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Chapter 3: Problem 15

Suppose that \(A\) and \(B\) are matrices having \(n\) rows. Prove that $M(A \mid B)=(M A \mid M B)\( for any \)m \times n\( matrix \)M$.

Short Answer

Expert verified
The problem involves proving that the expression \(M(A \mid B) = (M A \mid M B)\) holds true, for matrices \(A\), \(B\), and \(M\), where \(A\) and \(B\) have \(n\) rows and \(M\) is of size \(m \times n\). The vertical concatenation, \((A \mid B)\), represents a partitioned matrix obtained by combining \(A\) and \(B\). When we multiply \(M\) with the partitioned matrix \((A \mid B)\), we get a resulting matrix \(C\), with each element \(c_{ij}\) equivalent to an element from \(M A\) if \(j \leq p\) and an element from \(M B\) if \(j > p\). Defining two new matrices \(M_A = M A\) and \(M_B = M B\), we see that the way the elements of the matrix products \(M A\) and \(M B\) are calculated is identical to how the elements are calculated in the product of \(M\) with the partitioned matrix. Thus, it can be concluded that \(M(A \mid B) = (M A \mid M B)\). Essentially, the product of \(M\) and the partitioned matrix \((A \mid B)\) is identical to the partitioned matrix \((M A \mid M B)\).

Step by step solution

01

Understand the problem

To solve this problem, first, we need to understand the terms given in the problem statement: matrices \(A\), \(B\), and \(M\), as well as the vertical concatenation of matrices \((A \mid B)\). Matrix \(A\): An \(m \times n\) matrix with \(n\) rows Matrix \(B\): An \(m \times n\) matrix with \(n\) rows Matrix \(M\): An \(m \times n\) matrix \((A \mid B)\): This notation represents the partitioned matrix obtained by vertically concatenating matrices \(A\) and \(B\). If \(A\) has dimensions \(n \times p\) and \(B\) has dimensions \(n \times q\), then \((A \mid B)\) will have dimensions \(n \times (p+q)\).
02

Matrix multiplication

The next step is to understand matrix multiplication. Given two matrices \(X\) and \(Y\) with dimensions \(m \times n\) and \(n \times p\), respectively, their product \(Z = X \cdot Y\) will have dimensions \(m \times p\). Each element of \(Z\) is given by: \[z_{ij} = \sum_{k=1}^n x_{ik} y_{kj}\] Let us now multiply the matrix M by the partitioned matrix \((A \mid B)\).
03

Multiply M by the partitioned matrix

Now, we want to multiply \(M\) by \((A \mid B)\). Let \(C = M(A\mid B)\). Each element of the resulting partitioned matrix \(C\) will have the form: \[c_{ij} = \sum_{k=1}^n m_{ik} (a \mid b)_{kj}\] However, \(c_{ij}\) will be equal to an element from \(M A\) if \(j \leq p\) and an element from \(M B\) if \(j > p\). Therefore, we can rewrite this expression as follows: \[c_{ij} = \left\{ \begin{array}{ll} \sum_{k=1}^n m_{ik} a_{kj} & \mbox{if } j \leq p \\ \sum_{k=1}^n m_{ik} b_{k(j-p)} & \mbox{if } j > p \end{array} \right.\]
04

Comparing the result to the product of M with A and B

Finally, let's define two new matrices \(M_A = M A\) and \(M_B = M B\). We will show that the product of the partitioned matrix is equal to the product of the matrices separately. Now, let's examine how the same elements would be computed if we were performing the separate matrix multiplications \(M A\) and \(M B\). For the elements in \(M A\): \[(M A)_{ij} = \sum_{k=1}^n m_{ik} a_{kj} \textrm{ for } j \leq p\] And, for the elements in \(M B\): \[(M B)_{ij} = \sum_{k=1}^n m_{ik} b_{kj} \textrm{ for } j > p\]
05

Conclusion

Comparing step 3 and step 4, we see that the elements of the matrix products are identical. This proves that: \[M(A \mid B) = (M A \mid M B)\] Thus, we have proven that, for any \(m \times n\) matrix \(M\), the product of \(M\) and the partitioned matrix \((A \mid B)\) is equal to the partitioned matrix \((M A \mid M B)\).

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Most popular questions from this chapter

For each system of linear equations with the invertible coefficient matrix \(A\), (1) Compute \(A^{-1}\). (2) Use \(A^{-1}\) to solve the system. (a) \(\begin{aligned} x_{1}+3 x_{2} &=4 \\ 2 x_{1}+5 x_{2} &=3 \end{aligned}\) (b) $\begin{aligned} x_{1}+2 x_{2}-x_{3} &=5 \\ x_{2}+x_{3} &=1 \\ 2 x_{1}-2 x_{2}+x_{3} &=4 \end{aligned}$

Let \(B\) be an \(n \times m\) matrix with rank \(m\). Prove that there exists an $m \times n\( matrix \)A\( such that \)A B=I_{m}$.

For each of the following matrices, compute the rank and the inverse if it exists. (a) \(\left(\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right)\) (b) \(\left(\begin{array}{ll}1 & 2 \\ 2 & 4\end{array}\right)\) (c) $\left(\begin{array}{rrr}1 & 2 & 1 \\ 1 & 3 & 4 \\ 2 & 3 & -1\end{array}\right)$ (d) $\left(\begin{array}{rrr}0 & -2 & 4 \\ 1 & 1 & -1 \\ 2 & 4 & -5\end{array}\right)$ (e) $\left(\begin{array}{rrr}1 & 2 & 1 \\ -1 & 1 & 2 \\ 1 & 0 & 1\end{array}\right)$ (f) $\left(\begin{array}{lll}1 & 2 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{array}\right)$ (g) $\left(\begin{array}{rrrr}1 & 2 & 1 & 0 \\ 2 & 5 & 5 & 1 \\ -2 & -3 & 0 & 3 \\ 3 & 4 & -2 & -3\end{array}\right)$ (h) $\left(\begin{array}{rrrr}1 & 0 & 1 & 1 \\ 1 & 1 & -1 & 2 \\ 2 & 0 & 1 & 0 \\ 0 & -1 & 1 & -3\end{array}\right)$

Find the rank of the following matrices. (a) $\left(\begin{array}{lll}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{array}\right)$ (b) $\left(\begin{array}{lll}1 & 1 & 0 \\ 2 & 1 & 1 \\ 1 & 1 & 1\end{array}\right)$ (c) \(\left(\begin{array}{lll}1 & 0 & 2 \\ 1 & 1 & 4\end{array}\right)\) (d) \(\left(\begin{array}{lll}1 & 2 & 1 \\ 2 & 4 & 2\end{array}\right)\) (e) $\left(\begin{array}{rrrrr}1 & 2 & 3 & 1 & 1 \\ 1 & 4 & 0 & 1 & 2 \\ 0 & 2 & -3 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0\end{array}\right)$ (f) $\left(\begin{array}{rrrrr}1 & 2 & 0 & 1 & 1 \\ 2 & 4 & 1 & 3 & 0 \\ 3 & 6 & 2 & 5 & 1 \\ -4 & -8 & 1 & -3 & 1\end{array}\right)$ (g) $\left(\begin{array}{llll}1 & 1 & 0 & 1 \\ 2 & 2 & 0 & 2 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 0 & 1\end{array}\right)$

Prove Theorem 3.4: Consider the equation \(a x=b\) (i) If \(a \neq 0,\) then \(x=b / a\) is a unique solution of \(a x=b\) (ii) If \(a=0\) but \(b \neq 0,\) then \(a x=b\) has no solution. (iii) If \(a=0\) and \(b=0,\) then every scalar \(k\) is a solution of \(a x=b\)
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