Matrices & Properties Test - 1

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

Properties of matrix multiplication:

Associative property of matrix multiplication:

• A(BC) = (AB)C

Distributive property of multiplication:

• A(B + C) = AB + AC
• (A + B)C = AC + BC
• AI_n = I_nA = A,  I_n is the appropriate identity matrix
• c(AB)= (cA)B = A(cB)

Note:  AB ≠ BA in general, i.e. multiplication of matrices is not commutative even if the matrix is square and of the same order.

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:

Consider a matrix A is skew-symmetric, then A^T = −A
and A is symmetric, then A^T = A

Calculation:

Since, A is skew-symmetric.
A^T = −A
Since, A is symmetric.
A^T = A
⇒ −A = A
⇒2A = O
⇒A = O
Hence, A is a null matrix.

Hence, option (4) is correct.

Concept:

1. Diagonal matrix:

• A square matrix in which every element except the principle diagonal elements is zero, it is called a Diagonal Matrix

2. Scalar matrix:

• A Diagonal matrix whose diagonal elements all contains the same scalar is called Scalar matrix.
• A square matrix, in which all diagonal elements are equal to same scalar and all other elements are zero, is called a scalar matrix.

3. Symmetric Matrix:

• Square matrix A is said to be symmetric if the transpose of matrix A is equal to matrix A itself
• A^T = A or A’ = A

Where,

A^T or A’ denotes the transpose of matrix

• A square matrix A is said to be symmetric if a_ij = a_ji for all i and j

Where a_ij and a_ji is an element present in matrix.

4. Skew-Symmetric Matrix or Anti-symmetric:

• Square matrix A is said to be skew-symmetric if a_ij = −a_ji for all i and j.
• Square matrix A is said to be skew-symmetric if transpose of matrix A is equal to negative of matrix A ⇔ A^T = −A
• Note that all the main diagonal elements in the skew-symmetric matrix are zero.

5. Transpose of a Matrix:

• The new matrix obtained by interchanging the rows and columns of the original matrix is called as the transpose of the matrix.
• It is denoted by A′or A^T

Calculation:

Given: A is a skew-symmetric matrix

⇒ A^T = −A

Now take (A²) ^T = (A.A) ^T = A^T A^T = −A × −A = A²

⇒ (A²) ^T = A²

Hence A² is Symmetric.

Important points:

1. Properties of Symmetric Matrix:

• The sum and difference of two symmetric matrices is again symmetric
• The product of two symmetric matrices is symmetric (Not always true)
• If A and B are symmetric matrices, then AB is symmetric if and only if A and B commute (i.e., if AB = BA)
• If A is symmetric then Aⁿ is also symmetric for integer n.
• If A^-1 exists, it is symmetric i;f and only if A is symmetric.

2. Properties of Transpose Matrix

• The transpose of the transpose of a matrix is the matrix itself. ⇔ (A^T) ^T = A
• The transpose of a matrix times a scalar (k) is equal to the constant times the transpose of the matrix. ⇔ (kA) ^T = k A^T
• The transpose of the sum or difference of two matrices is equivalent to the sum or difference of their transposes. ⇔ (A ± B) ^T = A^T ± B^T
• The transpose of the product of two matrices is equivalent to the product of their transposes in reversed order. ⇔ (AB) ^T = B^T A^T
• The determinant of a square matrix is the same as the determinant of its transpose.

Concept:

• To multiply an m × n matrix by an n × p matrix, the n must be the same, and the result is an m × p matrix.
• If A be a matrix of order m × n than the order of transpose matrix is n × m

Calculation:

Given:

Order of A is 4 × 3, the order of B is 4 × 5 and the order of C is 7 × 3

The transpose of the matrix obtained by interchanging the rows and columns of the original matrix.

So, order of A^T is 3 × 4 and order of C^T is 3 × 7

Now,

A^TB = {3 × 4} {4 × 5} = 3 × 5

⇒ Order of A^TB is 3 × 5

Hence order of (A^TB) ^T is 5 × 3

Now order of (A^TB) ^T C ^T = {5 × 3} {3 × 7} = 5 × 7

Calculation:

Given: \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right]\)

\({{\rm{A}}^2} = {\rm{AA}} = \left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right]\)

\(\Rightarrow {{\rm{A}}^2} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {0 + 1}&{0 + 0}\\ {0 + 0}&{1 + 0} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

Now,

\(\Rightarrow {{\rm{A}}^4} = {{\rm{A}}^2}{{\rm{A}}^2} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

Hence Option 1^st is correct answer.

Concept:

Multiplication of matrices:

• The number of columns of the 1st matrix must equal the number of rows of the 2nd matrix.
• The result will have the same number of rows as the 1st matrix, and the same number of columns as the 2nd matrix.
• To multiply an m × n matrix by an n × p matrix, the n must be the same, and the result is an m × p matrix.

Calculation:

From statement 1∶

AB = A

We cannot find anything from this statement.

From statement 2∶

\(A\; = \;\left[ {\begin{array}{*{20}{c}} n&9\\ 2&1 \end{array}} \right] , B\; = \;\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

We cannot find anything from this statement.

Combining statement 1 and 2∶

\(AB\; = \;\left[ {\begin{array}{*{20}{c}} (n\times 1+9\times0)&(n\times0+9\times1)\\ (2\times1+1\times0)&(2\times0+1\times1) \end{array}} \right]\)

\(AB = \left[ {\begin{array}{*{20}{c}} n&9\\ 2&1 \end{array}} \right]\)

Also, \(A = \left[ {\begin{array}{*{20}{c}} n&9\\ 2&1 \end{array}} \right]\)

∴ We cannot find the value of n from both statements together.

Concept:

Orthogonal matrix: When the product of a matrix to its transpose gives identity matrix.

Suppose A is a square matrix with real elements and of n x n order and A^T or A’ is the transpose of A.

AA^T = I

Calculation:

Suppose A is a square matrix with real elements and of n x n order and A^T or A’ is the transpose of A.

Then according to the definition;

AA^T = I

Pre multiplication by A ^{- 1}

A ^{- 1} AA^T = A ^{- 1} I

IA^T = A ^{- 1}

A^T = A ^{- 1} or A’ = A ^{- 1}
then A is orthogonal matrix.

Concept:

• Symmetric Matrix: Any real square matrix A = (aij) is said to be a symmetric matrix if and only if aij = aji, ∀ i, and j in other words we can say that if A is a real square matrix such that A = A’ then A is said to be a symmetric matrix.
• Skew-symmetric Matrix: Any real square matrix A = (aij) is said to be a skew-symmetric matrix if and only if aij = - aji, ∀ I, and j or in other words we can say that if A is a real square matrix such that A =- A’ then A is said to be a skew-symmetric matrix.

• Any real square matrix says A, can be expressed as the sum of the symmetric and skew-symmetric matrix.

  i.e \(A = \frac{1}{2}\;\left( {A + A'} \right) + \frac{1}{2}\;\left( {A - A'} \right)\) where A + A’ is symmetric and A – A’ is a skew-symmetric matrix.

Calculation:

Given: \(A = \left[ {\begin{array}{*{20}{c}} 3&2&5\\ 4&1&3\\ 0&6&7 \end{array}} \right] = \frac{1}{2} \cdot (P + Q)\) where P is symmetric and Q is a skew-symmetric matrix

Here we have to find the matrix P and Q

As we know, any square matrix can be expressed as the sum of the symmetric and skew-symmetric matrices.

i.e If A is a square matrix then A can be expressed as

where A + A’ is symmetric and A – A’ is a skew-symmetric matrix.

By comparing \(A = \left[ {\begin{array}{*{20}{c}} 3&2&5\\ 4&1&3\\ 0&6&7 \end{array}} \right] = \frac{1}{2} \cdot (P + Q)\) with \(A = \frac{1}{2}\;\left( {A + A'} \right) + \frac{1}{2}\;\left( {A - A'} \right)\) we get,

⇒ P = A + A' and Q = A - A'

\(A = \left[ {\begin{array}{*{20}{c}} 3&2&5\\ 4&1&3\\ 0&6&7 \end{array}} \right] \Rightarrow A' = \left[ {\begin{array}{*{20}{c}} 3&4&0\\ 2&1&6\\ 5&3&7 \end{array}} \right]\)

\(\Rightarrow P = \;\;\left[ {\begin{array}{*{20}{c}} 3&2&5\\ 4&1&3\\ 0&6&7 \end{array}} \right] + \;\left[ {\begin{array}{*{20}{c}} 3&4&0\\ 2&1&6\\ 5&3&7 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 6&6&5\\ 6&2&9\\ 5&9&{14} \end{array}} \right]\)

Similarly,

 \(Q = \;\left[ {\begin{array}{*{20}{c}} 3&2&5\\ 4&1&3\\ 0&6&7 \end{array}} \right] - \;\left[ {\begin{array}{*{20}{c}} 3&4&0\\ 2&1&6\\ 5&3&7 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&{ - \;2}&5\\ 2&0&{ - \;3}\\ { - \;5}&3&0 \end{array}} \right]\)

Hence, \(P = \left[ {\begin{array}{*{20}{c}} 6&6&5\\ 6&2&9\\ 5&9&{14} \end{array}} \right]\;and\;Q = \;\left[ {\begin{array}{*{20}{c}} 0&{ - \;2}&5\\ 2&0&{ - \;3}\\ { - \;5}&3&0 \end{array}} \right]\)

Concept: 

A rectangular representation of mn numbers (complex or real) in the form of m rows and n columns is called a matrix of order m × n.

Any m × n matrix is represented as, \(A = \;{\left[ {\begin{array}{*{20}{c}} {{a_{11}}}& ⋅s &{{a_{1n}}}\\ \vdots & \ddots & \vdots \\ {{a_{m1}}}& ⋅s &{{a_{mn}}} \end{array}} \right]_{m × \;n}}\)

Calculation:

Let A be the matrix of order m × n such that it has p elements where p is a prime.

⇒ m ⋅ n = p

As we know that, prime factorization of a prime number is p = p × 1.

So, the possible orders of matrix A are: (p × 1), (1 × p)

Hence, the no. of possible orders it can have is 2.

Concept:

• Elementary matrix: An elementary matrix is a matrix which differs from the identity matrix by one single elementary row operation
• An elementary matrix has each diagonal element 1.

Calculation:

Let's check option b,

Let E = \(\left[ {\begin{array}{*{20}{c}} 1&5&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\)

Apply R₁ → R₁ – 5R₂

\( \Rightarrow {\rm{E\;}} = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right] = \;{{\rm{I}}_{3{\rm{\;}} \times 3}}\)

We can see that option B converted in to an identity matrix by one elementary operation.

Hence Option B is correct.

Shortcut Method:

We know that an elementary matrix has each diagonal element 1.

Only B has diagonal element is 1. (By Definition)

So, Option B is correct.