The element in the first row and third column of the inverse of the matrix `[(1,2,-3),(0,1,2),(0,0,1)]` is

Options

A. `-2`
B. 0
C. 1
D. None of these

Answer

To find the element in the first row and third column of the inverse of the matrix

\[ A = \begin{pmatrix} 1 & 2 & -3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix}, \]

we will follow these steps:

### Step 1: Calculate the Determinant of Matrix A

The determinant of a 3x3 matrix

\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

is calculated using the formula:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg). \]

For our matrix \( A \):

\[ \text{det}(A) = 1 \cdot (1 \cdot 1 - 2 \cdot 0) - 2 \cdot (0 \cdot 1 - 2 \cdot 0) + (-3) \cdot (0 \cdot 0 - 1 \cdot 0). \]

Calculating this gives:

\[ = 1 \cdot (1 - 0) - 2 \cdot (0 - 0) + (-3) \cdot (0 - 0) = 1 - 0 + 0 = 1. \]

### Step 2: Calculate the Adjoint of Matrix A

The adjoint of a matrix is found by calculating the cofactor matrix and then transposing it.

1. **Cofactor Calculation**:
- For element \( a_{11} = 1 \): Minor is \( \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1 \) → Cofactor = \( 1 \).
- For element \( a_{12} = 2 \): Minor is \( \begin{vmatrix} 0 & 2 \\ 0 & 1 \end{vmatrix} = 0 \) → Cofactor = \( -0 = 0 \).
- For element \( a_{13} = -3 \): Minor is \( \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} = 0 \) → Cofactor = \( 0 \).
- For element \( a_{21} = 0 \): Minor is \( \begin{vmatrix} 2 & -3 \\ 0 & 1 \end{vmatrix} = 2 \) → Cofactor = \( -2 \).
- For element \( a_{22} = 1 \): Minor is \( \begin{vmatrix} 1 & -3 \\ 0 & 1 \end{vmatrix} = 1 \) → Cofactor = \( 1 \).
- For element \( a_{23} = 2 \): Minor is \( \begin{vmatrix} 1 & 2 \\ 0 & 0 \end{vmatrix} = 0 \) → Cofactor = \( 0 \).
- For element \( a_{31} = 0 \): Minor is \( \begin{vmatrix} 2 & -3 \\ 1 & 2 \end{vmatrix} = 4 \) → Cofactor = \( 4 \).
- For element \( a_{32} = 0 \): Minor is \( \begin{vmatrix} 1 & -3 \\ 0 & 2 \end{vmatrix} = 2 \) → Cofactor = \( -2 \).
- For element \( a_{33} = 1 \): Minor is \( \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1 \) → Cofactor = \( 1 \).

2. **Cofactor Matrix**:
\[ C = \begin{pmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 4 & -2 & 1 \end{pmatrix}. \]

3. **Transpose the Cofactor Matrix** to get the adjoint:
\[ \text{adj}(A) = C^T = \begin{pmatrix} 1 & -2 & 4 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix}. \]

### Step 3: Calculate the Inverse of Matrix A

Using the formula for the inverse:

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A). \]

Since \( \text{det}(A) = 1 \):

\[ A^{-1} = 1 \cdot \begin{pmatrix} 1 & -2 & 4 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -2 & 4 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix}. \]

### Step 4: Find the Element in the First Row and Third Column

From the inverse matrix \( A^{-1} \):

\[ A^{-1} = \begin{pmatrix} 1 & -2 & 4 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix}. \]

The element in the first row and third column is \( 4 \).

### Final Answer

The element in the first row and third column of the inverse of the matrix is **4**.

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