1.

If the A = \(\begin{bmatrix} -1& 3 \\ 4 & -2 \end{bmatrix}\) satisfies equation A2 + 3A - 10I = 0, find A-1.1. \({1\over10}\begin{bmatrix} -2& -3 \\ 4 & 1 \end{bmatrix}\)2. \({1\over10}\begin{bmatrix} -2& 3 \\ 4 & -1 \end{bmatrix}\)3. \({1\over10}\begin{bmatrix} 2& 3 \\ 4 & 1 \end{bmatrix}\)4. \({1\over10}\begin{bmatrix} 2& -3 \\ -4 & 1 \end{bmatrix}\)

Answer» Correct Answer - Option 3 : \({1\over10}\begin{bmatrix} 2& 3 \\ 4 & 1 \end{bmatrix}\)

Calculation:

Given A satisfies the equation 

A2 + 3A - 10I = 0

Multiply the equation by A-1

⇒ A-1A2 + 3 A-1A - 10A-1I = 0

⇒ A + 3I - 10 A-1 = 0

⇒ \(\begin{bmatrix} -1& 3 \\ 4 & -2 \end{bmatrix}\) + 3 \(\begin{bmatrix} 1&0 \\ 0 & 1 \end{bmatrix}\) - 10 A-1 = \(\begin{bmatrix} 0& 0 \\ 0 & 0 \end{bmatrix}\)

⇒ \(\begin{bmatrix} -1+3&3+0\\4+0&-2+3\end{bmatrix}\) = 10 A-1

⇒ A-1 = \({1\over10}\begin{bmatrix} 2& 3 \\ 4 & 1 \end{bmatrix}\)



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