Question 9.18: (a) Find the inverse of the matrix D = (1 0 -2 2 2 3 1 3 2) ......

(a) Find the inverse of the matrix

D = $\left(\begin{matrix} 1 & 0 & -2 \\ 2 & 2 & 3 \\ 1 & 3 & 2 \end{matrix} \right)$

(b) Show that $DD^{−1} = I$.

Step-by-Step
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(a) Use the definition of the inverse given in equation (9.30) to determine the inverse:

$−bY + C = C_0$        (9.30)

Step 1: Evaluate |D|. This is calculated by multiplying each element in row 1 by its cofactor. If the table for calculating cofactors is set up, the required cofactors are the first three cofactors in the table. So step 1 is deferred to step 2.

Step 2: Set up the table to calculate all the cofactors.

Step 3: Replace each element in D by its cofactor. Then transpose.

$C^T = \left(\begin{matrix} -5 & -1 & 4 \\ -6 & 4 & -3 \\ 4 & -7 & 2 \end{matrix} \right)^T = \left(\begin{matrix} -5 & -6 & 4 \\ -1 & 4 & -7 \\ 4 & -3 & 2 \end{matrix} \right)$ = (adjoint of D)

Step 4: Multiply the adjoint matrix by

$\frac{1}{|D|}=\frac{1}{-13}$

$D^{−1} = \frac{1}{-13} \left(\begin{matrix} -5 & -6 & 4 \\ -1 & 4 & -7 \\ 4 & -3 & 2 \end{matrix} \right)$

Every element in this matrix could be multiplied by –(1/13), but this will introduce awkward fractions, so the scalar multiplication is usually left until a final single matrix is required.

(b) $DD^{−1} = D = \left(\begin{matrix} 1 & 0 & -2 \\ 2 & 2 & 3 \\ 1 & 3 & 2 \end{matrix} \right) × \frac{1}{-13} \left(\begin{matrix} -5 & -6 & 4 \\ -1 & 4 & -7 \\ 4 & -3 & 2 \end{matrix} \right)$

= $-\frac{1}{13} \left(\begin{matrix} -13 & 0 & 0 \\ 0 & -13 & 0 \\ 0 & 0 & -13 \end{matrix} \right) = \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right)$

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