Pregunta
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1. Given \( M=\left[\begin{array}{cccc}1 & 3 & -5 & 9 \\ -2 & 1 & 8 & -1 \\ 5 & 2 & -4 & -1 \\ 2 & -4 & 7 & 6\end{array}\right] \) and \( N=\left[\begin{array}{ccc}1 & -7 & -1 \\ 5 & 3 & 6 \\ -3 & 1 & 0 \\ 2 & 4 & -3\end{array}\right] \). (a) Without calculating the product \( M N \) determine (i) the entry in row 4 , column 3 of \( M N \), (Ans: -44\( ) \) (ii) the entries in column 2 of \( M N \), and \( \left[\begin{array}{c}33 \\ 21 \\ -37 \\ 5\end{array}\right] \) (Ans: (iii) the entries in row 3 of \( M N . \quad \) (Ans: \( \left[\begin{array}{ccc}25 & -37 & 10\end{array}\right] \) ) (b) Obtain the matrix \( N^{T} M . \quad \) (Ans: \( \left.\left[\begin{array}{cccc}-20 & -6 & 61 & 19 \\ 0 & -32 & 83 & -43 \\ -19 & 15 & 32 & -33\end{array}\right]\right) \)

Ask by Guzman Powell. in South Africa
Mar 09,2025

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**Part (a):** 1. **Entry in row 4, column 3 of \( MN \):** \(-44\) 2. **Entries in column 2 of \( MN \):** \[ \begin{bmatrix} 33 \\ 21 \\ -37 \\ 5 \end{bmatrix} \] 3. **Entries in row 3 of \( MN \):** \[ \begin{bmatrix} 25 & -37 & 10 \end{bmatrix} \] **Part (b):** \[ N^T M = \begin{bmatrix} -20 & -6 & 61 & 19 \\ 0 & -32 & 83 & -43 \\ -19 & 15 & 32 & -33 \end{bmatrix} \]

Solución

To solve the problem, we will analyze the matrices \( M \) and \( N \) and use properties of matrix multiplication to find the required entries without directly calculating the entire product \( MN \). ### Part (a) #### (i) Entry in row 4, column 3 of \( MN \) The entry in row \( i \) and column \( j \) of the product \( MN \) is calculated as the dot product of the \( i \)-th row of \( M \) and the \( j \)-th column of \( N \). For row 4 of \( M \) and column 3 of \( N \): - Row 4 of \( M \): \( [2, -4, 7, 6] \) - Column 3 of \( N \): \( [-1, 6, 0, -3] \) Calculating the dot product: \[ 2 \cdot (-1) + (-4) \cdot 6 + 7 \cdot 0 + 6 \cdot (-3) = -2 - 24 + 0 - 18 = -44 \] Thus, the entry in row 4, column 3 of \( MN \) is \(-44\). #### (ii) Entries in column 2 of \( MN \) The entries in column 2 of \( MN \) are obtained by taking the dot product of each row of \( M \) with column 2 of \( N \). Column 2 of \( N \): \( [-7, 3, 1, 4] \) Calculating each entry: 1. **Row 1 of \( M \)**: \[ 1 \cdot (-7) + 3 \cdot 3 + (-5) \cdot 1 + 9 \cdot 4 = -7 + 9 - 5 + 36 = 33 \] 2. **Row 2 of \( M \)**: \[ -2 \cdot (-7) + 1 \cdot 3 + 8 \cdot 1 + (-1) \cdot 4 = 14 + 3 + 8 - 4 = 21 \] 3. **Row 3 of \( M \)**: \[ 5 \cdot (-7) + 2 \cdot 3 + (-4) \cdot 1 + (-1) \cdot 4 = -35 + 6 - 4 - 4 = -37 \] 4. **Row 4 of \( M \)**: \[ 2 \cdot (-7) + (-4) \cdot 3 + 7 \cdot 1 + 6 \cdot 4 = -14 - 12 + 7 + 24 = 5 \] Thus, the entries in column 2 of \( MN \) are: \[ \begin{bmatrix} 33 \\ 21 \\ -37 \\ 5 \end{bmatrix} \] #### (iii) Entries in row 3 of \( MN \) The entries in row 3 of \( MN \) are obtained by taking the dot product of row 3 of \( M \) with each column of \( N \). Row 3 of \( M \): \( [5, 2, -4, -1] \) Calculating each entry: 1. **Column 1 of \( N \)**: \[ 5 \cdot 1 + 2 \cdot 5 + (-4) \cdot (-3) + (-1) \cdot 2 = 5 + 10 + 12 - 2 = 25 \] 2. **Column 2 of \( N \)**: \[ 5 \cdot (-7) + 2 \cdot 3 + (-4) \cdot 1 + (-1) \cdot 4 = -35 + 6 - 4 - 4 = -37 \] 3. **Column 3 of \( N \)**: \[ 5 \cdot (-1) + 2 \cdot 6 + (-4) \cdot 0 + (-1) \cdot (-3) = -5 + 12 + 0 + 3 = 10 \] Thus, the entries in row 3 of \( MN \) are: \[ \begin{bmatrix} 25 & -37 & 10 \end{bmatrix} \] ### Part (b) To obtain the matrix \( N^T M \), we first need to find the transpose of \( N \). The transpose \( N^T \) is: \[ N^T = \begin{bmatrix} 1 & 5 & -3 & 2 \\ -7 & 3 & 1 & 4 \\ -1 & 6 & 0 & -3 \end{bmatrix} \] Now we will calculate the product \( N^T M \). Calculating each entry of \( N^T M \): 1. **Row 1 of \( N^T \)** with each column of \( M \): - Column 1 of \( M \): \( [1, -2, 5, 2] \) \[ 1 \cdot 1 + 5 \cdot (-2) + (-3) \cdot 5 + 2 \cdot 2 = 1 - 10 - 15 + 4 = -20 \] - Column 2 of \( M \): \( [3, 1, 2, -4] \) \[ 1 \cdot 3 + 5 \cdot 1 + (-3) \cdot 2 + 2 \cdot (-4) = 3 + 5 - 6 - 8 = -6 \] - Column 3 of \( M \): \( [-5, 8, -4, 7] \) \[ 1 \cdot (-5) + 5 \cdot 8 + (-3) \cdot (-4) + 2 \cdot 7 = -5 + 40 + 12 + 14 = 61 \] - Column 4 of \( M \): \( [9, -1, -1, 6] \) \[ 1 \cdot 9 + 5 \cdot (-1) + (-3) \cdot (-1) + 2 \cdot 6 = 9 - 5 + 3 + 12 = 19 \] 2. **Row 2 of \( N^T \)** with each column of \( M \): - Column 1 of \( M \): \( [1, -2, 5, 2] \) \[ -

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To find the entry in row 4, column 3 of \( MN \), we only need to focus on the elements from row 4 of matrix \( M \) and column 3 of matrix \( N \). The elements in matrix \( M \), row 4 are \( [2, -4, 7, 6] \), and the corresponding elements in column 3 of matrix \( N \) are \( [-1, 6, 0, -3] \). So, we calculate: \[ 2 \cdot (-1) + (-4) \cdot 6 + 7 \cdot 0 + 6 \cdot (-3) = -2 - 24 + 0 - 18 = -44. \] Now for the entries in column 2 of \( MN \), we multiply each row of \( M \) by column 2 of \( N \), \( [-7, 3, 1, 4] \): 1. For row 1: \( 1 \cdot (-7) + 3 \cdot 3 + (-5) \cdot 1 + 9 \cdot 4 = -7 + 9 - 5 + 36 = 33 \). 2. For row 2: \( -2 \cdot (-7) + 1 \cdot 3 + 8 \cdot 1 + (-1) \cdot 4 = 14 + 3 + 8 - 4 = 21 \). 3. For row 3: \( 5 \cdot (-7) + 2 \cdot 3 + (-4) \cdot 1 + (-1) \cdot 4 = -35 + 6 - 4 - 4 = -37 \). 4. For row 4: \( 2 \cdot (-7) + (-4) \cdot 3 + 7 \cdot 1 + 6 \cdot 4 = -14 - 12 + 7 + 24 = 5 \). Thus, the entries in column 2 of \( MN \) are \( \left[\begin{array}{c}33 \\ 21 \\ -37 \\ 5\end{array}\right] \). For the entries in row 3 of \( MN \), similar calculations apply with elements from row 3 of \( M \) and all columns of \( N \): \[ 5 \cdot 1 + 2 \cdot 5 + (-4) \cdot (-3) + (-1) \cdot 2 = 5 + 10 + 12 - 2 = 25, \] \[ 5 \cdot (-7) + 2 \cdot 3 + (-4) \cdot 1 + (-1) \cdot 4 = -35 + 6 - 4 - 4 = -37, \] \[ 5 \cdot (-1) + 2 \cdot 6 + (-4) \cdot 0 + (-1) \cdot -3 = -5 + 12 + 0 + 3 = 10. \] Thus, the entries in row 3 of \( MN \) are \( \left[\begin{array}{ccc}25 & -37 & 10\end{array}\right] \). Lastly, to compute \( N^T M \), we first transpose \( N \) to obtain: \[ N^T = \left[\begin{array}{cccc}1 & 5 & -3 & 2 \\ -7 & 3 & 1 & 4 \\ -1 & 6 & 0 & -3\end{array}\right]. \] Now we multiply \( N^T \) by \( M \), verifying each element carefully through row-column multiplication. The final result is as follows: \[ N^T M = \left[\begin{array}{cccc}-20 & -6 & 61 & 19 \\ 0 & -32 & 83 & -43 \\ -19 & 15 & 32 & -33\end{array}\right]. \]

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