Solveeit Logo
Login

Question

Mathematics Question on Determinants

If A=[31−12]\begin{bmatrix}3&1\\\\-1&2\end{bmatrix},show that A2-5A+7I=0.Hence find A-1.

Answer

Given A=[31−12]\begin{bmatrix}3&1\\\\-1&2\end{bmatrix}

A2=A.A =\begin{bmatrix}3&1\\\\-1&2\end{bmatrix}$$\begin{bmatrix}3&1\\\\-1&2\end{bmatrix} =[9−13+2−3−2−1+4]\begin{bmatrix}9-1&3+2\\\\-3-2&-1+4\end{bmatrix}=[85−53]\begin{bmatrix}8&5\\\\-5&3\end{bmatrix}

therefore LHS=A2-5A+7I

⇒[85−53]\Rightarrow\begin{bmatrix}8&5\\\\-5&3\end{bmatrix}-5[31−12]\begin{bmatrix}3&1\\\\-1&2\end{bmatrix}+7[10\01]\begin{bmatrix}1&0\\\0&1\end{bmatrix}

=[85−53]\begin{bmatrix}8&5\\\\-5&3\end{bmatrix}-[155−510]\begin{bmatrix}15&5\\\\-5&10\end{bmatrix}+[70\07]\begin{bmatrix}7&0\\\0&7\end{bmatrix}

=[−70\0−7]\begin{bmatrix}-7&0\\\0&-7\end{bmatrix}+[70\07]\begin{bmatrix}7&0\\\0&7\end{bmatrix} =0

=RHS

So A2-5A+7I=0
therefore A. A-5A=-7I
⇒\Rightarrow A. A(A-1)-5A(A-1)=-7I(A-1) [post multiplying by A-1 as IAI≠0]
⇒\Rightarrow A(AA-1)-5I=-7I(A-1)
⇒\Rightarrow AI-5I=-7A-1
⇒\Rightarrow A-1=-17\frac{1}{7}(A-5I)
⇒\Rightarrow A-1=17\frac{1}{7}(5I-A)

=417\frac{41}{7} \bigg($$\begin{bmatrix}5&0\\\0&5\end{bmatrix}-\begin{bmatrix}3&1\\\\-1&2\end{bmatrix}$$\bigg)=17[2−1\13]\frac{1}{7}\begin{bmatrix}2&-1\\\1&3\end{bmatrix}

∴\therefore A-1=17[2−1\13]\frac{1}{7}\begin{bmatrix}2&-1\\\1&3\end{bmatrix}