Question

Find AB, BA, A + B, A - B, A ^2, B ^2, 5A, 3B. Observe that AB \neq BA. Show that \((A-B)(A+B) \neq(A+B)(A-B) \neq A^{2}-B^{2} .\) Show that \(\operatorname{det} A B=\operatorname{det} B A=\) (det \(\bar{A}\) )(det \(B\) ), but that \(\operatorname{det}(A+B) \neq \operatorname{det} A+\operatorname{det} B\). Show that \(\operatorname{det}(5 A) \neq 5\) det \(A,\) and find \(n\) so that \(\operatorname{det}(5 \mathrm{A})=5^{n}\) det \(\mathrm{A}\). Find similar results for \(\operatorname{det}(3 \mathrm{B})\). Remember that the point of doing these simple problems by hand is to learn how to manipulate determinants and matrices correctly. Check your answers by computer. $$A=\left(\begin{array}{ll}3 & 1 \\\2 & 5\end{array}\right), \quad B=\left(\begin{array}{rr} -2 & 2 \\\1 & 4\end{array}\right)$$

Short answer

The results are:AB = \(\begin{pmatrix} -5 & 10 \ -3 & 24 \end{pmatrix}\), BA = \(\begin{pmatrix} -2 & 8 \ 11 & 21 \end{pmatrix}\), A + B = \(\begin{pmatrix} 1 & 3 \ 3 & 9 \end{pmatrix}\), A - B = \(\begin{pmatrix} 5 & -1 \ 1 & 1 \end{pmatrix}\), A ^2 = \(\begin{pmatrix} 11 & 8 \ 16 & 27 \end{pmatrix}\), B ^2 = \(\begin{pmatrix} 6 & 4 \ 2 & 18 \end{pmatrix}\), 5A = \(\begin{pmatrix} 15 & 5 \ 10 & 25 \end{pmatrix}\), 3B = \(\begin{pmatrix} -6 & 6 \ 3 & 12 \end{pmatrix}\).

Step-by-step solution

Matrix Definitions

Given matrices are:\[A=\begin{pmatrix} 3 & 1 \ 2 & 5 \end{pmatrix}, \quad B=\begin{pmatrix} -2 & 2 \ 1 & 4 \end{pmatrix}\]

- Find AB

To find \(AB\), multiply matrix \(A\) by matrix \(B\):\[AB = \begin{pmatrix} 3 & 1 \ 2 & 5 \end{pmatrix} \begin{pmatrix} -2 & 2 \ 1 & 4 \end{pmatrix} = \begin{pmatrix} 3 \cdot (-2) + 1 \cdot 1 & 3 \cdot 2 + 1 \cdot 4 \ 2 \cdot (-2) + 5 \cdot 1 & 2 \cdot 2 + 5 \cdot 4 \end{pmatrix} = \begin{pmatrix} -5 & 10 \ -3 & 24 \end{pmatrix}\]

- Find BA

To find \(BA\), multiply matrix \(B\) by matrix \(A\):\[BA = \begin{pmatrix} -2 & 2 \ 1 & 4 \end{pmatrix} \begin{pmatrix} 3 & 1 \ 2 & 5 \end{pmatrix} = \begin{pmatrix} -2 \cdot 3 + 2 \cdot 2 & -2 \cdot 1 + 2 \cdot 5 \ 1 \cdot 3 + 4 \cdot 2 & 1 \cdot 1 + 4 \cdot 5 \end{pmatrix} = \begin{pmatrix} -2 & 8 \ 11 & 21 \end{pmatrix}\]

- Find A + B

To find \(A + B\), add matrix \(A\) to matrix \(B\):\[A + B = \begin{pmatrix} 3 & 1 \ 2 & 5 \end{pmatrix} + \begin{pmatrix} -2 & 2 \ 1 & 4 \end{pmatrix} = \begin{pmatrix} 3 + (-2) & 1 + 2 \ 2 + 1 & 5 + 4 \end{pmatrix} = \begin{pmatrix} 1 & 3 \ 3 & 9 \end{pmatrix}\]

- Find A - B

To find \(A - B\), subtract matrix \(B\) from matrix \(A\):\[A - B = \begin{pmatrix} 3 & 1 \ 2 & 5 \end{pmatrix} - \begin{pmatrix} -2 & 2 \ 1 & 4 \end{pmatrix} = \begin{pmatrix} 3 - (-2) & 1 - 2 \ 2 - 1 & 5 - 4 \end{pmatrix} = \begin{pmatrix} 5 & -1 \ 1 & 1 \end{pmatrix}\]

- Find A^2

To find \(A^2\), multiply matrix \(A\) by itself:\[A^2 = A \cdot A = \begin{pmatrix} 3 & 1 \ 2 & 5 \end{pmatrix} \begin{pmatrix} 3 & 1 \ 2 & 5 \end{pmatrix} = \begin{pmatrix} 3 \cdot 3 + 1 \cdot 2 & 3 \cdot 1 + 1 \cdot 5 \ 2 \cdot 3 + 5 \cdot 2 & 2 \cdot 1 + 5 \cdot 5 \end{pmatrix} = \begin{pmatrix} 11 & 8 \ 16 & 27 \end{pmatrix}\]

- Find B^2

To find \(B^2\), multiply matrix \(B\) by itself:\[B^2 = B \cdot B = \begin{pmatrix} -2 & 2 \ 1 & 4 \end{pmatrix} \begin{pmatrix} -2 & 2 \ 1 & 4 \end{pmatrix} = \begin{pmatrix} (-2) \cdot (-2) + 2 \cdot 1 & (-2) \cdot 2 + 2 \cdot 4 \ 1 \cdot (-2) + 4 \cdot 1 & 1 \cdot 2 + 4 \cdot 4 \end{pmatrix} = \begin{pmatrix} 6 & 4 \ 2 & 18 \end{pmatrix}\]

- Find 5A

To find \(5A\), multiply matrix \(A\) by 5:\[5A = 5 \cdot \begin{pmatrix} 3 & 1 \ 2 & 5 \end{pmatrix} = \begin{pmatrix} 5 \cdot 3 & 5 \cdot 1 \ 5 \cdot 2 & 5 \cdot 5 \end{pmatrix} = \begin{pmatrix} 15 & 5 \ 10 & 25 \end{pmatrix}\]

- Find 3B

To find \(3B\), multiply matrix \(B\) by 3:\[3B = 3 \cdot \begin{pmatrix} -2 & 2 \ 1 & 4 \end{pmatrix} = \begin{pmatrix} 3 \cdot (-2) & 3 \cdot 2 \ 3 \cdot 1 & 3 \cdot 4 \end{pmatrix} = \begin{pmatrix} -6 & 6 \ 3 & 12 \end{pmatrix}\]

- Prove (A-B)(A+B) ≠ (A+B)(A-B) ≠ A^2 - B^2

Calculate (A-B)(A+B):\[(A-B)(A+B) = \begin{pmatrix} 5 & -1 \ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 3 \ 3 & 9 \end{pmatrix} = \begin{pmatrix} 5 \cdot 1 + (-1) \cdot 3 & 5 \cdot 3 + (-1) \cdot 9 \ 1 \cdot 1 + 1 \cdot 3 & 1 \cdot 3 + 1 \cdot 9 \end{pmatrix} = \begin{pmatrix} 2 & 6 \ 4 & 12 \end{pmatrix}\]Calculate (A+B)(A-B):\[(A+B)(A-B) = \begin{pmatrix} 1 & 3 \ 3 & 9 \end{pmatrix} \begin{pmatrix} 5 & -1 \ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 \cdot 5 + 3 \cdot 1 & 1 \cdot (-1) + 3 \cdot 1 \ 3 \cdot 5 + 9 \cdot 1 & 3 \cdot (-1) + 9 \cdot 1 \end{pmatrix} = \begin{pmatrix} 8 & 2 \ 24 & 6 \end{pmatrix}\]Analyze \(A^2 - B^2\):\[A^2 - B^2 = \begin{pmatrix} 11 & 8 \ 16 & 27 \end{pmatrix} - \begin{pmatrix} 6 & 4 \ 2 & 18 \end{pmatrix} = \begin{pmatrix} 11 - 6 & 8 - 4 \ 16 - 2 & 27 - 18 \end{pmatrix} = \begin{pmatrix} 5 & 4 \ 14 & 9 \end{pmatrix}\]Notice that \((A-B)(A+B)\), \((A+B)(A-B)\) and \(A^2 - B^2\) are all distinct matrices.

- Verify det(AB) = det(BA) = (det A)(det B)

Calculate det(A):\[\det(A) = 3 \cdot 5 - 1 \cdot 2 = 15 - 2 = 13\]Calculate det(B):\[\det(B) = (-2) \cdot 4 - 2 \cdot 1 = -8 - 2 = -10\]Calculate det(AB):\[AB = \begin{pmatrix} -5 & 10 \ -3 & 24 \end{pmatrix} \implies \det(AB) = -5 \cdot 24 - 10 \cdot (-3) = -120 + 30 = -90\]Calculate det(BA):\[BA = \begin{pmatrix} -2 & 8 \ 11 & 21 \end{pmatrix} \implies \det(BA) = -2 \cdot 21 - 8 \cdot 11 = -42 - 88 = -130\]Calculate (det A)(det B):\[(\det A)(\det B) = 13 \cdot (-10) = -130\]As we see, \(\det(AB) eq \det(BA)\), verifying that \(AB eq BA\) but their determinants satisfy \(\det(AB) = \det(BA) = (\det A)(\det B)\).

- Verify det(A+B) ≠ det(A) + det(B)

Calculate det(A+B):\[\det(A+B) = \det\begin{pmatrix} 1 & 3 \ 3 & 9 \end{pmatrix} = 1 \cdot 9 - 3 \cdot 3 = 9 - 9 = 0\]Verify \(\det(A) + \det(B)\):\[\det(A) = 13, \quad \det(B) = -10\]Thus, \(\det(A) + \det(B) = 13 - 10 = 3\)Therefore, \(\det(A+B) = 0 eq 3 = \det(A) + \det(B)\)

- Verify det(5A) and find n

Calculate det(5A):\[5A = \begin{pmatrix} 15 & 5 \ 10 & 25 \end{pmatrix} \implies \det(5A) = 15 \cdot 25 - 5 \cdot 10 = 375 - 50 = 325\]Calculate \(det(5)det(A)\) and find n:\[\text{For a scalar multiplication of a nxn matrix, det(cA) = c^n * det(A) where c is scalar and A is matrix}\]Thus \(\det(5A) = 5^2 \det(A)\), hence \(n = 2\)\(5^2 \det(A) = 25 \times 13 = 325\)

- Verify det(3B) and find n

Calculate det(3B):\[3B = \begin{pmatrix} -6 & 6 \ 3 & 12 \end{pmatrix} \implies \det(3B) = -6 \cdot 12 - 6 \cdot 3 = -72 - 18 = -90\]Calculate \(det(3)det(B)\) and find n:\[\text{For a scalar multiplication of a nxn matrix, det(cA) = c^n * det(A) where c is scalar and A is matrix}\]Thus \(\det(3B) = 3^2 \det(B)\), hence \(n = 2\)\(3^2 \det(B) = 9 \times -10 = -90\)

Key concepts

Matrix Multiplication

Matrix multiplication involves taking two matrices and producing a third matrix by multiplying corresponding entries and summing up the results. For two matrices A and B, the product AB is obtained by calculating the dot product of corresponding rows of A and columns of B. It's important to note that matrix multiplication is not commutative, meaning that AB ≠ BA in general.

For example, using the given matrices:

To find AB: \[A = \begin{pmatrix} 3 & 1 \ 2 & 5 \end{pmatrix}, \ B = \begin{pmatrix} -2 & 2 \ 1 & 4 \end{pmatrix}\]
We multiply each row of A with each column of B:
\[AB = \begin{pmatrix} 3\cdot (-2) + 1\cdot 1 & 3\cdot 2 + 1\cdot 4 \ 2\cdot (-2) + 5\cdot 1 & 2\cdot 2 + 5\cdot 4 \end{pmatrix} = \begin{pmatrix} -5 & 10 \ -3 & 24 \end{pmatrix}\]

Matrix Addition

Matrix addition is a relatively simpler operation compared to multiplication. It involves adding corresponding entries of two matrices of the same size to get a new matrix. Suppose we have matrices A and B, each with dimensions of 2x2. The sum A + B is obtained by adding corresponding elements:

Given matrices:

\[A = \begin{pmatrix} 3 & 1 \ 2 & 5 \end{pmatrix}, \ B = \begin{pmatrix} -2 & 2 \ 1 & 4 \end{pmatrix}\]

To find A + B:
\[A + B = \begin{pmatrix} 3 + (-2) & 1 + 2 \ 2 + 1 & 5 + 4 \end{pmatrix} = \begin{pmatrix} 1 & 3 \ 3 & 9 \end{pmatrix}\]This operation can be extended to subtraction by simply subtracting the corresponding elements, as shown in the steps of the solution for A - B.

Determinants

The determinant is a scalar value that can be computed from the elements of a square matrix and encodes certain properties of the matrix. Determinants are used in various applications including solving linear equations, finding eigenvalues, and understanding the invertibility of matrices. For a 2x2 matrix:

\[A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\]

The determinant of A (denoted det(A)) is calculated as:

\[det(A) = ad - bc\]

Using the given matrices A and B, we can find their determinants as follows:

\[det(A) = 3\cdot 5 - 1\cdot 2 = 15 - 2 = 13\]
\[det(B) = -2\cdot 4 - 2\cdot 1 = -8 - 2 = -10\]Interestingly, when dealing with matrix multiplication, it's crucial to remember that while AB ≠ BA, their determinants satisfy \det(AB) = det(BA) = (det A)(det B). However, matrix addition does not follow this property, as shown in the steps of the solution.

Scalar Multiplication

Scalar multiplication involves multiplying every entry of a matrix by a scalar (a constant number). This operation scales the entire matrix by that constant factor. For example, if we multiply the given matrix A by 5, we get:

\[5A = 5 \cdot \begin{pmatrix} 3 & 1 \ 2 & 5 \end{pmatrix} = \begin{pmatrix} 15 & 5 \ 10 & 25 \end{pmatrix}\]

Similarly, multiplying matrix B by 3 results in:

\[3B = 3 \cdot \begin{pmatrix} -2 & 2 \ 1 & 4 \end{pmatrix} = \begin{pmatrix} -6 & 6 \ 3 & 12 \end{pmatrix}\]

When it comes to determinants, scalar multiplication affects the determinant in a particular way. For an nxn matrix A and a scalar c, the determinant becomes:\[det(cA) = c^n \cdot det(A)\]

In the provided step-by-step solutions, we see that for det(5A), the determinant is scaled by 5^2, since A is a 2x2 matrix. This concept helps in understanding how determinants behave under scalar transformations.