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Problem 157

# Let $\mathrm{A}=\begin{array}{ccc}\mid 1 & 1 & 1 \mid \\ \mid 0 & 1 & 0 \\\ \mid 0 & 1 & 1\end{array}$ Find the inverse of $$\mathrm{A}$$.

Expert verified
The inverse of matrix A is: A⁻¹ = $\begin{array}{ccc}\mid 1 & -1 & 1 \mid \\\ \mid 0 & 1 & 0 \\\ \mid 0 & -1 & 0\end{array}$
See the step by step solution

## Step 1: Calculate the determinant of A

First, we need to find the determinant of matrix A. For a 3x3 matrix, the determinant can be calculated using the diagonal method. $det(A) = a_{1,1}(a_{2,2}a_{3,3} - a_{2,3}a_{3,2}) - a_{1,2}(a_{2,1}a_{3,3} - a_{2,3}a_{3,1}) + a_{1,3}(a_{2,1}a_{3,2} - a_{2,2}a_{3,1})$ Plugging in the values from matrix A: det(A) = $$1(1\cdot1 - 0\cdot1) - 1(0\cdot1 - 0\cdot0) + 1(0\cdot0 - 1\cdot0) = 1$$ Since the determinant is non-zero (det(A) = 1), we can find the inverse of A.

## Step 2: Find the matrix of minors

To find the matrix of minors, replace each entry a_{i,j} in matrix A with its corresponding minor, which is the determinant of the 2x2 matrix formed by deleting the row i and column j in the given matrix. Matrix of minors(M) = $\begin{array}{ccc}\mid 1 & 0 & 0 \mid \\\ \mid 1 & 1 & 1 \\\ \mid 1 & 0 & 0\end{array}$

## Step 3: Find the matrix of cofactors

Next, find the matrix of cofactors by applying a checkerboard pattern of positive and negative signs to the matrix M: Cofactor matrix(C) = $\begin{array}{ccc}\mid 1 & 0 & 0 \mid \\\ \mid -1 & 1 & -1 \\\ \mid 1 & 0 & 0\end{array}$

## Step 4: Transpose the matrix of cofactors (adjugate matrix)

Transpose the matrix of cofactors to get the adjugate matrix. The adjugate matrix (adj(A)) is the transpose of the matrix of cofactors: adj(A) = $\begin{array}{ccc}\mid 1 & -1 & 1 \mid \\\ \mid 0 & 1 & 0 \\\ \mid 0 & -1 & 0\end{array}$

## Step 5: Multiply the adjugate matrix by the reciprocal of the determinant

Finally, to find the inverse of matrix A, multiply the adjugate matrix by the reciprocal of the determinant: A⁻¹ = $$\frac{1}{det(A)}$$ * adj(A) = $$\frac{1}{1}$$ * $\begin{array}{ccc}\mid 1 & -1 & 1 \mid \\\ \mid 0 & 1 & 0 \\\ \mid 0 & -1 & 0\end{array}$ A⁻¹ = $\begin{array}{ccc}\mid 1 & -1 & 1 \mid \\\ \mid 0 & 1 & 0 \\\ \mid 0 & -1 & 0\end{array}$ So, the inverse of matrix A is: A⁻¹ = $\begin{array}{ccc}\mid 1 & -1 & 1 \mid \\\ \mid 0 & 1 & 0 \\\ \mid 0 & -1 & 0\end{array}$

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