Example Question - scalar multiple
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Determining the Scalar Multiple of a Matrix
<p>To solve for \( \alpha \), we need to set up an equation based on the given matrix equality:</p> <p>\( \alpha^2 \times \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 3 & 1 & 2 \\ 0 & -10 & 2 \\ 0 & -6 & 2 \end{bmatrix} = \begin{bmatrix} 5 & -10 & 2 \\ 0 & -6 & 2 \end{bmatrix} + 3X \)</p> <p>We can see that the matrix \( X \) is also the matrix we are multiplying by \( \alpha^2 \). So, let's denote it as \( X \) and find \( 3X \):</p> <p>\( 3X = 3 \times \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 6 & 9 \\ 0 & 3 & 6 \\ 0 & 0 & 3 \end{bmatrix} \)</p> <p>Now, let's put \( 3X \) back into the matrix equality and compare the corresponding entries:</p> <p>\( \alpha^2 \times \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 3 & 1 & 2 \\ 0 & -10 & 2 \\ 0 & -6 & 2 \end{bmatrix} = \begin{bmatrix} 5 & -10 & 2 \\ 0 & -6 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 6 & 9 \\ 0 & 3 & 6 \\ 0 & 0 & 3 \end{bmatrix} \)</p> <p>Thus, we obtain the following matrix equation:</p> <p>\( \begin{bmatrix} \alpha^2 + 3 & 2\alpha^2 + 1 & 3\alpha^2 + 2 \\ 0 & \alpha^2 - 10 & 2\alpha^2 + 2 \\ 0 & 0 & \alpha^2 + 2 \end{bmatrix} = \begin{bmatrix} 5 + 3 & -10 + 6 & 2 + 9 \\ 0 & -6 + 3 & 2 + 6 \\ 0 & 0 & 2 + 3 \end{bmatrix} \)</p> <p>\( \begin{bmatrix} \alpha^2 + 3 & 2\alpha^2 + 1 & 3\alpha^2 + 2 \\ 0 & \alpha^2 - 10 & 2\alpha^2 + 2 \\ 0 & 0 & \alpha^2 + 2 \end{bmatrix} = \begin{bmatrix} 8 & -4 & 11 \\ 0 & -3 & 8 \\ 0 & 0 & 5 \end{bmatrix} \)</p> <p>By comparing the corresponding entries, we can now write a system of equations. For instance, by comparing the (1,1) entries, we have:</p> <p>\( \alpha^2 + 3 = 8 \)</p> <p>\( \alpha^2 = 5 \)</p> <p>\( \alpha = \pm \sqrt{5} \)</p> <p>Since \( \alpha \) has been determined from the first entry, we do not need to proceed with the rest, as it will be the same for all corresponding entries.</p> <p>The principal square root is often taken as the solution for \( \alpha \), thus:</p> <p>\( \alpha = \sqrt{5} \)</p>
Linear Dependence of Vectors in R^4
The given task is to find all pairs of numbers (a, b) with a, b in the set of real numbers R, such that the vectors x = (1, a, 3, 4) and y = (2, 3, b, 8) are linearly dependent in R^4. Two vectors x and y in R^4 are linearly dependent if there is a non-zero scalar c such that x = cy or y = cx, or equivalently, if at least one of the vectors can be written as a scalar multiple of the other. This implies that the corresponding components of the vectors must be proportional to each other. For vectors x and y to be linearly dependent: 1 * c = 2 → c = 2, this defines the scalar multiple; a * c = 3 → a * 2 = 3 → a = 3/2 → a = 1.5; 3 * c = b → 3 * 2 = b → b = 6; 4 * c = 8 → 4 * 2 = 8, this is true for c = 2. The system holds true for c = 2, so we find that a = 1.5 and b = 6 are the values that make the two vectors linearly dependent. Therefore, the pair of numbers (a, b) that makes x and y linearly dependent is (1.5, 6).
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