Solve this
Question:

If $A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]$, find $(A B)^{-1}$.

Solution:

We know that $(A B)^{-1}=B^{-1} A^{-1}$.

$B=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]$

$B^{-1}=\frac{1}{|B|}$ Adj. $B$

Now,

$|B|=\left|\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right|$

$=1(3+0)+1(2-4)$

$=1$

Now, to find Adj. $B$

$|B|=\left|\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right|$

$=1(3+0)+1(2-4)$

$=1$

Now, to find Adj. $B$

$B_{11}=(-1)^{1+1}(3)=3 \quad B_{21}=(-1)^{2+1}(2-4)=2 \quad B_{31}=(-1)^{3+1}(6)=6$

$B_{12}=(-1)^{1+2}(-1)=1 \quad B_{22}=(-1)^{2+2}(1)=1 \quad B_{32}=(-1)^{3+2}(-2)=2$

$B_{13}=(-1)^{1+3}(2)=2 \quad B_{23}=(-1)^{2+3}(-2)=2 \quad B_{33}=(-1)^{3+3}(3+2)=5$

Therefore,

Adj. $B=\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]$

Thus,

$B^{-1}=\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]$

$(A B)^{-1}=B^{-1} A^{-1}$

$=\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$

$=\left[\begin{array}{ccc}9-30+30 & -3+12-12 & 3-10+12 \\ 3-15+10 & -1+6-4 & 1-5+4 \\ 6-30+25 & -2+12-10 & 2-10+10\end{array}\right]$

$=\left[\begin{array}{ccc}9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2\end{array}\right]$

Hence, $(A B)^{-1}=\left[\begin{array}{ccc}9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2\end{array}\right]$

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