estion 7 \( (A+A)^{2} \) is: \( \left(\begin{array}{ccc}28 & -8 & -12 \\ -12 & 44 & 16 \\ -36 & 32 & 24\end{array}\right) \) For the above matrix \( A \), find the determinant. is the matrix \( A \) invertible? Explain.

To find the determinant of matrix \( A \), we first need to expand the expression \( (A+A)^{2} \) and then calculate the determinant of the resulting matrix. Given that \( (A+A)^{2} = \left(\begin{array}{ccc}28 & -8 & -12 \\ -12 & 44 & 16 \\ -36 & 32 & 24\end{array}\right) \), we can expand this expression as follows: \[ (A+A)^{2} = (A+A) \times (A+A) \] Expanding the multiplication, we get: \[ (A+A)^{2} = A \times A + A \times A + A \times A + A \times A \] Simplifying this expression, we have: \[ (A+A)^{2} = 2A \times A \] Now, we can calculate the determinant of the resulting matrix \( 2A \times A \). The determinant of a matrix is a scalar value that can be calculated using various methods depending on the size of the matrix. For a 3x3 matrix, the determinant can be calculated using the formula: \[ \text{det}(A) = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31} - a_{12}a_{21}a_{33} - a_{11}a_{23}a_{32} \] where \( a_{ij} \) represents the element in the \( i \)th row and \( j \)th column of matrix \( A \). Let's calculate the determinant of matrix \( A \) using the given values. Matrices by following steps: - step0: Solution: \(2\times \left[\begin{array}{rrr}{28}&{-8}&{-12}\\{-12}&{44}&{16}\\{-36}&{32}&{24}\end{array}\right]\times \left[\begin{array}{rrr}{28}&{-8}&{-12}\\{-12}&{44}&{16}\\{-36}&{32}&{24}\end{array}\right]\) - step1: Multiply the terms: \(\left[\begin{array}{rrr}{56}&{-16}&{-24}\\{-24}&{88}&{32}\\{-72}&{64}&{48}\end{array}\right]\times \left[\begin{array}{rrr}{28}&{-8}&{-12}\\{-12}&{44}&{16}\\{-36}&{32}&{24}\end{array}\right]\) - step2: Multiply the terms: \(\left[\begin{array}{rrr}{56\times 28-16\left(-12\right)-24\left(-36\right)}&{56\left(-8\right)-16\times 44-24\times 32}&{56\left(-12\right)-16\times 16-24\times 24}\\{-24\times 28+88\left(-12\right)+32\left(-36\right)}&{-24\left(-8\right)+88\times 44+32\times 32}&{-24\left(-12\right)+88\times 16+32\times 24}\\{-72\times 28+64\left(-12\right)+48\left(-36\right)}&{-72\left(-8\right)+64\times 44+48\times 32}&{-72\left(-12\right)+64\times 16+48\times 24}\end{array}\right]\) - step3: Evaluate: \(\left[\begin{array}{rrr}{2624}&{-1920}&{-1504}\\{-2880}&{5088}&{2464}\\{-4512}&{4928}&{3040}\end{array}\right]\) The determinant of matrix \( A \) is \( \left[ \begin{array}{rrr} 2624 & -1920 & -1504 \\ -2880 & 5088 & 2464 \\ -4512 & 4928 & 3040 \end{array} \right] \). To determine if the matrix \( A \) is invertible, we need to check if the determinant of matrix \( A \) is non-zero. If the determinant is non-zero, then the matrix is invertible. If the determinant is zero, then the matrix is not invertible. In this case, the determinant of matrix \( A \) is non-zero, so the matrix \( A \) is invertible.