Analyzing Matrix Inverse Properties

The question asks us to evaluate two statements related to matrix multiplication and inverse operations involving two given matrices A and B. We need to perform matrix calculations to check the validity of each statement.

The given matrices are:

\({\rm{A}} = \left[ {\begin{array}{c} 1&{ - 1}\\ 2&3 \end{array}} \right]{\rm{\;\;and\;B}} = \left[ {\begin{array}{c} 2&3\\ { - 1}&{ - 2} \end{array}} \right]\)

Evaluating Statement 1: AB(A⁻¹B⁻¹) is a unit matrix

To check this statement, we first need to find the inverses of matrices A and B (A⁻¹ and B⁻¹), then calculate AB and A⁻¹B⁻¹, and finally multiply the results AB and A⁻¹B⁻¹.

Calculating the Inverse of Matrix A (A⁻¹)

First, find the determinant of A:

\(\det(A) = (1)(3) - (-1)(2) = 3 + 2 = 5\)

Since the determinant is non-zero, the inverse exists.

The adjoint of A is:

\(\operatorname{adj}(A) = \left[ {\begin{array}{c} 3&{ - ( - 1)}\\ { - 2}&1 \end{array}} \right]^\top = \left[ {\begin{array}{c} 3&1\\ { - 2}&1 \end{array}} \right]\)

The inverse of A is:

\(A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A) = \frac{1}{5} \left[ {\begin{array}{c} 3&1\\ { - 2}&1 \end{array}} \right] = \left[ {\begin{array}{c} 3/5&1/5\\ { - 2/5}&1/5 \end{array}} \right]\)

Calculating the Inverse of Matrix B (B⁻¹)

First, find the determinant of B:

\(\det(B) = (2)(-2) - (3)(-1) = -4 + 3 = -1\)

Since the determinant is non-zero, the inverse exists.

The adjoint of B is:

\(\operatorname{adj}(B) = \left[ {\begin{array}{c} { - 2}&{ - 3}\\ { - ( - 1)}&2 \end{array}} \right]^\top = \left[ {\begin{array}{c} { - 2}&1\\ { - 3}&2 \end{array}} \right]\)

The inverse of B is:

\(B^{-1} = \frac{1}{\det(B)} \operatorname{adj}(B) = \frac{1}{-1} \left[ {\begin{array}{c} { - 2}&1\\ { - 3}&2 \end{array}} \right] = \left[ {\begin{array}{c} 2&{ - 1}\\ 3&{ - 2} \end{array}} \right]\)

Hold on, let's re-calculate the adjoint of B and inverse of B. \(B = \begin{bmatrix} 2 & 3 \\ -1 & -2 \end{bmatrix}\) The adjoint is found by swapping diagonal elements and negating off-diagonal elements: \(adj(B) = \begin{bmatrix} -2 & -3 \\ 1 & 2 \end{bmatrix}\) Then \(B^{-1} = \frac{1}{-1} \begin{bmatrix} -2 & -3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ -1 & -2 \end{bmatrix}\). It appears \(B^{-1} = B\). This is possible for some matrices.

Calculating Matrix Product AB

\(AB = \left[ {\begin{array}{c} 1&{ - 1}\\ 2&3 \end{array}} \right] \left[ {\begin{array}{c} 2&3\\ { - 1}&{ - 2} \end{array}} \right]\)

\(AB = \left[ {\begin{array}{c} (1)(2) + (-1)(-1) & (1)(3) + (-1)(-2)\\ (2)(2) + (3)(-1) & (2)(3) + (3)(-2) \end{array}} \right] = \left[ {\begin{array}{c} 2 + 1 & 3 + 2\\ 4 - 3 & 6 - 6 \end{array}} \right] = \left[ {\begin{array}{c} 3&5\\ 1&0 \end{array}} \right]\)

Calculating Matrix Product A⁻¹B⁻¹

\(A^{-1}B^{-1} = \left[ {\begin{array}{c} 3/5&1/5\\ { - 2/5}&1/5 \end{array}} \right] \left[ {\begin{array}{c} 2&3\\ { - 1}&{ - 2} \end{array}} \right]\)

\(A^{-1}B^{-1} = \left[ {\begin{array}{c} (3/5)(2) + (1/5)(-1) & (3/5)(3) + (1/5)(-2)\\ (-2/5)(2) + (1/5)(-1) & (-2/5)(3) + (1/5)(-2) \end{array}} \right] = \left[ {\begin{array}{c} 6/5 - 1/5 & 9/5 - 2/5\\ -4/5 - 1/5 & -6/5 - 2/5 \end{array}} \right] = \left[ {\begin{array}{c} 5/5 & 7/5\\ -5/5 & -8/5 \end{array}} \right] = \left[ {\begin{array}{c} 1&7/5\\ { - 1}&{ - 8/5} \end{array}} \right]\)

Calculating AB(A⁻¹B⁻¹)

\(AB(A^{-1}B^{-1}) = \left[ {\begin{array}{c} 3&5\\ 1&0 \end{array}} \right] \left[ {\begin{array}{c} 1&7/5\\ { - 1}&{ - 8/5} \end{array}} \right]\)

\(AB(A^{-1}B^{-1}) = \left[ {\begin{array}{c} (3)(1) + (5)(-1) & (3)(7/5) + (5)(-8/5)\\ (1)(1) + (0)(-1) & (1)(7/5) + (0)(-8/5) \end{array}} \right] = \left[ {\begin{array}{c} 3 - 5 & 21/5 - 40/5\\ 1 + 0 & 7/5 + 0 \end{array}} \right] = \left[ {\begin{array}{c} { - 2}&{ - 19/5}\\ 1&7/5 \end{array}} \right]\)

A unit matrix of order 2 is \(\left[ {\begin{array}{c} 1&0\\ 0&1 \end{array}} \right]\). The result we obtained is not a unit matrix.

Therefore, Statement 1 is incorrect.

Evaluating Statement 2: (AB)⁻¹ = A⁻¹B⁻¹

We know the general property for invertible matrices A and B is \((AB)^{-1} = B^{-1}A^{-1}\). The statement claims \((AB)^{-1} = A^{-1}B^{-1}\), which is generally not true unless AB = BA (A and B commute).

Let's calculate (AB)⁻¹ and compare it with A⁻¹B⁻¹ (which we already calculated).

Calculating the Inverse of Matrix AB ((AB)⁻¹)

We have \(AB = \left[ {\begin{array}{c} 3&5\\ 1&0 \end{array}} \right]\).

First, find the determinant of AB:

\(\det(AB) = (3)(0) - (5)(1) = 0 - 5 = -5\)

Since the determinant is non-zero, the inverse exists.

The adjoint of AB is:

\(\operatorname{adj}(AB) = \left[ {\begin{array}{c} 0&{ - 5}\\ { - 1}&3 \end{array}} \right]\)

The inverse of AB is:

\((AB)^{-1} = \frac{1}{\det(AB)} \operatorname{adj}(AB) = \frac{1}{-5} \left[ {\begin{array}{c} 0&{ - 5}\\ { - 1}&3 \end{array}} \right] = \left[ {\begin{array}{c} 0&1\\ 1/5&{ - 3/5} \end{array}} \right]\)

Comparing (AB)⁻¹ and A⁻¹B⁻¹

We found:

\((AB)^{-1} = \left[ {\begin{array}{c} 0&1\\ 1/5&{ - 3/5} \end{array}} \right]\)

\(A^{-1}B^{-1} = \left[ {\begin{array}{c} 1&7/5\\ { - 1}&{ - 8/5} \end{array}} \right]\)

Clearly, \((AB)^{-1} \neq A^{-1}B^{-1}\).

Therefore, Statement 2 is incorrect.

Summary of Findings

• Statement 1: AB(A⁻¹B⁻¹) is a unit matrix. Our calculation showed AB(A⁻¹B⁻¹) is \(\left[ {\begin{array}{c} { - 2}&{ - 19/5}\\ 1&7/5 \end{array}} \right]\), which is not a unit matrix. Statement 1 is incorrect.
• Statement 2: (AB)⁻¹ = A⁻¹B⁻¹. Our calculations showed \((AB)^{-1} = \left[ {\begin{array}{c} 0&1\\ 1/5&{ - 3/5} \end{array}} \right]\) and \(A^{-1}B^{-1} = \left[ {\begin{array}{c} 1&7/5\\ { - 1}&{ - 8/5} \end{array}} \right]\). These are not equal. Statement 2 is incorrect.

Both statements are incorrect.

Conclusion

Based on our calculations, neither statement 1 nor statement 2 is correct.

The correct answer is the option that states neither 1 nor 2 is correct.