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Chapter 1: Problem 17

Let \(A\) be a \(2 \times 2\) matrix with \(a_{11} \neq 0\) and let \(\alpha=a_{21} / a_{11} .\) Show that \(A\) can be factored into a product of the form $$\left(\begin{array}{ll} 1 & 0 \\ \alpha & 1 \end{array}\right)\left(\begin{array}{cc} a_{11} & a_{12} \\ 0 & b \end{array}\right)$$ What is the value of \(b ?\)

Short Answer

Expert verified
The matrix A can be factored into a product of the two given matrices, and the value of b is given by \(b = a_{22} - \frac{a_{21}}{a_{11}} a_{12}\).

Step by step solution

01

Analyze matrix A

: Let A be a 2x2 matrix, which is given as follows: \[A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\] We are also given that \(a_{11} \neq 0\) and \(\alpha = \frac{a_{21}}{a_{11}}\).
02

Define the given matrices

: We are given two matrices to multiply: \[(1) \begin{bmatrix} 1 & 0 \\ \alpha & 1 \end{bmatrix} \text{ and } (2) \begin{bmatrix} a_{11} & a_{12} \\ 0 & b \end{bmatrix}\]
03

Multiply the given matrices

: Multiply the two matrices, (1) and (2), to obtain a new matrix C: \( C = \begin{bmatrix} 1 & 0 \\ \alpha & 1 \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} \\ 0 & b \end{bmatrix} \) Calculate the multiplication: \[C_{11} = 1 * a_{11} + 0 * 0 = a_{11}\] \[C_{12} = 1 * a_{12} + 0 * b = a_{12}\] \[C_{21} = \alpha * a_{11} + 1 * 0 = \alpha a_{11}\] \[C_{22} = \alpha * a_{12} + 1 * b = \alpha a_{12} + b\] Therefore, the matrix C after multiplication is: \[C = \begin{bmatrix} a_{11} & a_{12} \\ \alpha a_{11} & \alpha a_{12} + b \end{bmatrix}\]
04

Compare matrix C with matrix A

: We need to show that the matrix C is the same as matrix A. Since \(\alpha = \frac{a_{21}}{a_{11}}\), we can replace \(\alpha a_{11}\) with \(a_{21}\) in matrix C: \[C = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & \alpha a_{12} + b \end{bmatrix}\] Now, we observe that the first three elements of matrix C are the same as those of matrix A. So, the only remaining task is to find the value of 'b' (which is in the last element of matrix C) so that the last element of both matrices is the same.
05

Find the value of b

: To find the value of b, we need to set the last element of matrix C equal to the last element of matrix A: \[\alpha a_{12} + b = a_{22}\] Now, we just need to solve for b, knowing that \(\alpha = \frac{a_{21}}{a_{11}}\): \[b = a_{22} - \alpha a_{12}\] \[b = a_{22} - \frac{a_{21}}{a_{11}} a_{12}\] Summing up, A can indeed be factored into a product of the two given matrices, and the value of b is given by \(b = a_{22} - \frac{a_{21}}{a_{11}} a_{12}\).

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Most popular questions from this chapter

Let \(A=\left[\begin{array}{ll}A_{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right] \quad\) and \(\quad A^{T}=\left[\begin{array}{cc}A_{11}^{T} & A_{21}^{T} \\ A_{12}^{T} & A_{22}^{T}\end{array}\right]\) Is it possible to perform the block multiplications of \(A A^{T}\) and \(A^{T} A ?\) Explain.

Given \\[ X=\left(\begin{array}{lll} 2 & 1 & 5 \\ 4 & 2 & 3 \end{array}\right) \quad Y=\left[\begin{array}{lll} 1 & 2 & 4 \\ 2 & 3 & 1 \end{array}\right] \\] (a) Compute the outer product expansion of \(X Y^{T}\) (b) Compute the outer product expansion of \(Y X^{T}\) How is the outer product expansion of \(Y X^{T}\) related to the outer product expansion of \(X Y^{T} ?\)

For each of the following pairs of matrices, find an elementary matrix \(E\) such that \(E A=B\) : (a) \(A=\left(\begin{array}{rr}2 & -1 \\ 5 & 3\end{array}\right), B=\left(\begin{array}{rr}-4 & 2 \\ 5 & 3\end{array}\right)\) (b) \(A=\left[\begin{array}{rrr}2 & 1 & 3 \\ -2 & 4 & 5 \\ 3 & 1 & 4\end{array}\right], B=\left[\begin{array}{rrr}2 & 1 & 3 \\ 3 & 1 & 4 \\ -2 & 4 & 5\end{array}\right]\) (c) \(A=\left(\begin{array}{rrr}4 & -2 & 3 \\ 1 & 0 & 2 \\ -2 & 3 & 1\end{array}\right)\) \(B=\left(\begin{array}{rrr}4 & -2 & 3 \\ 1 & 0 & 2 \\ 0 & 3 & 5\end{array}\right)\)

Nitric acid is prepared commercially by a series of three chemical reactions. In the first reaction, nitro\(\operatorname{gen}\left(\mathrm{N}_{2}\right)\) is combined with hydrogen \(\left(\mathrm{H}_{2}\right)\) to form ammonia \(\left(\mathrm{NH}_{3}\right) .\) Next, the ammonia is combined with oxygen \(\left(\mathrm{O}_{2}\right)\) to form nitrogen dioxide \(\left(\mathrm{NO}_{2}\right)\) and water. Finally, the \(\mathrm{NO}_{2}\) reacts with some of the water to form nitric acid (HNO \(_{3}\) ) and nitric oxide (NO). The amounts of each of the components of these reactions are measured in moles (a standard unit of measurement for chemical reactions). How many moles of nitrogen, hydrogen, and oxygen are necessary to produce 8 moles of nitric acid?

Let \(A\) and \(B\) be \(n \times n\) matrices and let \(C=A-B\) Show that if \(A \mathbf{x}_{0}=B \mathbf{x}_{0}\) and \(\mathbf{x}_{0} \neq \mathbf{0},\) then \(C\) must be singular.
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