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(2) Solve for \( p \) in the equation: (a) \( \left|\begin{array}{ccc}1 & p & -1 \\ -1 & 0 & 1 \\ p+1 & 1 & 1\end{array}\right|=0 \) (B) Determine fe cofactor of \( p \) and 0 (c) Determine the minox of \( (p+1) \)

Ask by Whittaker Henry. in South Africa
Jan 23,2025

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Tutor-Verified Answer

Answer

- **(a)** \( p = 0 \) or \( p = -2 \) - **(b)** Cofactor of \( p \) and \( 0 \) is \( p + 2 \) - **(c)** Minimum of \( p + 1 \) is **-1**

Solution

Let's address each part of the problem step by step. ### (a) Solve for \( p \) in the Determinant Equation We are given the matrix: \[ \begin{vmatrix} 1 & p & -1 \\ -1 & 0 & 1 \\ p+1 & 1 & 1 \\ \end{vmatrix} = 0 \] To find the determinant, we'll expand along the first row: \[ \text{Determinant} = 1 \cdot \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} - p \cdot \begin{vmatrix} -1 & 1 \\ p+1 & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} -1 & 0 \\ p+1 & 1 \end{vmatrix} \] Calculating each minor: \[ \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} = (0)(1) - (1)(1) = -1 \] \[ \begin{vmatrix} -1 & 1 \\ p+1 & 1 \end{vmatrix} = (-1)(1) - (1)(p+1) = -1 - p - 1 = -p - 2 \] \[ \begin{vmatrix} -1 & 0 \\ p+1 & 1 \end{vmatrix} = (-1)(1) - (0)(p+1) = -1 \] Substituting back into the determinant: \[ \text{Determinant} = 1 \cdot (-1) - p \cdot (-p - 2) + (-1) \cdot (-1) = -1 + p(p + 2) + 1 = p^2 + 2p \] Setting the determinant equal to zero: \[ p^2 + 2p = 0 \\ p(p + 2) = 0 \] **Solutions:** \[ p = 0 \quad \text{or} \quad p = -2 \] ### (b) Determine the Cofactors of \( p \) and \( 0 \) Assuming \( p \) is located at position (1,2) and \( 0 \) at position (2,2) in the matrix, we find their respective cofactors. **Cofactor of \( p \) (Element at (1,2)):** \[ C_{12} = (-1)^{1+2} \cdot \begin{vmatrix} -1 & 1 \\ p+1 & 1 \end{vmatrix} = -1 \cdot (-p - 2) = p + 2 \] **Cofactor of \( 0 \) (Element at (2,2)):** \[ C_{22} = (-1)^{2+2} \cdot \begin{vmatrix} 1 & -1 \\ p+1 & 1 \end{vmatrix} = 1 \cdot (1 + p + 1) = p + 2 \] **Cofactors:** \[ \text{Cofactor of } p = p + 2 \\ \text{Cofactor of } 0 = p + 2 \] ### (c) Determine the Minimum of \( (p + 1) \) From part (a), the possible values of \( p \) are 0 and -2. \[ p + 1 \quad \text{for} \quad p = 0 \quad \text{is} \quad 1 \\ p + 1 \quad \text{for} \quad p = -2 \quad \text{is} \quad -1 \] **Minimum Value:** \[ \min(p + 1) = -1 \] --- ### **Summary of Answers** - **(a)** The solutions are \( p = 0 \) and \( p = -2 \). - **(b)** Both the cofactor of \( p \) and the cofactor of \( 0 \) are \( p + 2 \). - **(c)** The minimum value of \( p + 1 \) is **-1**.

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Bonus Knowledge

To solve for \( p \) in the given equation, we first need to compute the determinant of the matrix and set it to zero. **(a)** Given the matrix: \[ \begin{bmatrix} 1 & p & -1 \\ -1 & 0 & 1 \\ p+1 & 1 & 1 \end{bmatrix} \] The determinant can be computed using cofactor expansion along the first row: \[ \text{Det} = 1 \cdot \left|\begin{array}{cc}0 & 1 \\ 1 & 1\end{array}\right| - p \cdot \left|\begin{array}{cc}-1 & 1 \\ p+1 & 1\end{array}\right| - 1 \cdot \left|\begin{array}{cc}-1 & 0 \\ p+1 & 1\end{array}\right| \] Calculating each of these 2x2 determinants: \[ \left|\begin{array}{cc}0 & 1 \\ 1 & 1\end{array}\right| = 0 \cdot 1 - 1 \cdot 1 = -1 \] \[ \left|\begin{array}{cc}-1 & 1 \\ p+1 & 1\end{array}\right| = (-1) \cdot 1 - (1)(p + 1) = -1 - (p + 1) = -p - 2 \] \[ \left|\begin{array}{cc}-1 & 0 \\ p+1 & 1\end{array}\right| = (-1)(1) - (0)(p + 1) = -1 \] Therefore, plugging these back into the determinant: \[ \text{Det} = 1(-1) - p(-p - 2) - 1(-1) \] \[ = -1 + p^2 + 2p + 1 = p^2 + 2p \] Setting the determinant equal to zero: \[ p^2 + 2p = 0 \] Factoring gives: \[ p(p + 2) = 0 \] Thus, the solutions are: \[ p = 0 \quad \text{or} \quad p = -2 \] **(b)** To find the cofactor of \( p \) and \( 0 \), we calculate the cofactor for the terms in \( p \): 1. **Cofactor of \( p \)**: The cofactor is determined from the determinant of the 2x2 matrix that remains when we remove the row and column containing \( p \): \[ \text{Remove row 1 and column 2:}\; \begin{vmatrix} -1 & 1 \\ p + 1 & 1 \end{vmatrix} = -1 \cdot 1 - (1)(p + 1) = -1 - p - 1 = -p - 2 \] Therefore, the cofactor is \( -(-p - 2) = p + 2 \). 2. **Cofactor of \( 0 \)**: The cofactor is calculated by removing the 2nd row and 2nd column: \[ \text{Remove row 2 and column 2:}\; \begin{vmatrix} 1 & -1 \\ p + 1 & 1 \end{vmatrix} = (1)(1) - (-1)(p + 1) = 1 + p + 1 = p + 2 \] The cofactor is again \( p + 2 \). **(c)** Lastly, to determine the minox (which I assume to mean "min/max") of \( (p + 1) \): \[ \text{Expression: } f(p) = p + 1 \] Since \( p + 1 \) is a linear function, it does not have a minimum or maximum but is continuously increasing. Thus, depending on the context or constraints for \( p \) within defined limits (such as given context from previously defined maxima or minima), we can provide a specific evaluation, but otherwise, it simply trends upwards as \( p \) increases towards positive or negative infinity.

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