Question

Find the products AB and BA to determine whether B is the multiplicative inverse of A.A=[123134143],B=[72−312−12012−121−12]A = \begin{bmatrix} 1 & 2 & 3 \ 1 & 3 & 4 \ 1 & 4 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} \frac{7}{2} & -3 & \frac{1}{2} \ -\frac{1}{2} & 0 & \frac{1}{2} \ -\frac{1}{2} & 1 & -\frac{1}{2} \end{bmatrix}

Accepted Answer

Step 1: Understand the problem. You are given two matrices $$A$$ and $$B$$, and you need to find the products $$AB$$ and $$BA to $$check if $$B is $$the multiplicative inverse of $$A. $$Recall that if $$B is $$the inverse of $$A$$, then both $$AB$$ and $$BA$$ should equal the identity matrix $$I of $$the same size. Step 2: Write down the matrices explicitly:

$$A = \begin{bmatrix} 1 & 2 & 3 \ 1 & 3 & 4 \ 1 & 4 & 3 \end{bmatrix}$$

$$B = \begin{bmatrix} \frac{7}{2} & -3 & \frac{1}{2} \ -\frac{1}{2} & 0 & \frac{1}{2} \ -\frac{1}{2} & 1 & -\frac{1}{2} \end{bmatrix}$$ Step 3: Calculate the product $$AB. To do $$this, multiply each row of $$A by $$each column of $$B$$ and sum the products. For example, the element in the first row and first column of $$AB is$$:

$$ (1)(\frac{7}{2}) + (2)(-\frac{1}{2}) + (3)(-\frac{1}{2}) .

$$Repeat this for all elements to form the matrix $$AB. $$Step 4: Calculate the product $$BA$$ similarly by multiplying each row of $$B by $$each column of $$A. $$For example, the element in the first row and first column of $$BA is$$:

$$ (\frac{7}{2})(1) + (-3)(1) + (\frac{1}{2})(1) .

$$Complete this for all elements to form the matrix $$BA. $$Step 5: Compare the resulting matrices $$AB$$ and $$BA to $$the identity matrix $$I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}. If $$both $$AB$$ and $$BA$$ equal $$I$$, then $$B is $$the multiplicative inverse of $$A. $$Otherwise, it is not.

Find the products AB and BA to determine whether B is the multiplicative inverse of A.
$A = [\begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 4 \\ 1 & 4 & 3 \end{matrix}],$

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주요 개념

Matrix Multiplication

Multiplicative Inverse of a Matrix

Identity Matrix