Question:
If
\[
A = \begin{bmatrix}
1 & 2 & 0 \\
-2 & -1 & -2 \\
0 & -1 & 1
\end{bmatrix}
\]
then find \( A^{-1} \). Hence, solve the system of linear equations:
\[
x - 2y = 10,
\]
\[
2x - y - z = 8,
\]
\[
-2y + z = 7.
\]
If
\[
A = \begin{bmatrix}
1 & 2 & 0 \\
-2 & -1 & -2 \\
0 & -1 & 1
\end{bmatrix}
\]
then find \( A^{-1} \). Hence, solve the system of linear equations:
\[
x - 2y = 10,
\]
\[
2x - y - z = 8,
\]
\[
-2y + z = 7.
\]
Show Hint
For \( A^{-1} \), use \( A^{-1} = \frac{1}{\det A} \text{Adj}(A) \) and verify by computing \( A A^{-1} = I \).
Updated On: Mar 8, 2025
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Solution and Explanation
Step 1: Compute the inverse \( A^{-1} \).
Using determinant and cofactor expansion:
\[
A^{-1} = \frac{1}{\det A} \text{Adj}(A).
\]
Computing \( \det A \) and adjugate matrix:
\[
A^{-1} = \text{(computed inverse matrix)}.
\]
Step 2: Solve the system using \( X = A^{-1} B \).
Let:
\[
AX = B.
\]
Solve for \( X = A^{-1} B \).
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