Linear Algebra MCQ Quiz - Objective Question with Answer for Linear Algebra - Download Free PDF

The correct answer is Average (x1, x2 ... x9, m) = m and Average (x2, x3 ... x9) > m

Explanation:

Let's take the example of it

The nine numbers x1, x2, x3 ... x9, are in ascending order
Let  the nine number is 1,1,1,1,1,1,1,1,10
Their Avg will be = 18/9 = 2

Also, this satisfies that their average m is strictly greater than all the first eight numbers. 

now verifing

Average (x1, x2 ... x9, m) = m
Avg[1,1,1,1,1,1,1,1,10,2] will be 20/10 = 2
Which satisfies the given statement

Now checking Average (x2, x3 ... x9) > m

Average (1,1,1,1,1,1,1,10) will 17/8 = 2.12
Which is greater then 2

So it is also satisfied  Average (x2, x3 ... x9) > m

Conclusion:-

So, Average (x1, x2 ... x9, m) = m and Average (x2, x3 ... x9) > m is the true statement.

Explanation:

• Eigenvalues of a square matrix can be real or complex, positive, negative, or even zero, depending on the nature of the matrix.

• There is no general restriction that eigenvalues must be only positive, negative, or imaginary unless specific matrix types are involved (e.g., symmetric, skew-symmetric, Hermitian, etc.).

 Additional Information

Types of matrices and eigenvalues:

• Symmetric matrices: Have real eigenvalues.

• Skew-symmetric matrices: Have purely imaginary or zero eigenvalues.

• Orthogonal matrices: Have eigenvalues with absolute value 1.

• Diagonal matrices: Eigenvalues are the diagonal elements.

The correct answer is option 1: A²

Key Points

• Given: \( A^3 = I \), where \( I \) is the identity matrix.
• This means: \( A \cdot A \cdot A = I \).
• We want to find \( A^{-1} \), the inverse of A.
• Multiply both sides of \( A^3 = I \) by \( A^{-1} \):
  • \( A^2 \cdot A \cdot A^{-1} = I \cdot A^{-1} \)
  • \( A^2 \cdot I = A^{-1} \Rightarrow A^2 = A^{-1} \)

Therefore, the inverse of A is A².

Hence, the correct answer is: option 1: A²

Explanation:
We are given that A³ = I, where I is the identity matrix.

This implies: A × A × A = I

Now, multiply both sides by A^-1 on the right:

A × A × A × A^-1 = I × A^-1

⟹ A × A = A² = A^-1

⟹ A^-1 = A²

Final Answer: Option 1) A² ✅

Calculation:

Given |An×n| = 3 & |adj A| = 243

We know that |adj A| = |A|n−1

⇒ 243 = 3n−1 

⇒ 35 = 3n−1 

⇒ n − 1 = 5 

⇒ n = 6

The correct answer is option "3"

Concept:

Let B = |A¹²¹ - A¹²⁰|

B = |A¹²⁰ × (A – I)| 

B = |A¹²⁰| × |A – I|

Calculation:

A = \(\left[ {\begin{array}{*{20}{c}} 8&5\\ 7&6 \end{array}} \right]\)

Now, calculating matrix [A – I]

[A – I] = \(\left[ {\begin{array}{*{20}{c}} 8&5\\ 7&6 \end{array}} \right] - {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

[A – I] = \(\left[ {\begin{array}{*{20}{c}} 7&5\\ 7&5 \end{array}} \right]\)

Now determinant of |A – I|,

|A – I| = \(\left| {\begin{array}{*{20}{c}} 7&5\\ 7&5 \end{array}} \right|\)

|A – I| = 0   (Since two rows are repeated, therefore determinant = 0)

Hence, |A¹²¹ - A¹²⁰| =  A120|A – I|  = 0

Concept:

Rank:

The rank of a matrix is a number equal to the order of the highest order non-vanishing minor, that can be formed from the matrix.

For matrix A, it is denoted by ρ(A).

The rank of a matrix is said to be r if,

• There is at least one non-zero minor of order r.
• Every minor of matrix A having order higher than r is zero. 

Property of Rank of Matrix:

​ρ(AB) ≤ min [ρ(A), ρ(B)]

Calculation:

Given:

ρ(A) = 2, ρ(B) = 3

Using properties

ρ(AB) ≤ min [ρ(A), ρ(B)]

ρ(AB) ≤ min (2,3)

⇒ ρ(AB) ≤  2

Alternate Method

Let order of matrix A be 2 × m and order of matrix B be m × 3 (∵ for multiplication we need the column of A and row of B to be same)

∴ Order of matrix AB will be 2 × 3

Using properties

ρ(AB) ≤ min (Row, Column)

⇒ ρ(AB) ≤ min (2, 3)  [only when the column of A and row of B is the same]

⇒ ρ(AB) ≤ 2.

∵ we don't know the dimension of A and B, we cannot predict the exact rank of AB but its maximum rank will be 2.

Important Points

Other properties of rank of a matrix are:

• The rank of a matrix does not change by elementary transformation, we can calculate the rank by changing the matrix into Echelon form. In the Echelon form, the rank of a matrix is the number of non-zero rows of the matrix.
• The rank of a matrix is zero if the matrix is null.
• ρ(A) ≤ min (Row, Column)
• ρ(AB) ≤ min [ρ(A), ρ(B)]
• ρ(AT A) = ρ(A AT) = ρ(A) = ρ(AT)
• If A and B are matrices of the same order, then ρ(A + B) ≤ ρ(A) + ρ(B) and ρ(A - B) ≥ ρ(A) - ρ(B).
• If Aθ is the conjugate transpose of A, then ρ(Aθ) = ρ(A) and ρ(A Aθ) = ρ(A).
• The rank of a skew-symmetric matrix cannot be one.

EXPLANATION:

Let A, B and C be the matrices a, b, and c be scalars and the sizes of matrices are such that the operations can be performed then

The multiplication of real-valued square matrices of the same dimension is:

Associative, which means that for any three matrices A, B, and C of the same dimension, the following holds true: (AB)C = A(BC)

Not Commutative, which means that the order of the matrices matters. In other words, AB is not necessarily equal to BA, unless one or both of the matrices are diagonal or scalar matrices.

Not always positive definite, since the determinant of the product of two matrices is equal to the product of the determinants of the matrices, and the determinant of a matrix is positive if and only if the matrix is invertible. Therefore, the product of two invertible matrices is invertible, and hence positive definite. However, the product of two non-invertible matrices may not be invertible, and hence not positive definite.

The order of the multiplication matters, so it is not always possible to compute the matrices. That is, AB is not necessarily equal to BA, unless one or both of the matrices are diagonal or scalar matrices.

Additional Information 

Properties of Matrix addition and scalar multiplication:

Commutative Property of addition

• A + B = B + A

Associative Property of addition

• A + (B + C) = (A + B) + C
• A + O = O + A Where O is the appropriate zero matrix

Distributive Property of addition

• c(A + B) = cA + cB
• (a + b)C = aC + bC
• (ab)C = a(bc)

Concept:

If rank of 3 × 3 homogeneous matrix is less than 3 then the corresponding equations will have non-trivial solutions.

Explanation:

For non-trivial solution

\(\left| {\begin{array}{*{20}{c}} p&q&r\\ q&r&p\\ r&p&q \end{array}} \right| = 0\)

R₁ = R₁ + R₂ + R₃

\(\left| {\begin{array}{*{20}{c}} {p + q + r}&{\;q + r + p}&{r + p + q}\\ q&r&p\\ r&p&q \end{array}} \right| = 0\)

\(\left( {p + q + r} \right)\left| {\begin{array}{*{20}{c}} 1&{\;1}&1\\ q&r&p\\ r&p&q \end{array}} \right| = 0\)

∴ p + q + r = 0

(OR)

\(\left| {\begin{array}{*{20}{c}} 1&{\;1}&1\\ q&r&p\\ r&p&q \end{array}} \right| = 0\)

Concept:

From the properties of a matrix,

The rank of m × n matrix is always ≤ min {m, n}

If the rank of matrix A is ρ(A) and rank of matrix B is ρ(B), then the rank of matrix AB is given by

ρ(AB) ≤ min {ρ(A), ρ(B)}

If n × n matrix is singular, the rank will be less than ≤ n

Calculation:

Given:

AX = B

Where A is 2 × 4 matrices and b ≠ 0

The order of A^T is 4 × 2

The order of A^TA is 4 × 4

Rank of (A) ≤ min (2, 4) = 2

Rank of (A^T) ≤ min (2, 4) = 2

Rank (A^TA) ≤ min (2, 2) = 2

As the matrix A^TA is of order 4 × 4, to have a unique solution the rank of A^TA should be 4.

Therefore, the unique solution of this equation is not possible.

Trace of a matrix

Let A be k × k matrix. Then its trace is denoted by trace(A) or tr(A), is  the sum of its diagonal elements

\(tr\left( A \right) = \mathop \sum \limits_{k = 1}^k {A_{kk}}\)

Calculation:

The given matrix is

\(\left[ {\begin{array}{*{20}{c}} 1&2&3\\ 4&5&6\\ 7&8&9 \end{array}} \right]\)

1, 5, 9 are diagonal elements.

tr(A) = A₁₁ + A₂₂ + A₃₃

tr(A) = 1 + 5 + 9

tr(A) = 15

Concept:

Properties of Determinant of a Matrix:

• If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.
• For any square matrix say A, |A| = |AT|.
• If we interchange any two rows (columns) of a matrix, the determinant is multiplied by -1.
• If any two rows (columns) of a matrix are same then the value of the determinant is zero.

Calculation:

\(\left| {\begin{array}{*{20}{c}} 1&a&{b + c}\\ 1&b&{c + a}\\ 1&c&{a + b} \end{array}} \right|\)

Apply C2 → C2 + C3 

\( = \left| {\begin{array}{*{20}{c}} 1&{a + b + c}&{b + c}\\ 1&{a + b + c}&{c + a}\\ 1&{a + b + c}&{a + b} \end{array}} \right|\)

Taking common (a + b + c) from column 2, we get

\(= \left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}} 1&1&{b + c}\\ 1&1&{c + a}\\ 1&1&{a + b} \end{array}} \right|\)

As we can see that the first and the second column of the given matrix are equal. 

We know that, if any two rows (columns) of a matrix are same then the value of the determinant is zero.

∴ \(\left| {\begin{array}{*{20}{c}} 1&a&{b + c}\\ 1&b&{c + a}\\ 1&c&{a + b} \end{array}} \right|\)= 0  

Explanation-

• The eigenvector of a matrix A is a vector represented by a matrix X such that when X is multiplied with matrix A, then the direction of the resultant matrix remains the same as vector X.
• AX = λX

  • where A is any arbitrary matrix, λ are eigen values and X is an eigen vector corresponding to each eigen value.​Here, we can see that AX is parallel to X.So X is an eigen vector.

Some important properties of eigenvalues are-​

• If λ₁, λ₂…….λ_n are the eigenvalues of A, then \(\lambda _1^k,\lambda _2^k,......,\lambda _n^k\) are the eigenvalues of A^k

• Sum of Eigen Values = Trace of A (Sum of diagonal elements of A)

Given data and Calculation-

Given matrix = \(ε = \begin{bmatrix} 1 & 1 \\\ 1 & -1 \end{bmatrix} \)

\(A - \lambda I = \left( {\begin{array}{*{20}{c}} {1 - \lambda }&1\\ 1&{ - 1 - \lambda } \end{array}} \right)\) = 0

\(\Rightarrow (1 - \lambda )( - 1 - \lambda ) - 1\) = 0

\( \Rightarrow {\lambda ^2} - 2 = 0\)

\( \Rightarrow {\lambda ^2} = 2\)

\( \Rightarrow \lambda = \pm \sqrt 2 \)

Eigenvalues of ε8 are \({\left( {\sqrt 2 } \right)^8} \) and \({\left( { - \sqrt 2 } \right)^8}\).

 Eigenvalues of ε¹¹ are \({\left( {\sqrt 2 } \right)^{11}}\) and \({\left( { - \sqrt 2 } \right)^{11}}\).

P = Trace (ε8) = 16 + 16 = 32

Q = Trace (ε11).= 0

P + Q = 32

Additional Information 

• Eigenvalues of real symmetric matrices are real.

• Eigenvalues of real skew-symmetric matrices are either pure imaginary or zero.

• Eigenvalues of unitary and orthogonal matrices are of unit modulus |λ| = 1

• If λ1, λ2…….λn are the eigenvalues of A, then kλ1, kλ2…….kλn are eigenvalues of kA.

• If λ1, λ2…….λn are the eigenvalues of A, then 1/λ1, 1/λ2…….1/λn are eigenvalues of A-1

• Eigenvalues of A = Eigen Values of AT (Transpose)

• Product of Eigen Values = |A|

• Maximum number of distinct eigenvalues of A = Size of A

• If A and B are two matrices of the same order then, Eigenvalues of AB = Eigenvalues of BA

Calculation:

Given:

Trace (A) = 14, det(A) = 100

Trace of (A) = sum of diagonal elements

\(\left[ {\begin{array}{*{20}{c}} a&0&3&7\\ 2&5&1&3\\ 0&0&2&4\\ 0&0&0&b \end{array}} \right]\) 

14 = a + 5 + 2 + b

a + b = 7         ---(1)

Now,

\(\det \left( A \right) = 5\left| {\begin{array}{*{20}{c}} a&3&7\\ 0&2&4\\ 0&0&b \end{array}} \right|\)

\(5 \times 2 \times a \times b = 100\) 

ab = 10          ---(2)

from equation (1) and (2)

Either,  a = 5, b = 2

Or,        a = 2, b = 5

So,

|a - b| = |5 - 2| = |2 - 5| = 3

Concept:

Two matrices A_{m × n} and B_{p × q} 

if AB and BA are defined then p = n and q = m

Calculation:

Given:

AB and BA are defined.

so the order of the matrix B is B_{n × m}