Suppose ADequalsISubscript m (the mtimesm identity matrix). Show that for any Bold b in set of real numbers R Superscript m,

Question

const2013 [10]

3 years ago

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Suppose ADequalsISubscript m (the mtimesm identity matrix). Show that for any Bold b in set of real numbers R Superscript m,

the equation ABold xequalsBold b has a solution. [Hint: Think about the equation ADBold bequalsBold b.] Explain why A cannot have more rows than columns.

Mathematics

1 answer:

horrorfan [7] · Answer 1 · 2022-09-14T10:29:13+03:00

Answer:

See Explanation

Step-by-step explanation:

(a)For matrices A and D, given that: $AD=I_m$ .

We want to show that $\forall b \in R^m$ , Ax=b has a solution.

If Ax=b

Multiply both sides by D

$(Ax)D=b \times D\\\implies (AD)x=bD$ (Recall: AD=I_m)\\\implies I_mx=Db $ (Since I_m$ is the m\times m$ identity matrix)\\\implies x=Db$

This means that the system Ax=b has a solution.

(b)Matrix A has a pivot position in each row where each pivot is a different column. Therefore, A must have at least as many columns as rows.

This means A cannot have more rows than columns.

Suppose ADequalsISubscript m ​(the mtimesm identity​ matrix). Show that for any Bold b in set of real numbers R Superscript m​,

Suppose ADequalsISubscript m (the mtimesm identity matrix). Show that for any Bold b in set of real numbers R Superscript m,