(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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