Let A=( 3 -1 -1 1 2 1 ), B=( -1 2 2 0 -3 4

Let A=( 3 -1 -1 1 2 1 ), B=( ...

Key Concepts

Matrix Transposition

Matrix transposition is an operation that flips a matrix over its diagonal, turning its rows into columns and vice versa. This concept is fundamental in linear algebra and helps in many calculations such as solving equations, finding symmetric matrices, and simplifying expressions involving matrices.

Transpose of a Product

The property (XY)? = Y?X? is a key rule used in linear algebra. It states that the transpose of the product of two matrices is equal to the product of their transposes in the reverse order. This property is highly useful for simplifying complex expressions involving matrix products without explicitly computing the product.

Matrix Multiplication

Matrix multiplication is an operation that combines two matrices to produce a new matrix, where each element is computed as the sum of products of corresponding elements from the rows of the first matrix and the columns of the second matrix. Understanding matrix multiplication is crucial for various applications in solving linear systems, transforming vectors, and more.

Transcript

00:01 In problem 13, it is given that matrix a is equal to 3 -1 and matrix b is equal to 2 -2 and we have to find a transpose times b, b transpose a, a b transpose and b a transpose.

01:06 So first we compute the transpose of the given two matrices.

01:13 That is by definition a transpose is equal to 3 -1.

01:27 As we know that by finding transpose of any given matrix, we change its columns into rows or rows into columns.

01:42 Here a is a column vector, so its transpose will be a row vector.

01:51 Similarly, we can find the transpose of the matrix b.

02:01 Since matrix b is a column vector, so the matrix b is a column vector, so its transpose will be a row vector to now first we find a transpose b which is equal to here a transpose is 3 1 times matrix b which is 2 as the dimensions of the first matrix are 1 by 2 and dimensions of the second matrix are 2 by 1 that is the two middle entries are equal so the matrices are comfortable for multiplication that is the number of columns of the first matrix are equal to the number of rows of the second matrix so the two matrices are comfortable for multiplication.

03:27 Now we are going to find the product by multiplying the first row of the first matrix with the first column of the second matrix.

03:43 So we get three times two is six.

03:52 Plus 2 times 1 is 2.

03:59 So the resulting matrix is of order 1 by 1.

04:09 So this is the product of a transpose.

04:17 Now we are going to find b transpose a as we find b transpose a as we find here, b transpose is the row vector 2 times the matrix a, which is the column vector 3, since the number of columns in the first matrix are equal to the number of rows in the second matrix.

05:08 So the two matrices are comfortable for multiplication.

05:14 Now to find the product we multiply the entries of the first row with the corresponding entries of the first column of the second matrix...

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