Chapter 3: Q10P (page 122)

Given
$A = (\begin{matrix} 1 & 2 \\ 3 & 6 \end{matrix}), C = (\begin{matrix} 7 & 6 \\ 2 & 3 \end{matrix}), D = (\begin{matrix} - 3 & 2 \\ 7 & 5 \end{matrix})$
Show that, $AC = AD, but C \neq D AND A \neq 0$ .

Short Answer

Expert verified

Find the products AC and AD of the given matrices to prove that AC=AD, but $C \neq D and A \neq 0 .$ .

Step by step solution

Definition of Matrix multiplication:

The element in row iand column j of the product matrix ABis equal to row iof Atimes column j of B. In index notation ${(AB)}_{ij} = \underset{k}{\sum A_{ik} B_{kj}}$ .

For example, if $A = (\begin{matrix} a & b \\ c & d \end{matrix}) and B = (\begin{matrix} e & f \\ c & d \end{matrix}) then, AB = (\begin{matrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{matrix})$ and then,

Multiply Matrices:

The give matrices are

$A = (\begin{matrix} 1 & 2 \\ 3 & 6 \end{matrix}), C = (\begin{matrix} 7 & 6 \\ 2 & 3 \end{matrix}) and D = (\begin{matrix} - 3 & 2 \\ 7 & 5 \end{matrix})$

It needs to be proved that $AC = AD, but C \neq D and, A \neq 0$

$AC = (\begin{matrix} 1 & 2 \\ 3 & 6 \end{matrix}) (\begin{matrix} 7 & 6 \\ 2 & 3 \end{matrix}) = (\begin{matrix} 7 + 4 & 6 + 6 \\ 21 + 12 & 18 + 18 \end{matrix}) = (\begin{matrix} 11 & 12 \\ 33 & 36 \end{matrix}) And AD = (\begin{matrix} 1 & 2 \\ 3 & 6 \end{matrix}) (\begin{matrix} - 3 & 2 \\ 7 & 5 \end{matrix}) = (\begin{matrix} - 3 + 14 & 2 + 10 \\ - 9 + 42 & 6 + 30 \end{matrix}) = (\begin{matrix} 11 & 12 \\ 33 & 36 \end{matrix})$

$Thus, AC = AD, while C \neq D and A \neq 0 .$

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Given
$A = (\begin{matrix} 1 & 2 \\ 3 & 6 \end{matrix}), C = (\begin{matrix} 7 & 6 \\ 2 & 3 \end{matrix}), D = (\begin{matrix} - 3 & 2 \\ 7 & 5 \end{matrix})$
Show that, $AC = AD, but C \neq D AND A \neq 0$ .

Short Answer

Step by step solution

Definition of Matrix multiplication:

Multiply Matrices:

One App. One Place for Learning.

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GivenA=(1236),C=(7623),D=(-3275)Show that,AC=AD,butC≠DANDA≠0.

Short Answer

Step by step solution

Definition of Matrix multiplication:

Multiply Matrices:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Solid State Physics

Atoms and Radioactivity

Energy Physics

Fields in Physics

Fluids

Translational Dynamics

Study anywhere. Anytime. Across all devices.

Given
$A = (\begin{matrix} 1 & 2 \\ 3 & 6 \end{matrix}), C = (\begin{matrix} 7 & 6 \\ 2 & 3 \end{matrix}), D = (\begin{matrix} - 3 & 2 \\ 7 & 5 \end{matrix})$
Show that, $AC = AD, but C \neq D AND A \neq 0$ .