Question

If A = $\left[\begin{array}{ccc}3&-3&4\\2&-3&4\\0&-1&1\end{array}\right]$, then show that A⁻¹ = A³

Answer 1

Answer:

Step-by-step explanation:

If $A =  \left[\begin{array}{ccc}3&-3&4\\2&-3&4\\0&-1&1\end{array}\right]$, then show that A⁻¹ = A³

for calculation of A³=> first calculate A²

$A\times A=\left[\begin{array}{ccc}3&-3&4\\2&-3&4\\0&-1&1\end{array}\right] \times\left[\begin{array}{ccc}3&-3&4\\2&-3&4\\0&-1&1\end{array}\right]\\\\\\A^{2} =\left[\begin{array}{ccc}3&-4&4\\4&5&6\\7&8&9\end{array}\right] \\\\\\=now\:\:for\:\:\\\\ A^{3}= A^{2}\times A\\\\=\left[\begin{array}{ccc}3&-3&4\\2&-3&4\\0&-1&1\end{array}\right]\:\times\:\left[\begin{array}{ccc}3&-4&4\\4&5&6\\7&8&9\end{array}\right] \\\\\\\\A^{3}=\left[\begin{array}{ccc}1&-1&0\\-2&3&-4\\-2&3&-3\end{array}\right]$----eq1

For LHS

$A^{-1} =\frac{{adj.A}}{|A|} \\\\\\|A|=3(-3+4)+3(2)+4(-2)\\\\=3+6-8\\\\=1\\\\[adjA]=\left[\begin{array}{ccc}1&-1&0\\-2&3&-4\\-2&3&-3\end{array}\right]\\\\\\A^{-1}=\left[\begin{array}{ccc}1&-1&0\\-2&3&-4\\-2&3&-3\end{array}\right]$----eq2

from eq1 and eq2 we can see that A⁻¹ = A³

hence prove