Question

If $A^{-1} = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 3 & 1 \\ 0 & 0 & -1/3 \end{bmatrix}$, then

• A. $|A| = -1$
• B. adj $A = \begin{bmatrix} -1 & 1 & -2 \\ 0 & -3 & -1 \\ 0 & 0 & 1/3 \end{bmatrix}$
• C. $A = \begin{bmatrix} 1 & 1/3 & 7 \\ 0 & 1/3 & 1 \\ 0 & 0 & -3 \end{bmatrix}$
• D. $A = \begin{bmatrix} 1 & 1/3 & -7 \\ 0 & -3 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Answer 1

Answer: (A), (B), (C)

Options (A), (B), and (C) are correct.

Answer 2

To solve the problem, we are given the inverse of matrix A, $A^{-1}$, and we need to determine which of the given options are correct.

Given: $A^{-1} = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 3 & 1 \\ 0 & 0 & -1/3 \end{bmatrix}$

Let's evaluate each option:

Option (A): $|A| = -1$

We know that for any invertible matrix A, $|A^{-1}| = \frac{1}{|A|}$.

The given matrix $A^{-1}$ is an upper triangular matrix. The determinant of an upper triangular matrix is the product of its diagonal elements.

$|A^{-1}| = (1) \times (3) \times (-1/3) = -1$.

Now, using the property $|A^{-1}| = \frac{1}{|A|}$:

$-1 = \frac{1}{|A|}$

This implies $|A| = -1$.

So, Option (A) is correct.

Option (B): adj $A = \begin{bmatrix} -1 & 1 & -2 \\ 0 & -3 & -1 \\ 0 & 0 & 1/3 \end{bmatrix}$

We know the formula relating $A^{-1}$, $|A|$, and adj A:

$A^{-1} = \frac{1}{|A|} \text{adj } A$

From Option (A), we found $|A| = -1$.

Substitute the value of $|A|$ into the formula:

$\text{adj } A = |A| A^{-1}$

$\text{adj } A = (-1) \begin{bmatrix} 1 & -1 & 2 \\ 0 & 3 & 1 \\ 0 & 0 & -1/3 \end{bmatrix}$

$\text{adj } A = \begin{bmatrix} -1 & 1 & -2 \\ 0 & -3 & -1 \\ 0 & 0 & 1/3 \end{bmatrix}$

So, Option (B) is correct.

Option (C): $A = \begin{bmatrix} 1 & 1/3 & 7 \\ 0 & 1/3 & 1 \\ 0 & 0 & -3 \end{bmatrix}$

To find matrix A, we need to find the inverse of $A^{-1}$, i.e., $A = (A^{-1})^{-1}$.

Let $B = A^{-1} = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 3 & 1 \\ 0 & 0 & -1/3 \end{bmatrix}$.

We need to calculate $B^{-1}$. The formula for the inverse of a matrix B is $B^{-1} = \frac{1}{|B|} \text{adj } B$.

We already found $|B| = |A^{-1}| = -1$.

Now, let's find the adjoint of B. The adjoint of B is the transpose of its cofactor matrix.

Let $C_{ij}$ be the cofactor of element $b_{ij}$ in matrix B.

$C_{11} = \begin{vmatrix} 3 & 1 \\ 0 & -1/3 \end{vmatrix} = 3(-1/3) - 1(0) = -1$

$C_{12} = - \begin{vmatrix} 0 & 1 \\ 0 & -1/3 \end{vmatrix} = -(0 - 0) = 0$

$C_{13} = \begin{vmatrix} 0 & 3 \\ 0 & 0 \end{vmatrix} = 0 - 0 = 0$

$C_{21} = - \begin{vmatrix} -1 & 2 \\ 0 & -1/3 \end{vmatrix} = -((-1)(-1/3) - 2(0)) = -(1/3) = -1/3$

$C_{22} = \begin{vmatrix} 1 & 2 \\ 0 & -1/3 \end{vmatrix} = (1)(-1/3) - 2(0) = -1/3$

$C_{23} = - \begin{vmatrix} 1 & -1 \\ 0 & 0 \end{vmatrix} = -(0 - 0) = 0$

$C_{31} = \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} = (-1)(1) - 2(3) = -1 - 6 = -7$

$C_{32} = - \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = -(1(1) - 2(0)) = -1$

$C_{33} = \begin{vmatrix} 1 & -1 \\ 0 & 3 \end{vmatrix} = 1(3) - (-1)(0) = 3$

The cofactor matrix of B is $C = \begin{bmatrix} -1 & 0 & 0 \\ -1/3 & -1/3 & 0 \\ -7 & -1 & 3 \end{bmatrix}$.

The adjoint of B is $\text{adj } B = C^T = \begin{bmatrix} -1 & -1/3 & -7 \\ 0 & -1/3 & -1 \\ 0 & 0 & 3 \end{bmatrix}$.

Now, calculate $A = B^{-1} = \frac{1}{|B|} \text{adj } B$:

$A = \frac{1}{-1} \begin{bmatrix} -1 & -1/3 & -7 \\ 0 & -1/3 & -1 \\ 0 & 0 & 3 \end{bmatrix}$

$A = \begin{bmatrix} 1 & 1/3 & 7 \\ 0 & 1/3 & 1 \\ 0 & 0 & -3 \end{bmatrix}$

So, Option (C) is correct.

Option (D): $A = \begin{bmatrix} 1 & 1/3 & -7 \\ 0 & -3 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Comparing this with the calculated matrix A from Option (C), this option is incorrect.

Based on the calculations, options (A), (B), and (C) are correct.