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Chapter 4: Chapter 4

Problem 120

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a) Show that: (i) \(\mathrm{A} 0=0\) (ii) \(0 \mathrm{~A}=0\) (iii) \(\mathrm{AI}=\mathrm{A}\) (iv) \(\mathrm{IA}=\mathrm{A}\) where 0 and I denote the zero and identity matrices respectively, and \(\begin{array}{lll}\mathrm{A} & =12 & 1 & 3 \mid\end{array}\) \(\begin{array}{lll}14 & -1 & -1 \mid\end{array}\) b) Give examples of the following rules: (i) if A has a row of zeros, the same row of \(\mathrm{AB}\) consists of zeros, (ii) if \(\mathrm{B}\) has a column of zeros, the same column of AB consists of zeros.

Short Answer

Expert verified
In this exercise, we proved the following properties: (i) A0=0, 0A=0 (ii) AI=A, IA=A We also provided examples illustrating the two rules: (i) If A has a row of zeros, the same row of AB consists of zeros. Example: \(A = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}\), \(B = \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix}\), and \(AB = \begin{bmatrix} 13 & 16 \\ 0 & 0 \end{bmatrix}\). (ii) If B has a column of zeros, the same column of AB consists of zeros. Example: \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\), \(B = \begin{bmatrix} 5 & 0 \\ 6 & 0 \end{bmatrix}\), and \(AB = \begin{bmatrix} 17 & 0 \\ 39 & 0 \end{bmatrix}\).
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TABLE OF CONTENTS :
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Step 1: (Property 1: A0=0)

Recall that zero matrix (0) has the same dimensions as the matrix it multiplies. In this case, the zero matrix has dimensions 2x2 with all its elements being 0. To calculate the matrix A0, perform the regular matrix multiplication and you should obtain a 2x2 matrix with all elements being 0. A = \(\begin{bmatrix} 12 & 1 & 3 \\ 14 & -1 & -1 \end{bmatrix}\) 0 = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{bmatrix}\) A0 = A × 0 = \(\begin{bmatrix} 12 & 1 & 3 \\ 14 & -1 & -1 \end{bmatrix}\) × \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{bmatrix}\) = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)

Step 2: (Property 2: 0A=0)

Similarly, calculate the product 0A using regular matrix multiplication and you should also obtain a 2x2 matrix with all elements being 0. 0A = 0 × A = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{bmatrix}\) × \(\begin{bmatrix} 12 & 1 & 3 \\ 14 & -1 & -1 \end{bmatrix}\) = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)

Step 3: (Property 3: AI=A)

Recall that the identity matrix (I) has 1s on its diagonal and 0s elsewhere. In this case, our I will be a 3x3 matrix. Calculate the product of AI using regular matrix multiplication, and you should obtain the matrix A. I = \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\) AI = A × I = \(\begin{bmatrix} 12 & 1 & 3 \\ 14 & -1 & -1 \end{bmatrix}\) × \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 12 & 1 & 3 \\ 14 & -1 & -1 \end{bmatrix}\) = A

Step 4: (Property 4: IA=A)

For this, the identity matrix (I) is multiplied to the left of A, that is I × A. Calculate the product IA using regular matrix multiplication, and you should also obtain the matrix A. IA = I × A = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) × \(\begin{bmatrix} 12 & 1 & 3 \\ 14 & -1 & -1 \end{bmatrix}\) = \(\begin{bmatrix} 12 & 1 & 3 \\ 14 & -1 & -1 \end{bmatrix}\) = A #b - Part: Rules for rows and columns with zeros#

Step 5: (Rule 1: if A has a row of zeros, the same row of AB consists of zeros)

Let matrix A and B be: A = \(\begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}\) B = \(\begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix}\) AB = \(\begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}\) × \(\begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix}\) = \(\begin{bmatrix} 13 & 16 \\ 0 & 0 \end{bmatrix}\) Notice that the second row in matrix A has all zeros, and the same row in the product AB also has all zeros.

Step 6: (Rule 2: if B has a column of zeros, the same column of AB consists of zeros )

Let matrix A and B be: A = \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) B = \(\begin{bmatrix} 5 & 0 \\ 6 & 0 \end{bmatrix}\) AB = \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) × \(\begin{bmatrix} 5 & 0 \\ 6 & 0 \end{bmatrix}\) = \(\begin{bmatrix} 17 & 0 \\ 39 & 0 \end{bmatrix}\) Notice that the second column in matrix B has all zeros, and the same column in the product AB also has all zeros.

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Most popular questions from this chapter

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Show that in matrix arithmetic we can have the following: (a) \(\quad \mathrm{AB} \neq \mathrm{BA}\). (b) \(\quad \mathrm{A} \neq 0, \mathrm{~B} \neq 0\) and yet, \(\mathrm{AB}=0\). (c) \(\quad A \neq 0\) and \(A^{2}=0\). (d) \(\quad \mathrm{A} \neq 0, \mathrm{~A}^{2} \neq 0\) and \(\mathrm{A}^{3}=0\). (e) \(\mathrm{A}^{2}=\mathrm{A}\) with \(\mathrm{A} \neq 0\) and $\mathrm{A} \neq \mathrm{I}$. (f) \(\quad A^{2}=\) I with \(A \neq I\) and \(A \neq-I\).
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