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(1 1 1,-1 0 1,4 2 6) diagonalisation method​

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Answered by 5589anuvrat
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Answer:

The Cofactor Expansion

In Section 2.4, we defined the determinant of a 2 \times 2 matrix

A = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]

as follows:

\begin{equation*} \func{det } A = \left| \begin{array}{cc} a & b \\ c & d \end{array} \right| = ad - bc \end{equation*}

and showed (in Example 2.4.4) that A has an inverse if and only if det A \neq 0. One objective of this chapter is to do this for any square matrix A. There is no difficulty for 1 \times 1 matrices: If A = \left[ a \right], we define \func{det} A = \func{det} \left[ a \right] = a and note that A is invertible if and only if a \neq 0.

If A is 3 \times 3 and invertible, we look for a suitable definition of \func{det } A by trying to carry A to the identity matrix by row operations. The first column is not zero (A is invertible); suppose the (1, 1)-entry a is not zero. Then row operations give

\begin{equation*} A = \left[ \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right] \rightarrow \left[ \begin{array}{ccc} a & b & c \\ ad & ae & af \\ ag & ah & ai \end{array} \right] \rightarrow \left[ \begin{array}{ccc} a & b & c \\ 0 & ae-bd & af-cd \\ 0 & ah-bg & ai-cg \end{array} \right] = \left[ \begin{array}{ccc} a & b & c \\ 0 & u & af-cd \\ 0 & v & ai-cg \end{array} \right] \end{equation*}

where u = ae - bd and v = ah - bg. Since A is invertible, one of u and v is nonzero (by Example 2.4.11); suppose that u \neq 0. Then the reduction proceeds

\begin{equation*} A \rightarrow \left[ \begin{array}{ccc} a & b & c \\ 0 & u & af-cd \\ 0 & v & ai-cg \end{array} \right] \rightarrow \left[ \begin{array}{ccc} a & b & c \\ 0 & u & af-cd \\ 0 & uv & u(ai-cg) \end{array} \right] \rightarrow \left[ \begin{array}{ccc} a & b & c \\ 0 & u & af-cd \\ 0 & 0 & w \end{array} \right] \end{equation*}

where w = u(ai - cg)- v(af - cd) = a(aei + bfg + cdh - ceg - afh - bdi). We define

(3.1) \begin{equation*} \func{det } A = aei + bfg + cdh - ceg - afh - bdi \end{equation*}

and observe that \func{det } A \neq 0 because \func{det } A = w \neq 0 (is invertible).

To motivate the definition below, collect the terms in Equation 3.1 involving the entries a, b, and c in row 1 of A:

\begin{align*} \func{det } A = \left| \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right| &= aei + bfg + cdh - ceg - afh - bdi \\ &= a (ei-fh) - b(di-fg) + c(dh-eg) \\ &= a \left| \begin{array}{cc} e & f \\ h & i \end{array} \right| - b \left| \begin{array}{cc} d & f \\ g & i \end{array} \right| + c \left| \begin{array}{cc} d & e \\ g & h \end{array} \right| \end{align*}

Step-by-step explanation:

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