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### Answers (1)

Answer:

Λ =1,   $A^{-1}= \frac{1}{2}\begin{bmatrix} -2 & 2\\ -4 & 3 \end{bmatrix}$

Hint:

Here, we use basic concept of determinant and inverse of matrix

$A^{-1}= \frac{1}{\left | A \right |}\times Adj\left ( A \right )$

Given:

$A= \begin{bmatrix} 3 &-2 \\ 4 & -2 \end{bmatrix}$

Solution:

$A^{2}=\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]\left[\begin{array}{cc} 3 & -2 \\ 4 & -2 \end{array}\right]=\left[\begin{array}{cc} 9-8 & -6+4 \\ 12-8 & -8+4 \end{array}\right]=\left[\begin{array}{cc} 1 & -2 \\ 4 & -4 \end{array}\right]$

Now,

\begin{aligned} &A^{2}=\lambda A-2 I \\ &\lambda A=A^{2}+2 I \\ &\lambda A=\left[\begin{array}{ll} 1 & -2 \\ 4 & -4 \end{array}\right]+\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]=\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right] \end{aligned}
\begin{aligned} &\lambda=\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right] \div\left[\begin{array}{cc} 3 & -2 \\ 4 & -2 \end{array}\right] \\ &\lambda=1 \\ &\text { So, } \\ &A^{2}=A-2 I \end{aligned}

Multiply $A^{-1}$  both side

\begin{aligned} &A^{-1} A \times A=A A^{-1}-2 L A^{-1} \\ &2 A^{-1}=I-A=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]=\left[\begin{array}{ll} -2 & 2 \\ -4 & 3 \end{array}\right] \\ &A^{-1}=\frac{1}{2}\left[\begin{array}{ll} -2 & 2 \\ -4 & 3 \end{array}\right] \end{aligned}

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