Practicing Success
Let $A=\begin{bmatrix}1&-1&1\\2&1&-3\\1&1&1\end{bmatrix}$ and $10B=\begin{bmatrix}4&2&2\\-5&0&α\\1&-2&3\end{bmatrix}$. If B is the inverse of A, then find the value of $α$. |
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5 |
Here, $A=\begin{bmatrix}1&-1&1\\2&1&-3\\1&1&1\end{bmatrix}$ $∴|A|=\begin{vmatrix}1&-1&1\\2&1&-3\\1&1&1\end{vmatrix}$ $= 1(1+3)+1(2+3)+1(2-1)$ $=4+5+1=10$ Now, $adj.A=\begin{bmatrix}4&-5&1\\2&0&-2\\2&5&3\end{bmatrix}^T=\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}$ $⇒B=A^{-1}=\frac{1}{10}\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}$ $⇒10B=\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}$ Hence, $α=5$. |