Mathematics Test - 14

Concept:

Matrices are added or subtracted only if they have the same number of rows and columns.

Product of matrices: only if the number of columns of the 1st matrix must equal the number of rows of the 2nd matrix.

A non-singular matrix is a square matrix whose determinant is not zero.

Calculation:

1. Here order of A = Order of B, so the sum and product of matrices A and B exit.

This statement is true.

2. Here, A and B are non-singular matrices, but their order can be different. So, this statement is not necessarily true.

∴ Only Statement I is true.

Concept:

For matrix A(p × r) and B(s × t), AB is defined only if r = s and the order of AB is p × t.

The transpose of a matrix is when columns become rows and rows become columns. 

Calculation:

Let the matrix B has an order p × q, i.e, 'p' rows and 'q' columns

Transpose of B is B' will have to order q × p

For matrix  AB^T = [A(2 × 3) × B^T(q × p)] is defined, so 

∴ q = 3

Given: ABT is a square matrix

AB^T = [A(2 × 3) × BT(3 × p)] is square matrix, so

∴ p = 2

So, the order of B = p × q = 2 × 3

Concept:

Transpose of a Matrix:

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

It is denoted by A′ or A^T

Equal matrix:

Two matrices are equal if they have the same dimension or order and the corresponding elements are identical.

Calculation:

Given \(\rm A = \left[ {\begin{array}{*{20}{c}} {\cos 2θ }&{ - \sin 2θ }\\ {\sin 2θ }&{\cos 2θ } \end{array}} \right]\)and A + AT = I

\(\rm A^T = \left[{\begin{array}{*{20}{c}} {\cos 2θ }&{ \sin 2θ }\\ {-\sin 2θ }&{\cos 2θ } \end{array}} \right]\)

A + A^T = \(\rm \left[ {\begin{array}{*{20}{c}} {\cos 2θ }&{ - \sin 2θ }\\ {\sin 2θ }&{\cos 2θ } \end{array}} \right]\) + \(\rm \left[ {\begin{array}{*{20}{c}} {\cos 2θ }&{ \sin 2θ }\\ {-\sin 2θ }&{\cos 2θ } \end{array}} \right]\) = I

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

∴ 2cos 2θ = 1

cos 2θ = \(1\over2\)

2θ = \(\pi\over3\)

θ = \(\boldsymbol{\pi\over6}\)

Concept:

Existence of inverse of matrix:

For any square matrix of order n×n A, if there exist a matrix A^-1 such that AA^-1=A^-1A=I where I is an identity matrix of the same order then we say that A^-1 is an inverse matrix of matrix A.

Existence of identity matrix:

For any square matrix of order n×n A, if there exist a matrix I of the same order such that AI = A = IA then the matrix I is known as identity matrix.

Necessary and sufficient condition for inverse:

For any square matrix A, the inverse A^-1 exist if and only if the matrix A is a square matrix with non-zero determinant. Such a matrix is known as non-singular matrix.

Calculation:

It is given that for square matrices A, B and C of same order the following relation holds:

AB = AC

If we premultiply by A^-1 then we will get the following:

\(\begin{align*} A^{-1}AB &= A^{-1}AC\\ IB &= IC\\ B&=C \end{align*}\)

Thus, to reach till this the only thing important is the existence of inverse.

Therefore, we can conclude B = C only if the inverse A^-1 exists, that means only if the matrix A is non-singular.

Given:

\( A =\begin{bmatrix} 1 & 2 \\\ 3 & 4 \end{bmatrix} \) and \(B =\begin{bmatrix} a & 0 \\\ 0 & b \end{bmatrix}\)

Concept:

Matrix Multiplication:

Multiplication is only possible when the number of columns of the first matrix is equal to the number of rows of the second matrix.

A m × n matrix multiplied by a n × p matrix results in a m × p matrix.

Calculation:

 \( AB =\begin{bmatrix} 1 & 2 \\\ 3 & 4 \end{bmatrix} \begin{bmatrix} a & 0 \\\ 0 & b \end{bmatrix}\)

⇒ \( AB =\begin{bmatrix} a & 2b \\\ 3a & 4b \end{bmatrix} \)    ----(1)

Similarly, 

\(BA =\begin{bmatrix} a & 0 \\\ 0 & b \end{bmatrix}\begin{bmatrix} 1 & 2 \\\ 3 & 4 \end{bmatrix} \)

\(⇒ BA =\begin{bmatrix} a & 2a \\\ 3b & 4b \end{bmatrix}\)    ----(2)

If AB = BA

⇒ \(\begin{bmatrix} a & 2b \\\ 3a & 4b \end{bmatrix} =\begin{bmatrix} a & 2a \\\ 3b & 4b \end{bmatrix}\)

This will be possible if

2b = 2q or 3a = 3b

⇒ a = b

∴ There exist infinitely many B's such that AB = BA

Concept:

Non – singular matrix: A matrix which has an inverse. Inverse of a matrix is possible when its determinant is non zero.

Explanation:

Given two matrices M₁ and M₂.

M₁*M₂ = 0

M₁ is non-singular.

Take inverse of M₁ both side:

M₁^-1 (M₁*M₂) = 0* M₁^-1

M₂ = 0

As, MM^-1 = 0 and when we multiply a matrix with null matrix result is zero.

Concept:

Properties of identity matrix:

If A is the square matrix of order n × n

• AI = IA = A
• Iⁿ = I           (Where n ∈ N)

Calculation:

Given

A² = I

Now, A3 + (A + I)2 - 9A - I² - A²

= A². A + A² + I² + 2AI - 9A - I² - A²

= I. A + I + I + 2AI - 9A - I - I           [∵ A2 = I and AI = IA = A]

= AI + 2AI - 9A 

= 3AI - 9A

= 3A - 9A

= - 6A

Multiplication of the matrix is not commutative because if the product of MN Exists, then it is not necessary that the product of NM will also exist.

Example:

Let us consider two 2 × 2 Matrices (same dimension) as shown:

\(M=\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ \end{matrix} \right]\)

\(N=\left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right]\)

M × N gives:

\(M× N =\left[ \begin{matrix} (1)(2)+(2)(1) & (1)( 1)+(2)( 3) \\ (3)( 2)+(4)( 1) & (3)( 1)+(4)(3) \\ \end{matrix} \right]\)

\(M× N =\left[ \begin{matrix} 4 & 7 \\ 10 & 15 \\ \end{matrix} \right]\)

Similarly, N × M gives:

\(N× M =\left[ \begin{matrix} (2)(1)+(1)(3) & (2)(2)+(1)(4) \\ (1)(1)+(3)(3) & (1)(2)+(3)(4) \\ \end{matrix} \right]\)

\(N× M =\left[ \begin{matrix} 5 & 8 \\ 10 & 14 \\ \end{matrix} \right]\)

We observe that (M × N)_2×2  ≠ (N × M)_2×2, even if the dimensions of the two matrices are equal.

But if we take two 2 × 2 Identity Matrices (same dimension), the product will be commutative, i.e. if:

\(M=\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\) and

\(N=\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\)

(M × N)2×2  = (N × M)2×2

We, therefore, conclude that (M × N)2×2  IS NOT ALWAYS EQUAL TO (N × M)2×2

Note: N_3×4 × M _2×3  also does not exists, since they are not compatible with multiplication.

Concept:

• Symmetric Matrix: Any real square matrix A = (a_ij) is said to be symmetric matrix if and only if a_ij = a_ji, ∀ 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 symmetric matrix.
• Skew-symmetric Matrix: Any real square matrix A = (a_ij) is said to be a skew-symmetric matrix if and only if a_ij = - a_ji, ∀ 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 say A, can be expressed as the sum of symmetric and skew-symmetric matrices.

  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: 

\(A = \left[ {\begin{array}{*{20}{c}} 2&3\\ { - \;1}&2 \end{array}} \right] = \frac{1}{2}\;\left( {P + Q} \right)\)

Where P is symmetric and Q is a skew-symmetric matrix

Here we have to find the matrix P and Q

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

i.e If A is a square matrix then A can be expressed as where A + A’ is symmetric and A – A’ is skew-symmetric matrix.

By comparing

 \(A = \left[ {\begin{array}{*{20}{c}} 2&3\\ { - \;1}&2 \end{array}} \right] = \frac{1}{2}\;\left( {P + Q} \right)\) 

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'

\(A = \;\left[ {\begin{array}{*{20}{c}} 2&3\\ { - \;1}&2 \end{array}} \right] ⇒ A' = \left[ {\begin{array}{*{20}{c}} 2&{ - \;1}\\ 3&2 \end{array}} \right]\)

⇒ \(P = \;\left[ {\begin{array}{*{20}{c}} 2&3\\ { - \;1}&2 \end{array}} \right] + \;\left[ {\begin{array}{*{20}{c}} 2&{ - \;1}\\ 3&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 4&2\\ 2&4 \end{array}} \right]\)

Similarly, 

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

Hence,

 \(P = \left[ {\begin{array}{*{20}{c}} 4&2\\ 2&4 \end{array}} \right]\)

\(Q = \;\left[ {\begin{array}{*{20}{c}} 0&4\\ { - \;4}&0 \end{array}} \right]\)

AB = 0 ⇒ then A = 0 or B = 0 satisfies but it is not a mandatory case.

\(A = \left[ {\begin{array}{*{20}{c}} 0&0\\ 1&0 \end{array}} \right]{\rm{and}}\) \(B = \left[ {\begin{array}{*{20}{c}} 0&0\\ 0&1 \end{array}} \right]\) then also \({\rm{AB\;}} = {\rm{\;}}0\)