Verify by direct multiplication that the following matrices...

Verify by direct multiplication that the following matrices are inverses, i.e, both conditions in (1.37) hold: (a) $A=\left(\begin{array}{rr}2 & 3 \\ -1 & -1\end{array}\right), A^{-1}=\left(\begin{array}{rr}-1 & -3 \\ 1 & 2\end{array}\right)$; (b) $A=\left(\begin{array}{lll}2 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 1 & 2\end{array}\right)$ $A^{-1}=\left(\begin{array}{rrr}3 & -1 & -1 \\ -4 & 2 & 1 \\ -1 & 0 & 1\end{array}\right)$ (c) $A=\left(\begin{array}{rrr}-1 & 3 & 2 \\ 2 & 2 & -1 \\ -2 & 1 & 3\end{array}\right), A^{-1}=\left(\begin{array}{r}-1 \\ \frac{4}{7} \\ -\frac{6}{7}\end{array}\right.$ $\left.\begin{array}{rr}1 & 1 \\ -\frac{1}{7} & -\frac{3}{7} \\ \frac{5}{7} & \frac{8}{7}\end{array}\right)$

Verify by direct multiplication that the following matrices are inverses, i.e, both
conditions in (1.37) hold: (a) $A=\left(\begin{array}{rr}2 & 3 \\ -1 & -1\end{array}\right), A^{-1}=\left(\begin{array}{rr}-1 & -3 \\ 1 & 2\end{array}\right)$;
(b) $A=\left(\begin{array}{lll}2 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 1 & 2\end{array}\right)$
$A^{-1}=\left(\begin{array}{rrr}3 & -1 & -1 \\ -4 & 2 & 1 \\ -1 & 0 & 1\end{array}\right)$
(c) $A=\left(\begin{array}{rrr}-1 & 3 & 2 \\ 2 & 2 & -1 \\ -2 & 1 & 3\end{array}\right), A^{-1}=\left(\begin{array}{r}-1 \\ \frac{4}{7} \\ -\frac{6}{7}\end{array}\right.$
$\left.\begin{array}{rr}1 & 1 \\ -\frac{1}{7} & -\frac{3}{7} \\ \frac{5}{7} & \frac{8}{7}\end{array}\right)$

Key Concepts

Matrix Multiplication

Matrix multiplication involves taking the dot product of rows of the first matrix with columns of the second matrix to produce the elements of the product matrix. This operation is essential for verifying whether two matrices are inverses, as the results must equal the identity matrix.

Direct Multiplication Verification

Verifying a matrix inverse by direct multiplication means computing both A × A?¹ and A?¹ × A, and checking that each product equals the identity matrix. This method confirms the inverse relationship and is a practical way to demonstrate the correctness of the inverse matrix.

Inverse Matrix

An inverse matrix, denoted as A?¹, exists for a square matrix A when there is another matrix such that the product A × A?¹ and A?¹ × A both yield the identity matrix. This concept is fundamental in solving systems of linear equations, as it allows one to 'undo' the effect of the matrix A.

Identity Matrix

The identity matrix is a square matrix where all the elements on the main diagonal are one and all other elements are zero. It functions as the multiplicative identity in matrix multiplication, meaning that any matrix multiplied by an identity matrix of compatible dimensions remains unchanged.

Transcript

00:01 A square matrix a is invertible if there exists a matrix b, such that ab equals ba, which equals i, where i is the identity matrix.

00:23 And the identity matrix is a square matrix where there's ones down the diagonal and zeros everywhere else.

00:31 So if i had a matrix a, one, two, four, zero, negative one, one, two, three, eight, and this is a square matrix.

00:47 And if i wanted to show that matrix b is its inverse, and matrix b is negative 11, negative four, six, two, zero one, oops, negative one, two, one, negative one.

01:03 So if i wanted to show that b is the inverse of a, what i would need to do is to multiply them together to see if i get the identity matrix.

01:13 I know this multiplication is defined because they're both square matrices of size 3 by 3.

01:19 So when i multiply ab, again, the resulting matrix is also going to be 3 by 3.

01:27 And remember when we multiply matrices, we multiply the row of the first times the columns of the second.

01:34 So 1 times negative 11 is negative 11 plus 0 times negative 4 is 0 plus 2 times 6 is 12.

01:53 First row second column we would have 1 times 2 equals 2 plus 0 times 0 is 0 plus 0 times 0 is 0 plus 2.

02:08 Times negative 1 is negative 2.

02:12 First row third column, 1 times 2 is 2 plus 0 times 1 is 0, plus 2 times negative 1 is negative 2.

02:36 2 row first column 2 times negative 11, negative 22 plus negative 1 times negative 4 is plus 4 plus 3 times 6, which is 18...

Please give Ace some feedback