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Matrices - Properties of transpose of the matrices

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The transpose of a matrix AA is formed by interchanging its rows and columns. It is denoted by ATA^T or A′A'. If A=[aij]m×nA = [a_{ij}]_{m \times n}, then AT=[aji]n×mA^T = [a_{ji}]_{n \times m}.

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A square matrix AA is said to be a Symmetric Matrix if AT=AA^T = A. In this case, aij=ajia_{ij} = a_{ji} for all possible values of ii and jj.

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A square matrix AA is said to be a Skew-Symmetric Matrix if AT=−AA^T = -A. In this case, aij=−ajia_{ij} = -a_{ji} for all i,ji, j. This implies that the diagonal elements aiia_{ii} must be zero because aii=−aii  ⟹  2aii=0a_{ii} = -a_{ii} \implies 2a_{ii} = 0.

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Any square matrix AA can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix.

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The operation of transpose follows specific algebraic properties including the Reversal Law for multiplication: (AB)T=BTAT(AB)^T = B^T A^T.

📐Formulae

(AT)T=A(A^T)^T = A

(kA)T=kAT (where k is a constant)(kA)^T = kA^T \text{ (where } k \text{ is a constant)}

(A+B)T=AT+BT(A + B)^T = A^T + B^T

(AB)T=BTAT(AB)^T = B^T A^T

A=12(A+AT)+12(A−AT)A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)

💡Examples

Problem 1:

If A=[3412]A = \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}, verify that (A+AT)(A + A^T) is a symmetric matrix.

Solution:

Given A=[3412]A = \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}, the transpose is AT=[3142]A^T = \begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix}.

Now, calculate P=A+ATP = A + A^T: P=[3412]P = \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix} + [3142]\begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix} = [3+34+11+42+2]\begin{bmatrix} 3+3 & 4+1 \\ 1+4 & 2+2 \end{bmatrix} = [6554] \begin{bmatrix} 6 & 5 \\ 5 & 4 \end{bmatrix}

To check if PP is symmetric, find PTP^T: PT=[6554]P^T = \begin{bmatrix} 6 & 5 \\ 5 & 4 \end{bmatrix} Since PT=PP^T = P, the matrix (A+AT)(A + A^T) is symmetric.

Explanation:

To verify a matrix is symmetric, we compute its transpose and check if it remains identical to the original matrix. Here, adding a matrix to its transpose always results in a symmetric matrix.

Problem 2:

For the matrices A=[1−43]A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} and B=[−121]B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}, verify (AB)T=BTAT(AB)^T = B^T A^T.

Solution:

First, find ABAB: AB=[1−43]AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} [−121]\begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = [(1)(−1)(1)(2)(1)(1)(−4)(−1)(−4)(2)(−4)(1)(3)(−1)(3)(2)(3)(1)]\begin{bmatrix} (1)(-1) & (1)(2) & (1)(1) \\ (-4)(-1) & (-4)(2) & (-4)(1) \\ (3)(-1) & (3)(2) & (3)(1) \end{bmatrix} = [−1214−8−4−363]\begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix} Then (AB)T=[−14−32−861−43](AB)^T = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}.

Now, find BTATB^T A^T: BT=[−121]B^T = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}, A^T = [1−43]\begin{bmatrix} 1 & -4 & 3 \end{bmatrix} BTAT=[(−1)(1)(−1)(−4)(−1)(3)(2)(1)(2)(−4)(2)(3)(1)(1)(1)(−4)(1)(3)]B^T A^T = \begin{bmatrix} (-1)(1) & (-1)(-4) & (-1)(3) \\ (2)(1) & (2)(-4) & (2)(3) \\ (1)(1) & (1)(-4) & (1)(3) \end{bmatrix} = [−14−32−861−43]\begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix} Thus, (AB)T=BTAT(AB)^T = B^T A^T.

Explanation:

This example demonstrates the Reversal Law of Transposes, which states that the transpose of a product of matrices is equal to the product of their transposes taken in the reverse order.