselves. Eigen values are therefore, $𝜆1=1,𝜆2=2,𝜆3=3.$λ₁ = 1, λ₂ = 2, λ₃ = 3. 
Now, the eigen value problem is $[𝐴-𝜆𝐼]𝑥^=0$[A − λI] Putting λ₁ = 1, we get the eigen vector corresponding to this eigen value,
Which gives the equations
The solution is 𝑥₂ = 0, 𝑥₃ = 0, 𝑥₁= 𝑘
So, one eigen vector is
Since, none of the eigen vectors given in choises matches with this, ratio we need to proceed further and find the other eigen vectors corresponding to the other eigen values.
Now, corresponding to $𝜆2=2,$λ₂ = 2, we get by substituting $𝜆=2,$λ = 2, in the eigen value problem, the following set of equations, 
Which gives the equations,
Solution is $𝑥3=0,𝑥1=𝑘,𝑥2=𝑘$x₃ = 0, x₁ = k, x₂ = k
Since none of the eigen vectors given in the choises is of this ratio, we need to proceed further and find 3rd eigen vector also.
By putting $𝜆=3,$λ = 3, in the eigen value problem, we get 
Only the eigen vector geven in choise  is in this ratio.

Q34: The trace and determinant of a 2 x 2 matrix are known to be -2 and -35 respectively. Its eigenvalues are     (2009)
(a) -30 and -5
(b) -37 and -1
(c) -7 and 5
(d) 17.5 and -2
Ans: (c)
Sol: Trace = Sum of principal diagonal element.

Q35: Let P be a 2x2 real orthogonal matrix and  is a real vector   Then, which one of the following statements is correct?    (2008)
(a)
(b)
(c)
(d) No relationship can be established between
Ans: (b)
Sol: Let on orthogonal matrixProperly of orthogonal matrix A

Q36: A is  m × n full rank matrix with m > n and I is identity matrix. Let matrix A⁺ = (A^TA)⁻¹A^T. Then, which one of the following statement is FALSE ?    (2008)
(a) $𝐴𝐴+𝐴=𝐴$AA⁺A =  A  
(b) (AA⁺)²  = AA⁺
(c) A⁺A = I
(d) AA ⁺ A = A ⁺
Ans: (b)
Sol: Choice (A) AA⁺A = A is corect

Q37: If the rank of a (5 x 6) matrix Q is 4, then which one of the following statement is correct ?    (2008)
(a) Q will have four linearly independent rows and four linearly independent columns
(b) Q will have four linearly independent rows and five linearly independent columns
(c) QQ^T will be invertible
(d) Q^TQ will be invertible
Ans: (a)
Sol: If rank of (5 x 6) matrix is 4, then surely it must have exactly 4 linearly independent rows as will as 4 linearly independent columns.

Q38: The characteristic equation of a (3x3) matrix P is defined as
a(λ)∣λI−P∣ = λ³+λ²+2λ+1 = 0
If I denotes identity matrix, then the inverse of matrix P will be     (2008)
(a) $(𝑃2+𝑃+2𝐼)$(P²+P+2I)
(b) (P²+P+I)
(c) −(P²+P+I)
(d) −(P²+P+2I)
Ans: (d)
Sol: If characteristic equation is
λ³ + λ² + 2λ + 1 = 0
Then by cayley- hamilton theorem,
P³ + P² + 2P + I = 0
I = −P³ − P² − 2P
Multiplying by P⁻¹ on both sides,
𝑃⁻¹ = −𝑃² − 𝑃 − 2𝐼 = −(𝑃² + 𝑃 + 2𝐼)  

Q39: Cayley-Hamilton Theorem states that a square matrix satisfies its own characteristic equation. Consider a matrix  A⁹ equals      (2007)
(a) $511𝐴+510𝐼$511A + 510I
(b) $309𝐴+104𝐼$309A + 104I
(c) 154A + 155I
(d) exp(9A)
Ans: a
Sol: To calculate A⁹
start from A² + 3A + 2I = 0 which has been derived above

Q40: Cayley-Hamilton Theorem states that a square matrix satisfies its own characteristic equation. Consider a matrix A satisfies the relation      (2007)
(a) A + 3I + 2A⁻¹ = 0
(b) A² + 2A + 2I = 0
(c) (A + I)(A + 2I)
(d) exp(A) = 0
Ans: (a)
Sol: A will satisfy this equation according to cayley hamition theorem
i.e. A² + 3A + 2I = 0
multiplying by A⁻¹ on both sides we get

Q41: The linear operation L(x) is defined by the cross product are three dimensional vectors. The 3 x 3 matrix M of this operations satisfies
Then the eigenvalues of M are     (2007)
(a) 0, +1, -1
(b) 1, -1, 1
(c) i, -i, 1
(d) i, -i, 0
Ans: (c)
Sol: The eigen values of M are (i, −i, 1)

Q42: Let x and y be two vectors in a 3 dimensional space and <x, y> denote their dot product. Then the determinant       (2007)(a) is zero when x and y are linearly independent
(b) is positive when x and y are linearly independent
(c) is non-zero for all non-zero x and y
(d) is zero only when either x or y is zero
Ans: (b)

Q43: x = [x₁ x₂ ... x_n]^T is an n-tuple nonzero vector. The nxn matrix V = xx^T       (2007)
(a) has rank zero
(b) has rank 1
(c) is orthogonal
(d) has rank n
Ans: (b)
Sol: The nxn matrix, V = XX^T has rank 1.

Q44: If then top row of R⁻¹ is      (2005)
(a) [5   6   4]
(b) [5   -3 1]
(c) [2  0  -1]
(d) [2  1   1/2]
Ans: (b)
Sol:
Since we need only the top row of $𝑅-1$R⁻¹, we need to find only first column of (R) which after transpose will become first row of Adj(A). 

Q45: For the matrix one of the eigen values is equal to -2. Which of the following is an eigen vector ?      (2005)
(a)
(b)
(c)
(d)
Ans: (d)
Sol: Since matrix is triangular, the eigen values are the diagonal elements themselves namely λ = 3,−2 and 1.
Corresponding to eigen value λ = −2 let us find the eigen vector
Which gives the equations,
Since equation (ii) and (iii) are same we have
Eigen vectors are of the form
 is an eigen vector of matrix p.

Q46: In the matrix equation Px = q, which of the following is a necessary condition for the existence of at least on solution for the unknown vector x:     (2005)
(a) Augmented matrix [Pq] must have the same rank as matrix P
(b) Vector q must have only non-zero elements
(c) Matrix P must be singular
(d) Matrix P must be square
Ans: (a)
Sol: rank [Pq] = rank [P] is necessary for existence of at least one solution to Px = q.

Q47: The eigen values of the system represented by  X are     (2002)
(a) 0, 0, 0, 0
(b) 1, 1, 1, 1
(c) 0, 0, 0, -1
(d) 1, 0, 0, 0
Ans: (d)

Q48: The determinant of the matrix  is:     (2002)
(a) 100
(b) 200
(c) 1
(d) 300
Ans: (c)
Sol: The given matrix is alower traingular matrix
Therefore, |A|= Product of its leading diagonal elemetns = (1)(1)(1)(1) = 1